Electrochemistry - Class 12 Chemistry - Chapter 2 - Notes, NCERT Solutions & Extra Questions
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Notes - Electrochemistry | Class 12 NCERT | Chemistry
Ultimate Guide to Electrochemistry Class 12 Notes: Detailed Explanation and Key Concepts
Introduction to Electrochemistry
Electrochemistry, a significant field in chemistry, explores the interaction between chemical reactions and electrical energy. The principles of electrochemistry are crucial for various applications, including powering electronic devices, metal production, and developing eco-friendly technologies.
Chemical Reactions and Electrical Energy
Electrochemistry demonstrates that chemical reactions can produce electrical energy, and conversely, electrical energy can drive non-spontaneous chemical reactions. This dual capability is the foundation for many practical and industrial processes.
Electrochemical Cells
Electrochemical cells play a pivotal role in electrochemistry by converting chemical energy into electrical energy. Let's delve into two major types of these cells.
Galvanic (Voltaic) Cells
A galvanic cell, also known as a voltaic cell, is an electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy.
Example: Daniell Cell The Daniell Cell utilises zinc and copper as electrodes. The chemical reaction within the cell can be represented as: [ \text{Zn (s) + Cu}^{2+} \text{(aq)} \rightarrow \text{Zn}^{2+} \text{(aq) + Cu (s)} ]
At the anode, zinc undergoes oxidation: [ \text{Zn (s) → Zn}^{2+} \text{(aq) + 2e}^- ]
At the cathode, copper undergoes reduction: [ \text{Cu}^{2+} \text{(aq) + 2e}^- → \text{Cu (s)} ]
Electrons flow from the zinc rod to the copper rod, generating an electrical current that can be harnessed to power electrical devices.
Oxidation and Reduction Reactions
In electrochemical cells, oxidation and reduction reactions occur simultaneously. These are referred to as redox reactions.
Oxidation: Loss of electrons (e.g., zinc dissolving at the anode)
Reduction: Gain of electrons (e.g., copper depositing at the cathode)
Electrode Potentials
Electrode potential is the potential difference between an electrode and its electrolyte. It is crucial for determining the feasibility and direction of redox reactions in electrochemical cells.
Standard Electrode Potential: Defined with respect to the standard hydrogen electrode (SHE), which is assigned a potential of zero volts.
The Nernst Equation
The Nernst Equation relates the electrode potential to the concentration of the involved species: [ E = E^{\circ} - \frac{RT}{nF} \ln Q ] where,
( E ) is the electrode potential.
( E^{\circ} ) is the standard electrode potential.
( R ) is the gas constant.
( T ) is the temperature.
( n ) is the number of moles of electrons exchanged.
( F ) is the Faraday constant.
( Q ) is the reaction quotient.
Faraday’s Laws of Electrolysis
Faraday's laws quantify the relationship between the amount of substance deposited or dissolved during electrolysis and the quantity of electricity passed through the electrolyte.
First Law: The amount of chemical reaction is proportional to the quantity of electricity.
Second Law: The amounts of different substances liberated by the same quantity of electricity are proportional to their chemical equivalent weights.
Electrolytic Cells
Electrolytic cells use electrical energy to drive non-spontaneous chemical reactions. These cells are essential in industries for electroplating, purification of metals, and chemical synthesis.
Conductance of Electrolytic Solutions
The conductance of electrolytic solutions depends on:
Nature of the electrolyte
Size of the ions
Solvent properties and viscosity
Electrolyte concentration
Temperature
Measurement Techniques: Conductivity meters and specially designed conductivity cells are used to measure the resistance and conductivity of electrolytic solutions.
Practical Applications of Electrochemical Cells
Batteries: Primary vs. Secondary
Primary Batteries: Single-use batteries where the chemical reactions are irreversible (e.g., dry cells used in torches).
Secondary Batteries: Rechargeable batteries where the chemical reactions are reversible (e.g., lead-acid batteries used in automobiles).
Fuel Cells
Fuel cells are galvanic cells where reactants are continuously supplied, and products are continuously removed. They efficiently convert the chemical energy of fuels like hydrogen directly into electrical energy.
Hydrogen Fuel Cell Reaction
[ 2 \text{H}_2 (\text{g}) + \text{O}_2 (\text{g}) \rightarrow 2 \text{H}_2 \text{O (l) } ]
Corrosion Prevention
Corrosion is the degradation of metals due to electrochemical reactions with the environment. Preventative measures include:
Painting: Applying protective coatings.
Galvanisation: Coating with a more reactive metal like zinc.
Sacrificial Anodes: Using more reactive metals to protect the primary metal object.
Summary
Electrochemistry is an integral part of Class 12 Chemistry that deals with the interplay between electrical energy and chemical reactions. Understanding the principles of electrochemical cells, electrode potentials, Faraday’s laws, and practical applications like batteries and fuel cells is essential for academic success and real-world applications.
Frequently Asked Questions
How to Prepare Electrochemistry Notes?
Break down the chapter into key sections: Introduction, Types of Cells, Reactions, Potentials, Laws, Applications.
Summarise each section with concise points.
Make diagrams to illustrate cell setups and reactions.
Important Topics for Class 12 Exams
Construction and working of Daniell Cell
Calculation of standard electrode potentials
Application of Nernst Equation
Faraday’s laws and their applications
By following these notes and strategies, students can master the essentials of electrochemistry and excel in their Class 12 Chemistry exams.
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Notes - Flashcards - Electrochemistry | Class 12 NCERT | Chemistry
NCERT Solutions - Electrochemistry | NCERT | Chemistry | Class 12
Arrange the following metals in the order in which they displace each other from the solution of their salts.
$\mathrm{Al}, \mathrm{Cu}, \mathrm{Fe}, \mathrm{Mg}$ and $\mathrm{Zn}$.
To arrange the metals in the order in which they displace each other from the solution of their salts, we need to refer to their standard electrode potentials. A metal with a more negative standard electrode potential can displace a metal with a less negative (or more positive) standard electrode potential from its salt solution.
Let's list the standard electrode potentials for the given metals:
Al (Aluminum): $ -1.66$ V
Cu (Copper): $ +0.34 $ V
Fe (Iron): $-0.44 $ V
Mg (Magnesium): $ -2.37 $ V
Zn (Zinc): $-0.76 $ V
The order of displacement from solution based on their standard electrode potentials from most negative to most positive is:
$$ \mathrm{Mg} > \mathrm{Al} > \mathrm{Zn} > \mathrm{Fe} > \mathrm{Cu} $$
Therefore, the metals will displace each other from their salts in the following order:
$$ \mathbf{Mg, Al, Zn, Fe, Cu} $$
In this order, Magnesium (Mg) will displace all the others, Aluminum (Al) will displace Zinc, Iron, and Copper, and so on. Copper (Cu) will not displace any of the others as it has the most positive standard electrode potential.
Given the standard electrode potentials,
$\mathrm{K}^{+} / \mathrm{K}=-2.93 \mathrm{~V}, \mathrm{Ag}^{+} / \mathrm{Ag}=0.80 \mathrm{~V}$
$\mathrm{Hg}^{2+} / \mathrm{Hg}=0.79 \mathrm{~V}$
$\mathrm{Mg}^{2+} / \mathrm{Mg}=-2.37 \mathrm{~V}, \mathrm{Cr}^{3+} / \mathrm{Cr}=-0.74 \mathrm{~V}$
Arrange these metals in their increasing order of reducing power.
To arrange the metals in the order of their increasing reducing power, we need to understand the relationship between the standard electrode potentials and reducing power.
Metals with higher (less negative) standard electrode potentials are weaker reducing agents because they are less likely to lose electrons than metals with lower (more negative) standard electrode potentials.
Given the standard electrode potentials:
$$ \mathrm{K}^{+} / \mathrm{K} = -2.93 , \mathrm{V} $$
$$ \mathrm{Ag}^{+} / \mathrm{Ag} = 0.80 , \mathrm{V} $$
$$ \mathrm{Hg}^{2+} / \mathrm{Hg} = 0.79 , \mathrm{V} $$
$$ \mathrm{Mg}^{2+} / \mathrm{Mg} = -2.37 , \mathrm{V} $$
$$ \mathrm{Cr}^{3+} / \mathrm{Cr} = -0.74 , \mathrm{V} $$
Arrange these metals in increasing order of reducing power:
$$ \mathrm{Ag} \quad (0.80 , \mathrm{V}) $$
$$ \mathrm{Hg} \quad (0.79 , \mathrm{V}) $$
$$ \mathrm{Cr} \quad (-0.74 , \mathrm{V}) $$
$$ \mathrm{Mg} \quad (-2.37 , \mathrm{V}) $$
$$ \mathrm{K} \quad (-2.93 , \mathrm{V}) $$
Conclusion:
The increasing order of reducing power is: $$ \mathrm{Ag} < \mathrm{Hg} < \mathrm{Cr} < \mathrm{Mg} < \mathrm{K} $$
Depict the galvanic cell in which the reaction
$\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
To depict the galvanic cell in which the reaction
$$ \mathrm{Zn}(\mathrm{s}) + 2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq}) + 2 \mathrm{Ag}(\mathrm{s}) $$
takes place, let's break it down into the required elements:
Electronic Cell Depiction
Anode (oxidation): Made of Zinc (Zn)
Cathode (reduction): Made of Silver (Ag)
Salt Bridge: Completes the circuit by maintaining electrical neutrality with ions.
The cell can be represented as: $$ \mathrm{Zn}(\mathrm{s}) \mid \mathrm{Zn}^{2+}(\mathrm{aq}) | \mathrm{Ag}^{+}(\mathrm{aq}) \mid \mathrm{Ag}(\mathrm{s}) $$
(i) Which of the electrode is negatively charged?
The anode where oxidation takes place is negatively charged. In this case, the Zinc (Zn) electrode loses electrons: $$ \mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2 \mathrm{e}^{-} $$
(ii) The carriers of the current in the cell
The carriers of the current within the electrolytic solution are ions:
Zn (Zinc): Becomes $\mathrm{Zn}^{2+}$ ions which move into the solution.
Ag (Silver): $\mathrm{Ag}^{+}$ ions in solution move towards the cathode to get reduced to $\mathrm{Ag}$ atoms.
The external circuit carriers of the current are electrons:
Flow from the Zinc anode (negative) to the Silver cathode (positive).
(iii) Individual reaction at each electrode
Anode (Oxidation) Reaction:
$$ \mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq}) + 2 \mathrm{e}^{-} $$
Cathode (Reduction) Reaction:
$$ 2 \mathrm{Ag}^{+}(\mathrm{aq}) + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Ag}(\mathrm{s}) $$
Putting it all together, the overall cell reaction is: $$ \mathrm{Zn}(\mathrm{s}) + 2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq}) + 2 \mathrm{Ag}(\mathrm{s}) $$
In summary:
Negative Electrode: Zinc
Current Carriers:
Inside the cell: Ions ($\mathrm{Zn}^{2+}$ and $\mathrm{Ag}^{+}$)
Outside the cell: Electrons
Anode Reaction: $\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+} + 2 \mathrm{e}^{-}$
Cathode Reaction: $2 \mathrm{Ag}^{+} + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Ag}$
Calculate the standard cell potentials of galvanic cell in which the following reactions take place:
(i) $2 \mathrm{Cr}(\mathrm{s})+3 \mathrm{Cd}^{2+}(\mathrm{aq}) \rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{Cd}$
(ii) $\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$
Calculate the $\Delta_{\mathrm{r}} G^{\circ}$ and equilibrium constant of the reactions.
For the reaction involving chromium and cadmium:
Standard reduction potential for $\mathrm{Cr}^{3+}/\mathrm{Cr}$ is approximately $-0.74 \, \text{V}$
Standard reduction potential for $\mathrm{Cd}^{2+}/\mathrm{Cd}$ is approximately $-0.40 \, \text{V}$
For the reaction involving iron and silver:
Standard reduction potential for $\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}$ is approximately $+0.77 \, \text{V}$
Standard reduction potential for $\mathrm{Ag}^{+}/\mathrm{Ag}$ is approximately $+0.80 \, \text{V}$
Now, we can use these values to calculate the standard cell potentials, Gibbs free energy change, and equilibrium constants.
(i) $2 \mathrm{Cr}(\mathrm{s}) + 3 \mathrm{Cd}^{2+}(\mathrm{aq}) \rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq}) + 3 \mathrm{Cd}$
Oxidation: $\mathrm{Cr} \rightarrow \mathrm{Cr}^{3+} + 3\mathrm{e}^-$ ( E^{\circ}_\text{oxidation} = +0.74 \text{ V})
Reduction: $\mathrm{Cd}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Cd}$ ( E^{\circ}_\text{reduction} = -0.40 \text{ V} )
[ E^{\circ}_\text{cell} = E^{\circ}_\text{cathode} - E^{\circ}_\text{anode} = -0.40 \text{ V} - (-0.74 \text{ V}) = +0.34 \text{ V} ]
(ii) $\mathrm{Fe}^{2+}(\mathrm{aq}) + \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq}) + \mathrm{Ag}(\mathrm{s})$
Oxidation: $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{e}^-$ ( E^{\circ}_\text{oxidation} = -0.77 \text{ V} )
Reduction: $\mathrm{Ag}^{+} + \mathrm{e}^- \rightarrow \mathrm{Ag}$ ( E^{\circ}_\text{reduction} = +0.80 \text{ V} )
[ E^{\circ}_\text{cell} = E^{\circ}_\text{cathode} - E^{\circ}_\text{anode} = +0.80 \text{ V} - (+0.77 \text{ V}) = +0.03 \text{ V} ]
Next, we calculate $\Delta_{\mathrm{r}} G^{\circ}$ and the equilibrium constant ( K ):
Gibbs free energy change:
[ \Delta_{\mathrm{r}} G^{\circ} = - n F E^{\circ}_\text{cell} ]
Equilibrium constant ( K ):
[ E^{\circ}_\text{cell} = \frac{0.059}{n} \log K ] [ \log K = \frac{n E^{\circ}_\text{cell}}{0.059} ]
Plugging in these values:
(i) For $2 \mathrm{Cr}(\mathrm{s}) + 3 \mathrm{Cd}^{2+}(\mathrm{aq}) \rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq}) + 3 \mathrm{Cd}$
Number of electrons transferred (n): 6 [ \Delta_{\mathrm{r}} G^{\circ} = - 6 \times 96485 \, \text{C mol}^{-1} \times 0.34 \, \text{V} ] [ \Delta_{\mathrm{r}} G^{\circ} = -196829.4 \, \text{J mol}^{-1} ]
[ \log K = \frac{6 \times 0.34}{0.059} = 34.57 ]
[ K = 10^{34.57} ]
[ K =3.71 \times 10^{34} ]
(ii) For $\mathrm{Fe}^{2+}(\mathrm{aq}) + \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq}) + \mathrm{Ag}(\mathrm{s})$
Number of electrons transferred (n): 1 [ \Delta_{\mathrm{r}} G^{\circ} = - 1 \times 96485 \, \text{C mol}^{-1} \times 0.03 \, \text{V} ] [ \Delta_{\mathrm{r}} G^{\circ} = -2894.55 \, \text{J mol}^{-1} ]
[ \log K = \frac{1 \times 0.03}{0.059} = 0.51 ] [ K = 10^{0.51} \approx 3.24 ]
These calculations give the standard cell potentials, Gibbs free energy changes, and equilibrium constants for the given reactions.
Write the Nernst equation and emf of the following cells at $298 \mathrm{~K}$ :
(i) $\mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(0.001 \mathrm{M}) \| \mathrm{Cu}^{2+}(0.0001 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})$
(ii) $\mathrm{Fe}(\mathrm{s})\left|\mathrm{Fe}^{2+}(0.001 \mathrm{M}) \| \mathrm{H}^{+}(1 \mathrm{M})\right| \mathrm{H}_{2}$ (g))(1bar) $\mid \mathrm{Pt}(\mathrm{s})$
(iii) $\operatorname{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(0.050 \mathrm{M}) \| \mathrm{H}^{+}(0.020 \quad \mathrm{M})\right| \mathrm{H}_{2}$ (g) (1 bar) $\mid \mathrm{Pt}(\mathrm{s})$
(iv) $\mathrm{Pt}(\mathrm{s})\left|\mathrm{Br}^{-}(0.010 \mathrm{M})\right| \mathrm{Br}_{2}(\mathrm{l}) \| \mathrm{H}^{+}(0.030 \mathrm{M}) \mid \mathrm{H}_{2}$ (g) (1 bar) $\mid \mathrm{Pt}(\mathrm{s})$.
Let's outline the standard reduction potentials from typical values found in textbooks:
$\mathrm{Mg}^{2+} / \mathrm{Mg}$: $-2.37 \, \text{V}$
$\mathrm{Cu}^{2+} / \mathrm{Cu}$: $+0.34 \, \text{V}$
$\mathrm{Fe}^{2+} / \mathrm{Fe}$: $-0.44 \, \text{V}$
$\mathrm{Sn}^{2+} / \mathrm{Sn}$: $-0.14 \, \text{V}$
$\mathrm{Br}_2 / \mathrm{Br}^-$: $+1.07 \, \text{V}$
$\mathrm{H}^{+} / \mathrm{H}_2$ (Standard Hydrogen Electrode): $0 \, \text{V}$
These standard reduction potentials provide the necessary base for the calculations using the Nernst equation:
[ E_{\text{{cell}}} = E_{\text{{cell}}}^{\circ} - \frac{0.059}{n} \log Q ]
where
( E_{\text{{cell}}}^{\circ} ) is the standard cell potential.
( n ) is the number of moles of electrons transferred.
( Q ) is the reaction quotient.
(i) $\mathrm{Mg}(\mathrm{s})|\mathrm{Mg}^{2+}(0.001 \mathrm{M}) | \mathrm{Cu}^{2+}(0.0001 \mathrm{M})|\mathrm{Cu}(\mathrm{s})$
Standard cell potential: [ E_{\text{{cell}}}^{\circ} = E_{\text{{Cu}}^{2+}/\text{{Cu}}}^{\circ} - E_{\text{{Mg}}^{2+}/\text{{Mg}}}^{\circ} = 0.34 \, \text{V} - (-2.37 \, \text{V}) = 2.71 \, \text{V} ]
Reaction quotient ( Q ): [ Q = \frac{[\text{{Mg}}^{2+}]}{[\text{{Cu}}^{2+}]} = \frac{0.001}{0.0001} = 10 ]
Number of electrons ( n = 2 )
[ E_{\text{{cell}}} = 2.71 - \frac{0.059}{2} \log 10 ] [ E_{\text{{cell}}} = 2.71 - \frac{0.059}{2} \times 1 ] [ E_{\text{{cell}}} = 2.71 - 0.0295 = 2.68 \, \text{V} ]
(ii) $\mathrm{Fe}(\mathrm{s})|\mathrm{Fe}^{2+}(0.001 \mathrm{M}) | \mathrm{H}^{+}(1 \mathrm{M}) | \mathrm{H}_2(\text{{g}},1\text{{bar}}) | \mathrm{Pt}(\mathrm{s})$
Standard cell potential: [ E_{\text{{cell}}}^{\circ} = E_{\text{{H}}^{+}/\text{{H}}_2}^{\circ} - E_{\text{{Fe}}^{2+}/\text{{Fe}}}^{\circ} = 0 \, \text{V} - (-0.44 \, \text{V}) = 0.44 \, \text{V} ]
Reaction quotient ( Q ): [ Q = \frac{[\text{{Fe}}^{2+}]}{[\text{{H}}^{+}]} = \frac{0.001}{1} = 0.001 ]
Number of electrons ( n = 2 )
[ E_{\text{{cell}}} = 0.44 - \frac{0.059}{2} \log 0.001 ] [ E_{\text{{cell}}} = 0.44 - \frac{0.059}{2} \times (-3) ] [ E_{\text{{cell}}} = 0.44 + 0.0885 = 0.53 \, \text{V} ]
(iii) $\mathrm{Sn}(\mathrm{s})|\mathrm{Sn}^{2+}(0.050 \mathrm{M}) | \mathrm{H}^{+}(0.020 \, \mathrm{M}) | \mathrm{H}_2(\text{{g}},1\text{{bar}}) | \mathrm{Pt}(\mathrm{s})$
Standard cell potential: [ E_{\text{{cell}}}^{\circ} = E_{\text{{H}}^{+}/\text{{H}}_2}^{\circ} - E_{\text{{Sn}}^{2+}/\text{{Sn}}}^{\circ} = 0 \, \text{V} - (-0.14 \, \text{V}) = 0.14 \, \text{V} ]
Reaction quotient ( Q ): [ Q = \frac{[\text{{Sn}}^{2+}]}{[\text{{H}}^{+}]^2} = \frac{0.050}{(0.020)^2} = 125 ]
Number of electrons ( n = 2 )
[ E_{\text{{cell}}} = 0.14 - \frac{0.059}{2} \log 125 ] [ E_{\text{{cell}}} = 0.14 - \frac{0.059}{2} \times 2.097 ] [ E_{\text{{cell}}} = 0.14 - 0.0618 = 0.0782 \, \text{V} ]
(iv) $\mathrm{Pt}(\mathrm{s})|\mathrm{Br}^{-}(0.010 \mathrm{M})|\mathrm{Br}_2(\mathrm{l}) | \mathrm{H}^{+}(0.030 \, \mathrm{M}) | \mathrm{H}_2(\text{{g}},1\text{{bar}}) | \mathrm{Pt}(\mathrm{s})$
Standard cell potential: [ E_{\text{{cell}}}^{\circ} = E_{\text{{H}}^{+}/\text{{H}}_2}^{\circ} - E_{\text{{Br}}_2/\text{{Br}}^-}^{\circ} = 0 \, \text{V} - 1.07 \, \text{V} = 1.07 \, \text{V} ]
Reaction quotient ( Q ): [ Q = \frac{[\text{{Br}}^-]^2}{[\text{{H}}^{+}]^2} ] [ Q = \frac{(0.010)^2}{(0.030)^2} = \left( \frac{0.01}{0.03} \right)^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9} ]
Number of electrons ( n = 2 )
[ E_{\text{{cell}}} = 1.07 - \frac{0.059}{2} \log \left( \frac{1}{9} \right) ] [ E_{\text{{cell}}} = 1.07 - \left( \frac{0.059}{2} \times (-0.954) \right) ] [ E_{\text{{cell}}} = 1.07 + 0.0281 = -1.0981 \, \text{V} ]
In the button cells widely used in watches and other devices the following reaction takes place:
$\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq})$
Determine $\Delta_{r} G^{0}$ and $E^{0}$ for the reaction.
Standard reduction potential of $\mathrm{Zn}^{2+}/\mathrm{Zn}$: $-0.76 , \mathrm{V}$
Standard reduction potential for $\mathrm{Ag}_2\mathrm{O} + \mathrm{H}_2\mathrm{O} + 2e^- \rightarrow 2\mathrm{Ag} + 2\mathrm{OH}^-: +0.34, \mathrm{V}$
To find $\Delta_r G^{0}$ and $E^{0}$ for the reaction:
$$ \mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq}) $$
We use the relationship:
$$ E^{0}_{\text{cell}} = E^{0}_{\text{cathode}} - E^{0}_{\text{anode}} $$
Cathode reaction: $\mathrm{Ag}_{2}\mathrm{O}(\mathrm{s}) + \mathrm{H}_{2}\mathrm{O}(\mathrm{l}) + 2e^- \rightarrow 2\mathrm{Ag}(\mathrm{s}) + 2\mathrm{OH}^- (\mathrm{aq})$
$E^{0}_{\text{cathode}} = +0.34 , \mathrm{V}$
Anode reaction: $\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+} (\mathrm{aq}) + 2e^{-}$
$E^{0}_{\text{anode}} = -(-0.76 , \mathrm{V}) = +0.76 , \mathrm{V}$
Now, combining these:
$$ E^{0}_{\text{cell}} = 0.34 , \mathrm{V} - (-0.76 , \mathrm{V}) = 1.10 , \mathrm{V} $$
For $\Delta_{r} G^{0}$, use the equation:
$$ \Delta_{r} G^{0} = -nFE^{0}_{\text{cell}} $$
where $n = 2$ (number of moles of electrons), $F = 96485 , \mathrm{C , mol^{-1}}$ (Faraday's constant), and $E^{0}_{\text{cell}} = 1.10, \mathrm{V}$.
So:
$$ \Delta_{r} G^{0} = -2 \cdot 96485 , \mathrm{C , mol^{-1}} \cdot 1.10 , \mathrm{V} $$
Using this in a computation:
$$ \Delta_{r} G^{0} = -2 \cdot 96485 , \mathrm{C , mol^{-1}} \cdot 1.10 , \mathrm{J , C^{-1}} = -212,867 , \mathrm{J , mol^{-1}} $$
Or in kilojoules:
$$ \Delta_{r} G^{0} = -212.87 , \mathrm{kJ , mol^{-1}} $$
Therefore,
Standard Cell Potential ($E^{0}$): $1.10 , \mathrm{V}$
Standard Gibbs Free Energy Change ($\Delta_{r} G^{0}$): $-212.87 , \mathrm{kJ , mol^{-1}}$
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Conductivity ($\kappa$)
Conductivity of an electrolytic solution refers to its ability to conduct electric current. It is the measure of how easily ions can move through the solution when an electric field is applied. Conductivity depends on the concentration of ions in the solution, the nature of the ions (their charge and size), the solvent, and temperature.
Conductivity is given by the formula: $$ \kappa = \frac{1}{\rho} $$ where $\rho$ is the resistivity of the solution.
Units:The SI unit of conductivity is Siemens per meter (S m(^-1)). It can also be expressed in Siemens per centimeter (S cm(^-1)).
Molar Conductivity ($\Lambda_m$)
Molar conductivity ($\Lambda_m$) is the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte. It gives the conductance of the volume of solution containing one mole of electrolyte when placed between two electrodes 1 cm apart.
Molar conductivity is calculated by: $$ \Lambda_{m} = \frac{\kappa}{c} $$ where $\kappa$ is the conductivity and $c$ is the concentration of the solution in moles per liter (mol L(^-1)).
Units:The SI unit of molar conductivity is Siemens meter squared per mole (S m² mol(^-1)), but it is often expressed in Siemens centimeter squared per mole (S cm² mol(^-1)).
Variation with Concentration
Conductivity ($\kappa$)
For Strong Electrolytes:
Conductivity decreases with a decrease in concentration. This is because, although dilution increases the number of ions dissociating into the solution, the total number of ions per unit volume of the solution decreases as the solution gets more diluted.
For Weak Electrolytes:
Conductivity also decreases with a decrease in concentration due to the lower degree of ionization at higher concentrations. As the solution is diluted, more molecules ionize, increasing the number of ions but still showing a net decrease in conductivity because of the significantly lower ion concentration per unit volume.
Molar Conductivity ($\Lambda_m$)
For Strong Electrolytes:
Molar conductivity increases slowly with a decrease in concentration. This is due to the decrease in ion-ion interactions as the ions are more widely spaced out in the solution upon dilution.
For Weak Electrolytes:
Molar conductivity increases steeply with a decrease in concentration. In very dilute solutions, weak electrolytes ionize more completely, leading to a significant increase in the number of ions per mole of electrolyte.
Kohlrausch's Law:
For strong electrolytes, Kohlrausch's Law of Independent Migration of Ions states: $$ \Lambda_{m} = \Lambda_{m}^{\circ} - A \sqrt{c} $$ where $\Lambda_{m}^{\circ}$ is the limiting molar conductivity (at infinite dilution), and A is a constant depending on the nature of the electrolyte.
The conductivity of $0.20 \mathrm{M}$ solution of $\mathrm{KCl}$ at $298 \mathrm{~K}$ is $0.0248 \mathrm{~S} \mathrm{~cm}^{-1}$. Calculate its molar conductivity.
To find the molar conductivity, $\Lambda_m$, of the given solution, we use the formula:
\[ \Lambda_m = \frac{1000 \times \kappa}{c} \]
where:
\( \kappa \) is the conductivity of the solution in $\mathrm{S} \mathrm{cm}^{-1}$
\( c \) is the concentration in $\mathrm{mol \, L^{-1}}$
Given:
\[ \kappa = 0.0248 \, \mathrm{S \, cm^{-1}} \]
\[ c = 0.20 \, \mathrm{mol \, L^{-1}} \]
Using the formula:
\[ \Lambda_m = \frac{1000 \times 0.0248 \, \mathrm{S \, cm^{-1}}}{0.20 \, \mathrm{mol \, L^{-1}}} = 124 \, \mathrm{S \, cm^2 \, mol^{-1}} \]
Thus, the molar conductivity of the $0.20 \, \mathrm{M}$ solution of $\mathrm{KCl}$ at $298 \, \mathrm{K}$ is \( \mathbf{124 \, \mathrm{S \, cm^2 \, mol^{-1}}} \).
The resistance of a conductivity cell containing $0.001 \mathrm{M} \mathrm{KCl}$ solution at 298 $\mathrm{K}$ is $1500 \Omega$. What is the cell constant if conductivity of $0.001 \mathrm{M} \mathrm{KCl}$ solution at $298 \mathrm{~K}$ is $0.146 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1}$.
The cell constant (\( G^{*} \)) can be calculated using the formula:
\[ G^{*} = \kappa \times R \]
where:
\( \kappa \) is the conductivity of the solution in $\mathrm{S \, cm}^{-1}$
\( R \) is the resistance in $\Omega$
Given:
\[ \kappa = 0.146 \times 10^{-3} \, \mathrm{S \, cm}^{-1} \]
\[ R = 1500 \, \Omega \]
Plugging in the values:
\[ G^{*} = 0.146 \times 10^{-3} \, \mathrm{S \, cm}^{-1} \times 1500 \, \Omega \]
\[ G^{*} = 0.146 \times 10^{-3} \times 1500 \]
\[ G^{*} = 0.219 \, \mathrm{cm^{-1}} \]
Therefore, the cell constant is \( \mathbf{0.219 \, \mathrm{cm^{-1}}} \).
The conductivity of sodium chloride at $298 \mathrm{~K}$ has been determined at different concentrations and the results are given below:
Concentration/M | 0.001 | 0.010 | 0.020 | 0.050 | 0.100 |
---|---|---|---|---|---|
$10^{2} \times \kappa / \mathrm{S} \mathrm{m}^{-1}$ | 1.237 | 11.85 | 23.15 | 55.53 | 106.74 |
Calculate $\Lambda_{m}$ for all concentrations and draw a plot between $\Lambda_{m}$ and $\mathrm{c}^{1 / 2}$. Find the value of $\Lambda_{m}^{0}$.
To calculate the molar conductivity, $\Lambda_m$, for each concentration, we use the formula:
[ \Lambda_m = \frac{\kappa}{c} ]
Given values:
Concentration (M) | ( 10^{2} \times \kappa ) (S m^{-1}) | ( \kappa ) (S m^{-1}) |
---|---|---|
0.001 | 1.237 | 0.01237 |
0.010 | 11.85 | 0.1185 |
0.020 | 23.15 | 0.2315 |
0.050 | 55.53 | 0.5553 |
0.100 | 106.74 | 1.0674 |
Calculations:
For each molarity:
[ \Lambda_m = \frac{\kappa}{c} ]
For ( 0.001 \, \text{M} ): [ \Lambda_m = \frac{0.01237}{0.001} = 12.37 \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} ]
For ( 0.010 \, \text{M} ): [ \Lambda_m = \frac{0.1185}{0.010} = 11.85 \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} ]
For ( 0.020 \, \text{M} ): [ \Lambda_m = \frac{0.2315}{0.020} = 11.575 \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} ]
For ( 0.050 \, \text{M} ): [ \Lambda_m = \frac{0.5553}{0.050} = 11.106 \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} ]
For ( 0.100 \, \text{M} ): [ \Lambda_m = \frac{1.0674}{0.100} = 10.674 \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} ]
Plotting $\Lambda_m$ vs $ \sqrt{c} $:
To plot $\Lambda_m$ against $\sqrt{c}$, we need to calculate $\sqrt{c}$ for each concentration:
Concentration (M) | ( c^{1/2} ) $M(^{1/2}$) | ( \Lambda_m ) $S m^2 mol^{-1}$ |
---|---|---|
0.001 | 0.0316 | 12.37 |
0.010 | 0.1000 | 11.85 |
0.020 | 0.1414 | 11.575 |
0.050 | 0.2236 | 11.106 |
0.100 | 0.3162 | 10.674 |
Using these values, you can plot $\Lambda_m$ (y-axis) against $ \sqrt{c} $ (x-axis).
Linear Fit to Find $\Lambda_m^0$:
To determine $\Lambda_m^0$, plot the values on a graph and fit a straight line to the data points. The y-intercept of the line will give you $\Lambda_m^0$.
Below is the plot of $\Lambda_{m}$ versus $\sqrt{c}$:
Plot Values and Analysis
$c^{1/2}$ (M(^{1/2})) | $\Lambda_m$ (S m² mol⁻¹) |
---|---|
0.0316 | 12.37 |
0.1000 | 11.85 |
0.1414 | 11.575 |
0.2236 | 11.106 |
0.3162 | 10.674 |
To obtain $\Lambda_m^0$, extend the linear relationship (line of best fit) to where it intersects the y-axis ($c^{1/2} = 0$) which is $124.1 \text{S}~\text{cm}^2 / \text{mol}$
Conductivity of $0.00241 \mathrm{M}$ acetic acid is $7.896 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}$. Calculate its molar conductivity. If $\Lambda_{m}^{0}$ for acetic acid is $390.5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, what is its dissociation constant?
To solve this problem, we'll follow these steps:
Calculate the molar conductivity using the formula:
$$ \Lambda_{m} = \frac{\kappa}{c} \times 1000 $$
where:
( \kappa = 7.896 \times 10^{-5} \ \mathrm{S \ cm^{-1}} )
( c = 0.00241 \ \mathrm{mol \ L^{-1}} )
Calculate the degree of dissociation ( \alpha ) using the formula:
$$ \alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{0}} $$
where:
( \Lambda_{m}^{0} = 390.5 \ \mathrm{S \ cm^{2} \ mol^{-1}} )
Calculate the dissociation constant ( K_a ) using the formula:
$$ K_a = \frac{c \alpha^2}{1 - \alpha} $$
Step 1: Calculate Molar Conductivity
$$ \Lambda_{m} = \frac{7.896 \times 10^{-5} \ \mathrm{S \ cm^{-1}}}{0.00241 \ \mathrm{mol \ L^{-1}}} \times 1000 $$
Step 2: Calculate the Degree of Dissociation ( \alpha )
$$ \alpha = \frac{\Lambda_{m}}{390.5 \ \mathrm{S \ cm^{2} \ mol^{-1}}} $$
Step 3: Calculate the Dissociation Constant ( K_a )
$$ K_a = \frac{0.00241 \times \alpha^2}{1 - \alpha} $$
Step 1: Calculate Molar Conductivity
$$ \Lambda_{m} = \frac{7.896 \times 10^{-5} \ \mathrm{S \ cm^{-1}}}{0.00241 \ \mathrm{mol \ L^{-1}}} \times 1000 \approx 32.76 \ \mathrm{S \ cm^{2} \ mol^{-1}} $$
Step 2: Calculate the Degree of Dissociation (( \alpha ))
$$ \alpha = \frac{\Lambda_{m}}{390.5 \ \mathrm{S \ cm^{2} \ mol^{-1}}} \approx 0.0839 $$
Step 3: Calculate the Dissociation Constant (( K_a ))
Using the formula: $$ K_a = \frac{c \alpha^2}{1 - \alpha} $$ where:
( c = 0.00241 \ \mathrm{mol \ L^{-1}} )
( \alpha \approx 0.0839 )
$$ K_a \approx 0.00001852 \ \mathrm{mol \ L^{-1}} $$
Thus, the dissociation constant ( K_a ) for the acetic acid is ( \mathbf{1.852 \times 10^{-5} \ \mathrm{mol \ L^{-1}}} ).
How much charge is required for the following reductions:
(i) $1 \mathrm{~mol}$ of $\mathrm{Al}^{3+}$ to $\mathrm{Al}$ ?
(ii) $1 \mathrm{~mol}$ of $\mathrm{Cu}^{2+}$ to $\mathrm{Cu}$ ?
(iii) $1 \mathrm{~mol}$ of $\mathrm{MnO}_{4}^{-}$to $\mathrm{Mn}^{2+}$ ?
Reduction of ( 1 \mathrm{~mol} ) of ( \mathrm{Al}^{3+} ) to ( \mathrm{Al} )
Charge required: 289,461 C
Reduction of ( 1 \mathrm{~mol} ) of ( \mathrm{Cu}^{2+} ) to ( \mathrm{Cu} )
Charge required: 192,974 C
Reduction of ( 1 \mathrm{~mol} ) of ( \mathrm{MnO}_{4}^{-} ) to ( \mathrm{Mn}^{2+} )
Charge required: 482,435 C
These charges are proportional to the number of moles of electrons required for each reduction process.
How much electricity in terms of Faraday is required to produce
(i) $20.0 \mathrm{~g}$ of $\mathrm{Ca}$ from molten $\mathrm{CaCl}_{2}$ ?
(ii) $40.0 \mathrm{~g}$ of $\mathrm{Al}$ from molten $\mathrm{Al}_{2} \mathrm{O}_{3}$ ?
To calculate the amount of electricity required in terms of Faraday (F), we need to consider the stoichiometry of the electrolysis reactions for calcium and aluminum:
(i) Production of $ \mathrm{Ca} $ from molten $ \mathrm{CaCl}_{2} $
The half-reaction for the production of calcium is: $$ \mathrm{Ca}^{2+} + 2 \mathrm{e}^{-} \rightarrow \mathrm{Ca} $$ Each calcium ion requires 2 electrons (or 2 Faraday of charge) to be reduced to calcium metal.
1 Faraday (F) = 96487 C/mol of electrons
Calculations:
Molar mass of $ \mathrm{Ca} = 40.08 ; \mathrm{g/mol} $
Mass to moles conversion: $$ \text{Moles of } \mathrm{Ca} = \frac{20.0 ; \mathrm{g}}{40.08 ; \mathrm{g/mol}} = 0.499 ; \mathrm{mol} $$
Faradays required: $$ \text{Faradays required} = 0.499 ; \mathrm{mol} \times 2 ; \mathrm{F/mol} = 0.998 ; \mathrm{F} $$
(ii) Production of $ \mathrm{Al} $ from molten $ \mathrm{Al}_{2} \mathrm{O}_{3} $
The half-reaction for the production of aluminum is: $$ \mathrm{Al}^{3+} + 3 \mathrm{e}^{-} \rightarrow \mathrm{Al} $$ Each aluminum ion requires 3 electrons (or 3 Faraday of charge) to be reduced to aluminum metal.
Calculations:
Molar mass of $ \mathrm{Al} = 26.98 ; \mathrm{g/mol} $
Mass to moles conversion: $$ \text{Moles of } \mathrm{Al} = \frac{40.0 ; \mathrm{g}}{26.98 ; \mathrm{g/mol}} = 1.482 ; \mathrm{mol} $$
Faradays required: $$ \text{Faradays required} = 1.482 ; \mathrm{mol} \times 3 ; \mathrm{F/mol} = 4.446 ; \mathrm{F} $$
Summary:
To produce $ 20.0 ; \mathrm{g} $ of $ \mathrm{Ca} $: $ 0.998 ; \mathrm{F} $
To produce $ 40.0 ; \mathrm{g} $ of $ \mathrm{Al} $: $ 4.446 ; \mathrm{F} $
How much electricity is required in coulomb for the oxidation of
(i) $1 \mathrm{~mol}$ of $\mathrm{H}_{2} \mathrm{O}$ to $\mathrm{O}_{2}$ ?
(ii) $1 \mathrm{~mol}$ of $\mathrm{FeO}$ to $\mathrm{Fe}_{2} \mathrm{O}_{3}$ ?
Let's calculate the required electricity for both reactions step by step.
(i) Oxidation of $1 \ \mathrm{mol}$ of $\mathrm{H}_{2} \mathrm{O}$ to $\mathrm{O}_{2}$
The half-reaction for the oxidation of water to oxygen is:
$$ 2 \ \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{O}_{2} + 4 \ \mathrm{H}^{+} + 4 \ \mathrm{e}^{-} $$
This indicates that 2 moles of $\mathrm{H}_{2} \mathrm{O}$ require 4 moles of electrons.
Thus, for 1 mole of $\mathrm{H}_{2} \mathrm{O}$, the moles of electrons required are:
$$ \text{Moles of electrons} = \frac{4}{2} = 2 \ \text{moles of electrons} $$
Since 1 mole of electrons is equivalent to 1 Faraday (\( F = 96487 \ \mathrm{C} \)), the total charge required is:
$$ \text{Charge} = 2 \ \text{moles of electrons} \times 96487 \ \mathrm{C \ per \ mol} $$
(ii) Oxidation of $1 \ \mathrm{mol}$ of $\mathrm{FeO}$ to $\mathrm{Fe}_{2} \mathrm{O}_{3}$
The oxidation states change as follows:
In $\mathrm{FeO}$, Fe is in \( +2 \) oxidation state.
In $\mathrm{Fe}_{2} \mathrm{O}_{3}$, Fe is in \( +3 \) oxidation state.
The balanced equations are:
$$ 2 \ \mathrm{FeO} \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3} + 2 \ e^{-} $$
This indicates that 2 moles of $\mathrm{FeO}$ lose 2 moles of electrons to form $\mathrm{Fe}_{2} \mathrm{O}_{3}$.
Thus, for 1 mole of $\mathrm{FeO}$, the moles of electrons required are:
$$ \text{Moles of electrons} = 1 \ \text{moles of electrons} $$
Since 1 mole of electrons is equivalent to 1 Faraday (\( F = 96487 \ \mathrm{C} \)), the total charge required is:
$$ \text{Charge} = 1 \ \text{mole of electrons} \times 96487 \ \mathrm{C \ per \ mol} $$
Computation for Charge
1. Oxidation of $1 \ \text{mol}$ of $\mathrm{H}_2 \mathrm{O}$:
$$ 2 \ \mathrm{mol} \times 96487 \ \mathrm{C \ per \ mol} = 192974 \ \mathrm{C} $$
2. Oxidation of $1 \ \text{mol}$ of $\mathrm{FeO}$:
$$ 1 \ \mathrm{mol} \times 96487 \ \mathrm{C \ per \ mol} = 96487 \ \mathrm{C} $$
Thus, the required electricity is:
For $1 \ \mathrm{mol}$ of $\mathrm{H}_{2} \mathrm{O}$: \( \mathbf{192974 \ \mathrm{C}} \)
For $1 \ \mathrm{mol}$ of $\mathrm{FeO}$: \( \mathbf{96487 \ \mathrm{C}} \)
A solution of $\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}$ is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of $\mathrm{Ni}$ is deposited at the cathode?
The mass of nickel ($\mathrm{Ni}$) deposited at the cathode is approximately 1.824 grams.
Here are the detailed steps:
The charge passed through the solution: $$ Q = 5 , \text{A} \times 1200 , \text{s} = 6000 , \text{C} $$
Number of moles of electrons: $$ n_{e} = \frac{6000 , \text{C}}{96485 , \text{C/mol}} = 0.0622 , \text{mol of electrons} $$
Number of moles of nickel: $$ n_{\mathrm{Ni}} = \frac{0.0622 , \text{mol}}{2} = 0.0311 , \text{mol} $$
Mass of nickel: $$ \text{mass} = 0.0311 , \text{mol} \times 58.69 , \text{g/mol} = 1.824 , \text{g} $$
Thus, the mass of nickel deposited is approximately 1.824 grams.
Three electrolytic cells A,B,C containing solutions of $\mathrm{ZnSO}_{4}, \mathrm{AgNO}_{3}$ and $\mathrm{CuSO}_{4}$, respectively are connected in series. A steady current of 1.5 amperes was passed through them until $1.45 \mathrm{~g}$ of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
1. Time Calculation
Given:
Mass of Ag deposited: $1.45$ g
Molar mass of Ag: $107.87$ g/mol
Steady current: $1.5$ A
First, calculate the moles of Ag deposited: [ n_{\mathrm{Ag}} = \frac{1.45 \text{ g}}{107.87 \text{ g/mol}} \approx 0.01344 \text{ mol} ]
Next, calculate the total charge required: [ Q = 0.01344 \text{ mol} \times 96500 \text{ C/mol} \approx 1297 \text{ C} ]
Finally, calculate the time: [ t = \frac{Q}{I} = \frac{1297 \text{ C}}{1.5 \text{ A}} \approx 860 \text{ seconds} ] [ t \approx 14 \text{ minutes} ]
2. Mass of Copper and Zinc Deposited
For Copper ($\mathrm{Cu}$):
Moles of $e^{-}$ required for $1$ mol Cu: $2$ moles
Molar mass of $\mathrm{Cu}$: $63.55$ g/mol
[ n_{\mathrm{Cu}} = \frac{1297 \text{ C}}{2 \times 96500 \text{ C/mol}} \approx 0.00672 \text{ mol} ] [ m_{\mathrm{Cu}} = n_{\mathrm{Cu}} \times 63.55 \text{ g/mol} \approx 0.427 \text{ g} ]
For Zinc ((\mathrm{Zn})):
Moles of (e^{-}) required for (1) mol Zn: (2) moles
Molar mass of ( \mathrm{Zn} ): ( 65.38 ) g/mol
[ n_{\mathrm{Zn}} = \frac{1297 \text{ C}}{2 \times 96500 \text{ C/mol}} \approx 0.00672 \text{ mol} ] [ m_{\mathrm{Zn}} = n_{\mathrm{Zn}} \times 65.38 \text{ g/mol} \approx 0.439 \text{ g} ]
Summary
Time the current flowed: $\mathbf{860 \text{ seconds}}$ (or $\mathbf{14 \text{ minutes}})$
Mass of copper deposited: $\mathbf{0.427 \text{ g}}$
Mass of zinc deposited: $\mathbf{0.439 \text{ g}}$
Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) $\mathrm{Fe}^{3+}(\mathrm{aq})$ and $\mathrm{I}^{-}(\mathrm{aq})$
(ii) $\mathrm{Ag}^{+}$(aq) and $\mathrm{Cu}(\mathrm{s})$
(iii) $\mathrm{Fe}^{3+}$ (aq) and $\mathrm{Br}^{-}$(aq)
(iv) $\mathrm{Ag}(\mathrm{s})$ and $\mathrm{Fe}^{3+}$ (aq)
(v) $\mathrm{Br}_{2}$ (aq) and $\mathrm{Fe}^{2+}$ (aq).
$\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}$: +0.77 V
$\mathrm{I_2}/\mathrm{I}^-$: +0.54 V
$\mathrm{Ag}^{+}/\mathrm{Ag}$: +0.80 V
$\mathrm{Cu}^{2+}/\mathrm{Cu}$: +0.34 V
$\mathrm{Br}_2/\mathrm{Br}^-$: +1.07 V
$\mathrm{Fe}^{2+}/\mathrm{Fe}$: -0.44 V
Now, let's evaluate each reaction.
(i) $\mathrm{Fe}^{3+}(\mathrm{aq})$ and $\mathrm{I}^{-}(\mathrm{aq})$
Reduction: $\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}$ ($E^\circ$ = +0.77 V)
Oxidation: $\mathrm{2I}^- \rightarrow \mathrm{I}_2 + 2e^-$ ($E^\circ$ = -0.54 V)
Since the reduction potential of $\mathrm{Fe}^{3+}(\mathrm{aq})$ is higher than the oxidation potential of $\mathrm{I}^{-}(\mathrm{aq})$, the reaction is feasible.
(ii) $\mathrm{Ag}^{+}$(aq) and $\mathrm{Cu}(\mathrm{s})$
Reduction: $\mathrm{Ag}^{+} + e^- \rightarrow \mathrm{Ag}$ ($E^\circ$ = +0.80 V)
Oxidation: $\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+} + 2e^-$ ($E^\circ$ = -0.34 V)
Since the reduction potential of $\mathrm{Ag}^{+}$ is higher than the oxidation potential of $\mathrm{Cu}$, the reaction is feasible.
(iii) $\mathrm{Fe}^{3+}$ (aq) and $\mathrm{Br}^{-}$(aq)
Reduction: $\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}$ ($E^\circ$ = +0.77 V)
Oxidation: $\mathrm{2Br}^- \rightarrow \mathrm{Br}_2 + 2e^-$ ($E^\circ$ = -1.07 V)
Since the reduction potential of $\mathrm{Fe}^{3+}$ is lower than the oxidation potential of $\mathrm{Br}^{-}$, the reaction is not feasible.
(iv) $\mathrm{Ag}(\mathrm{s})$ and $\mathrm{Fe}^{3+}$ (aq)
Reduction: $\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}$ ($E^\circ$ = +0.77 V)
Oxidation: $\mathrm{Ag} \rightarrow \mathrm{Ag}^{+} + e^-$ ($E^\circ$ = -0.80 V)
Since the reduction potential of $\mathrm{Fe}^{3+}$ is lower than the oxidation potential of $\mathrm{Ag}$, the reaction is not feasible.
(v) $\mathrm{Br}_{2}$ (aq) and $\mathrm{Fe}^{2+}$ (aq)
Reduction: $\mathrm{Br}_2 + 2e^- \rightarrow \mathrm{2Br}^-$ ($E^\circ$ = +1.07 V)
Oxidation: $\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^-$ ($E^\circ$ = -0.77 V)
Since the reduction potential of $\mathrm{Br}_2$ is higher than the oxidation potential of $\mathrm{Fe}^{2+}$, the reaction is feasible.
To summarize:
Feasible
Feasible
Not feasible
Not feasible
Feasible
Each reaction will proceed as indicated if the condition is met that the potential for the reduction half reaction is higher than that for oxidation when reversed.
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of $\mathrm{AgNO}_{3}$ with silver electrodes.
(ii) An aqueous solution of $\mathrm{AgNO}_{3}$ with platinum electrodes.
(iii) A dilute solution of $\mathrm{H}_{2} \mathrm{SO}_{4}$ with platinum electrodes.
(iv) An aqueous solution of $\mathrm{CuCl}_{2}$ with platinum electrodes.
To predict the products of electrolysis, we need to consider the nature of the electrodes (reactive or inert), the electrolyte solution, and the possible reactions at the cathode and anode based on standard electrode potentials.
(i) An aqueous solution of $\mathrm{AgNO}_{3}$ with silver electrodes.
Silver electrodes are reactive and will participate in the electrolysis.
Anode reaction (oxidation): $$ \mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Ag}^{+}(\mathrm{aq}) + \mathrm{e}^{-} $$ Silver metal dissolves, forming $\mathrm{Ag}^{+}$ ions.
Cathode reaction (reduction): $$ \mathrm{Ag}^{+}(\mathrm{aq}) + \mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) $$ $\mathrm{Ag}^{+}$ ions are reduced to silver metal.
Overall Reaction: $$ \mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Ag}^{+}(\mathrm{aq}) + \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Ag}(\mathrm{s}) $$ Effectively, silver is transferred from the anode to the cathode.
(ii) An aqueous solution of $\mathrm{AgNO}_{3}$ with platinum electrodes.
Platinum electrodes are inert and do not participate in the electrolysis.
Anode reaction (oxidation): $$ \mathrm{2H_2O}(\mathrm{l}) \rightarrow \mathrm{O_2}(\mathrm{g}) + 4 \mathrm{H}^{+}(\mathrm{aq}) + 4 \mathrm{e}^{-} $$ Water is oxidized, releasing oxygen gas and producing $\mathrm{H}^{+}$ ions.
Cathode reaction (reduction): $$ \mathrm{Ag}^{+}(\mathrm{aq}) + \mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) $$ Silver ions are reduced to silver metal.
Overall Reaction: $$ \mathrm{2H_2O}(\mathrm{l}) + 2 \mathrm{AgNO_3}(\mathrm{aq}) \rightarrow \mathrm{O_2}(\mathrm{g}) + 4 \mathrm{HNO_3}(\mathrm{aq}) + 2 \mathrm{Ag}(\mathrm{s}) $$
(iii) A dilute solution of $\mathrm{H}_{2} \mathrm{SO}_{4}$ with platinum electrodes.
Platinum electrodes are inert, so they do not participate in the electrolysis.
Anode reaction (oxidation): $$ 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{O}_{2}(\mathrm{g}) + 4 \mathrm{H}^{+}(\mathrm{aq}) + 4 \mathrm{e}^{-} $$ Water is oxidized, releasing oxygen gas and $\mathrm{H}^{+}$ ions.
Cathode reaction (reduction): $$ 2 \mathrm{H}^{+}(\mathrm{aq}) + 2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}(\mathrm{g}) $$ Hydrogen ions are reduced to hydrogen gas.
Overall Reaction: $$ 2 \mathrm{H}_2\mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{H}_2(\mathrm{g}) + \mathrm{O}_2(\mathrm{g}) $$
(iv) An aqueous solution of $\mathrm{CuCl}_{2}$ with platinum electrodes.
Platinum electrodes are inert, so they do not participate in the electrolysis.
Anode reaction (oxidation): $$ 2 \mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cl}_2(\mathrm{g}) + 2\mathrm{e}^{-} $$ Chloride ions are oxidized to chlorine gas.
Cathode reaction (reduction): $$ \mathrm{Cu}^{2+}(\mathrm{aq}) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) $$ Copper ions are reduced to copper metal.
Overall Reaction: $$ \mathrm{CuCl}_{2}(\mathrm{aq}) \rightarrow \mathrm{Cu}(\mathrm{s}) + \mathrm{Cl}_{2}(\mathrm{g}) $$
In conclusion, the significant products for each electrolysis are:
Aqueous $\mathrm{AgNO}_{3}$ with silver electrodes: Ag dissolves at anode, Ag deposits at cathode.
Aqueous $\mathrm{AgNO}_{3}$ with platinum electrodes: O₂ gas at anode, Ag deposits at cathode.
Dilute $\mathrm{H}_{2} \mathrm{SO}_{4}$ with platinum electrodes: O₂ gas at anode, H₂ gas at cathode.
Aqueous $\mathrm{CuCl}_{2}$ with platinum electrodes: Cl₂ gas at anode, Cu deposits at cathode.
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Ask Chatterbot AIExtra Questions - Electrochemistry | NCERT | Chemistry | Class 12
The pH of $0.5$ L of a $1.0$ M $\text{NaCl}$ solution after electrolysis for $965$ seconds using $5.0$ A current (100% efficiency) is:
Option A) 1
Option B) 13
Option C) 12.7
Option D) 1.3
To find the pH of the solution after the electrolysis, we need to understand what happens during the electrolysis of a sodium chloride (NaCl) solution (in this case which can also be referred to as brine).
Electrolysis Process:
1. Anode (oxidation): At the anode, chloride ions $\mathrm{Cl}^{-}$ are oxidized to chlorine gas $\mathrm{Cl}_2$.
$$ \text{2Cl}^- \rightarrow \text{Cl}_2 + 2e^-$$
2. Cathode (reduction): At the cathode, water molecules are reduced to hydrogen gas $H_2$ and hydroxide ions $\mathrm{OH}^{⁻}$.
$$ \text{2H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2OH^- $$
Amount of Electricity (Coulombs, Q):
$$ Q = \text{current} \times \text{time} = 5.0 \, \text{A} \times 965 \, \text{s} = 4825 \, \text{C} $$
Moles of Electrons (using Faraday's Constant, F = 96500 C/mol e⁻):
Number of moles of electrons transferred can be calculated by dividing the total charge by Faraday’s constant.
$$ n = \frac{Q}{F} = \frac{4825}{96500} \approx 0.05 \, \text{mol} $$
Moles of H₂ and Cl₂ generated:
Since 2 moles of electrons produce 1 mole of Cl₂ (from 2Cl⁻) and 1 mole of H₂ (from 2H⁺ derived from 2H₂O), the moles of Cl₂ and H₂ generated are half of the moles of electrons:
$$ \text{Moles of Cl}_2 \text{ and H}_2 = \frac{0.05}{2} = 0.025 \, \text{mol} $$
Moles of OH⁻ generated:
The same amount of OH⁻ is produced as H₂ at the cathode:
$$ \text{Moles of OH}^- = 0.05 \, \text{mol} $$
Concentration of OH⁻ in the Solution:
The new volume of the solution might not change significantly, so estimated concentration of OH⁻ will be:
$$ [\text{OH}^-] = \frac{0.05 \, \text{mol}}{0.5 \, \text{L}} = 0.1 \, \text{M} $$
pOH calculation:
$$ \text{pOH} = -\log[\text{OH}^-] = -\log[0.1] = 1 $$
pH Calculation:
$$ \text{pH} + \text{pOH} = 14 $$
$$ \text{pH} = 14 - 1 = 13 $$
Thus, the pH of the solution after electrolysis is 13, which is option B.
The oxidant in the reaction is: $\mathrm{SO}{2} + \mathrm{N}{2} \mathrm{O}{5} \rightarrow \mathrm{SO}{3} + 2 \mathrm{NO}_{2}$
A. $\mathrm{SO}{2}$ B. $\mathrm{N}{2} \mathrm{O}{5}$ C. $\mathrm{SO}{3}$ D. $2 \mathrm{NO}_{2}$
The correct answer is
B. $\mathrm{N}{2} \mathrm{O}{5}$.
To determine the oxidant from the given reaction:
$$ \mathrm{SO}{2} + \mathrm{N}{2} \mathrm{O}{5} \rightarrow \mathrm{SO}{3} + 2 \mathrm{NO}_{2} $$
We need to look at the changes in oxidation states:
In $\mathrm{N}{2} \mathrm{O}{5}$, nitrogen has an oxidation state of +5.
In $\mathrm{NO}_{2}$, nitrogen is at an oxidation state of +4.
This means that during the reaction, $\mathrm{N}{2} \mathrm{O}{5}$ behaves as an oxidizing agent (it loses oxygen and gets reduced), thereby making it the oxidant in the reaction. Hence, option B is correct.
Which of the below given options are correct for the electrolysis of aqueous copper sulphate solution with a platinum electrode?
A. Copper is oxidized.
B. Oxygen is evolved at the cathode.
C. Platinum is deposited.
D. Oxygen is evolved at the anode.
The correct option is D: Oxygen is evolved at the anode.
During the electroysis of an aqueous copper sulfate solution using a platinum electrode, the ions present in the solution include $\mathrm{Cu}^{2+}$, $\mathrm{H}^{+}$, $\mathrm{SO}_4^{2-}$, and $\mathrm{OH}^-$. Here are the reactions occurring at the cathode and anode:
At the Cathode: The reduction reaction involves the copper ions gaining electrons to form copper metal: $$ \mathrm{Cu}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Cu} $$
At the Anode: The hydroxide ions lose electrons to eventually produce oxygen: $$ 4\mathrm{OH}^- \rightarrow 2\mathrm{H}_2\mathrm{O} + \mathrm{O}_2 + 4\mathrm{e}^- $$
From this, we identify that oxygen gas is evolved at the anode, thus making choice D the correct answer.
In the electrolytic refining process of zinc, it is given that the crude metal is made anode while the electrolyte used is $\mathrm{H}{2} \mathrm{SO}{4}$ (dilute) $+\mathrm{ZnSO}_{4}$. The cathode should be:
A. Pure Zinc
B. $\mathrm{Zn}-\mathrm{Hg}$ Amalgam
C. Aluminum
D. All of the above
Solution
The correct answer is C. Aluminum.
During the electrolytic refining process of zinc, the cathode plays a crucial role. When using a dilute $\mathrm{H}_2\mathrm{SO}_4 + \mathrm{ZnSO}_4$ solution as the electrolyte, using pure zinc as the cathode would result in its dissolution due to the acidic environment. This makes using pure zinc impractical. On the other hand, aluminum does not dissolve under these electrolytic conditions, making it a suitable choice for the cathode material. Hence, the best option for the cathode in this setting is Aluminum.
Differentiate between primary and secondary cells, citing some examples for each.
Following are the key differences between primary and secondary cells along with examples for each:
-
Electrochemical Reaction: In a primary cell, the electrochemical reaction is not reversible, meaning the chemical components cannot convert back to their original form after usage. Conversely, in a secondary cell, the electrochemical reaction is reversible, allowing the cell to be recharged and reused.
-
Usability and Rechargeability: A primary cell is intended for single use and is disposed of after its charge is depleted. On the other hand, a secondary cell can be recharged and used multiple times by reversing the chemical reaction through the application of electrical current.
-
Cost Implications: Primary cells generally have a lower initial cost compared to secondary cells, which have a higher initial cost due to their complexity and rechargeability.
-
Examples:
- Primary Cells: Examples include Leclanché cells, Daniell cells, and dry cells.
- Secondary Cells: Examples include lead-acid accumulators, nickel-cadmium (NiCd), nickel-metal hydride (NiMH), and lithium-ion (Li-ion) batteries.
Standard potential of two half-cells E_{1}^{o} and E_{2}^{o} can not be added directly because they are
A intensive properties
B extensive properties
C non-directional
Solution
The correct option is A: intensive properties.
Standard electrode potentials are considered intensive properties. This means they do not depend on the amount of substance present and hence, cannot be simply added together. This concept is similar to how mixing two solutions doesn't allow for a straightforward addition of their densities to find the final density.
Although when two half-cells are combined, it might appear as if we are adding their standard potentials, what we actually add are the changes in Gibbs free energy ($ \Delta \mathrm{G} $) of the reactions occurring in each half-cell.
A student has to connect 4 cells of $1.5 \mathrm{~V}$ each in order to form a battery of voltage $6 \mathrm{~V}$. Which of the following highlights the correct arrangement of cells? (a) $$
- \text { HH. H上 } $$ (b) $$ -\mid н H_{H}- $$ (c) $$ -\mid H H H+ $$ (d) $$ -H H H \longrightarrow $$
A $\mathrm{A}$
B $\mathrm{B}$
C $\mathrm{C}$
D $\mathrm{D}$
The correct answer is A.
In order to achieve a total voltage of $6 \mathrm{~V}$, the cells, each rated at $1.5 \mathrm{~V}$, must be connected in series. This arrangement accumulates the voltage of each cell, leading to an overall voltage that is the sum of the individual cell voltages. In option A, the cells are depicted as connected end-to-end in a series format, which correctly sums up their voltages to $6 \mathrm{~V}$.
"How to find the number of electrons?"
The process to determine the number of electrons in an atom involves a few straightforward steps:
-
Mass Number: This is the sum of the neutrons and protons in an atom.
-
Atomic Number: This represents the total number of protons in the atom.
-
For atoms that are electrically neutral, the number of electrons is equal to the atomic number. This is because the number of electrons must balance out the positive charge of the protons to maintain electrical neutrality.
Calculate the emf of the following cell at 25°C: Ag(s)|Ag⁺$1.0×10^-3 M$ || Cu²⁺$10^-1 M$|Cu(s) given $E_cell° = +0.46 V$ and log $10^n = n$.
For the given cell arrangement, the reduction-oxidation reaction can be described as:
$$ 2 \mathrm{Ag(s)} + \mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow 2 \mathrm{Ag}^{+}(\mathrm{aq}) + \mathrm{Cu(s)} $$
From the reaction, it's evident we have 2 electrons transferred, thus number of electrons, $ n = 2 $.
The Nernst equation that relates standard cell potential ($ E^\circ_{\text{cell}} $), temperature, concentration, and number of electrons transferred is provided by:
$$ E_{\text{cell}} = E_{\text{cell}}^\circ - \frac{0.059}{n} \log \frac{[\mathrm{Ag}^+]^2}{[\mathrm{Cu}^{2+}]} $$
Substituting in the values from our problem, we calculate the emf as follows:
$$ E_{\text{cell}} = 0.46 - \frac{0.059}{2} \log \frac{(1.0 \times 10^{-3})^2}{10^{-1}} $$ $$ = 0.46 - \frac{0.059}{2} \log \left(10^{-6}/10^{-1}\right) $$ $$ = 0.46 - \frac{0.059}{2} \log 10^{-5} $$ $$ = 0.46 - 0.0295 \times (-5) $$ $$ = 0.46 + 0.1475 = 0.6075 , \text{V} $$
Thus, the calculated emf of the cell at 25°C is 0.6075 V.
The correct statement regarding an electrophile is:
(A) Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from a nucleophile.
(B) Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from an electrophile.
(C) Electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile.
(D) An electrophile can be either a neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile.
The correct answer is (D): An electrophile can be either a neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile.
In organic chemistry, an electrophile is defined as an electron pair acceptor. Electrophiles are typically either positively charged or neutral species. They have vacant orbitals and tend to be attracted to electron-rich centers, known as nucleophiles. During a chemical reaction, an electrophile accepts a pair of electrons from a nucleophile to form a bond. This characterization helps in understanding the nature of reactions involving electrophiles and their interaction with nucleophiles.
Read the given statements and choose the correct option: i) Electrolytic cell is a device which converts electrical energy into chemical energy by a process of electrolysis. ii) Electrochemical cell is a device which converts chemical energy into electrical energy through a redox reaction.
A) Only (i) is correct.
B) Only (ii) is correct.
C) Both (i) and (ii) are correct.
D) Neither of (i) and (ii) are correct.
The correct answer is C) Both (i) and (ii) are correct.
-
Electrolytic cell: This type of cell converts electrical energy into chemical energy through the process of electrolysis. An example is the Hydrogen-oxygen fuel cell.
-
Electrochemical cell: This cell converts chemical energy into electrical energy via a redox reaction. An example of this is the Galvanic cell.
Both statements (i) and (ii) accurately describe the functioning of these cells, confirming that option C is correct.
Which ion will be migrating towards the anode during the electrolysis of concentrated $\mathrm{NaCl}$ solution?
OR
What will be the reaction taking place at the anode during the electrolysis of concentrated $\mathrm{NaCl}$ solution?
During the electrolysis of concentrated $\mathrm{NaCl}$ solution, both $\mathrm{NaCl}$ (sodium chloride) and $\mathrm{H_2O}$ (water) are present and dissociate into their respective ions. The dissociation reactions are:
- $\mathrm{NaCl} \rightarrow \mathrm{Na^+} + \mathrm{Cl^-}$
- $\mathrm{H_2O} \rightarrow \mathrm{H^+} + \mathrm{OH^-}$
At the anode, competition occurs between the $\mathrm{Cl^-}$ ions from sodium chloride and the $\mathrm{OH^-}$ ions from water. Each seeks to lose electrons (oxidation process). The relevant reactions are:
- $\mathrm{2Cl^-} \rightarrow \mathrm{Cl_2} + 2e^-$, with a standard electrode potential $E^\circ = +1.36, \text{V}$
- $\mathrm{4OH^-} \rightarrow \mathrm{O_2} + 2\mathrm{H_2O} + 4e^-$, with a standard electrode potential $E^\circ = +1.23, \text{V}$
While reaction involving the $\mathrm{OH^-}$ ions would normally be preferred due to its lower standard electrode potential (lower energy requirement for oxidation), the overpotential effect influences the actual preference at the anode. In reality, it is the $\mathrm{Cl^-}$ that is more readily oxidized over $\mathrm{OH^-}$ ions, leading to the production of $\mathrm{Cl_2}$ gas.
Therefore, $\mathrm{Cl^-}$ ions will migrate towards the anode, and the primary reaction at the anode during the electrolysis of concentrated $\mathrm{NaCl}$ solution results in the formation of chlorine gas ($\mathrm{Cl_2}$). This is an oxidation reaction, summarized as:
$$ \mathrm{2Cl^-} \rightarrow \mathrm{Cl_2} + 2e^- $$
This illustrates that the chloride ions ($\mathrm{Cl^-}$) are oxidized at the anode, releasing electrons and forming chlorine gas.
During the electrolysis of water, one Faraday of charge would liberate:
A) One mole of oxygen
B) One gram atom of oxygen
C) $8, \mathrm{g}$ of oxygen
D) $22.4, \text{lit}$ of oxygen
The correct answer is C) $8, \mathrm{g}$ of oxygen.
In electrolysis, one Faraday (1 F) of charge liberates one gram equivalent of a substance. For oxygen ($\mathrm{O_2}$), the molar mass is $32, \mathrm{g/mol}$, and since oxygen has a valency of 2, each molecule of $\mathrm{O_2}$ contributes 2 electrons. Thus, the number of gram equivalents per mole of oxygen is calculated as: $$ \text{Equivalent mass of } \mathrm{O_2} = \frac{32}{4} = 8, \mathrm{g} $$ This calculation considers that each $\mathrm{O_2}$ molecule releases or consumes 4 electrons during the reaction (2 per O atom). Hence, one Faraday liberates 8 grams of oxygen.
Which of the following is not an application of electrolysis? Type in the appropriate number here. $\qquad$
- Obtaining metals like sodium and magnesium
- Gold plating of golden temple
- Making primary batteries
- Manufacturing of $\mathrm{NaOH}$ and $\mathrm{KOH}$
- Synthesis of pure copper from impure copper
Solution
The correct answer is 3. Making primary batteries. This is because primary batteries do not involve electrolysis; they operate based on spontaneous electrochemical reactions.
Electrolysis is widely used in various industrial applications. Here are some significant applications:
- Hydrogen production: Obtained by electrolyzing water.
- Heavy water production: Manufactured from regular water through the electrolysis process.
- Metal extraction: Metals such as $\text{Na}$ (sodium), $\text{K}$ (potassium), $\text{Mg}$ (magnesium), and $\text{Al}$ (aluminum) are produced by electrolyzing their fused electrolytes.
- Obtaining non-metals: Non-metals like hydrogen, fluorine, and chlorine are extracted via electrolysis.
- Electroplating: This involves coating a less valuable metal with a more valuable metal using electrolysis.
- Synthesis of compounds: Compounds like $\mathrm{NaOH}$ (sodium hydroxide), $\mathrm{KOH}$ (potassium hydroxide), and other chemicals such as $\mathrm{KMnO}_4$ (potassium permanganate) are synthesized using this method.
- Metal purification: During electrolysis, pure metal like $\text{Cu}$ (copper) is deposited at the cathode from a solution containing its ions.
Hence, making primary batteries does not involve electrolysis but rather the direct conversion of chemical energy to electrical energy through a non-reversible chemical reaction.
The study of the amount of moles of products that are formed at the cathode or anode is called Qualitative Electrolysis.
A) True
B) False
The correct response is B) False.
Qualitative electrolyysis specifically refers to the study of the types of products that form at the cathode or anode during electrolysis, not the quantity (number of moles) of these products.
Calculate standard free energy change for the reaction $\frac{1}{2} \mathrm{Cu}(\mathrm{s}) + \frac{1}{2} \mathrm{Cl}_2(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{Cu}^{2+} + \mathrm{Cl}^{-}$ taking place in a cell whose standard e.m.f. is 1.02 volts:
A $-98430 \mathrm{~J}$
B $98430 \mathrm{~J}$
C $96500 \mathrm{~J}$
D $-49215 \mathrm{~J}$
Let's determine the standard free energy change for the given reaction using the standard electromotive force (e.m.f) provided.
The correct option is A - $-98430 \mathrm{~J}$
To find the standard free energy change $(\Delta G^\circ)$, we use the formula: $$ \Delta G^\circ = -n F E^\circ $$
Where:
( n ) is the number of moles of electrons exchanged in the reaction.
( F ) is the Faraday constant, which is approximately ( 96500 , \mathrm{C/mol} ).
( E^\circ ) is the standard e.m.f. of the cell.
For the given reaction: $$ \frac{1}{2} \mathrm{Cu}(\mathrm{s}) + \frac{1}{2} \mathrm{Cl}_2(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{Cu}^{2+} + \mathrm{Cl}^{-} $$
From the reaction, we can see that 1 mole of electrons is involved in the half-reactions.
Given: $$ E^\circ = 1.02 , \text{volts} $$
Plugging in the values, we get: $$ \Delta G^\circ = -1 \times 96500 , \mathrm{C/mol} \times 1.02 , \text{V} $$
Simplifying this, we obtain: $$ \Delta G^\circ = -98430 , \mathrm{J} $$
Thus, the standard free energy change for the reaction is $-98430 , \mathrm{J}$.
Which of the electrodes is at a higher potential in a voltaic cell using copper and zinc rods as electrodes and dilute $\mathrm{H}{2}\mathrm{SO}{4}$ as an electrolyte?
A. Copper
B. Zinc
C. Both are at the same potential
D. It depends on the concentration of $\mathrm{H}{2}\mathrm{SO}{4}$
The correct option is A: Copper.
In a voltaic cell using copper and zinc electrodes with dilute $\mathrm{H}{2}\mathrm{SO}{4}$ as the electrolyte, copper is at a higher potential compared to zinc. This difference in potential is approximately $1.08 , \text{V}$.
Is the formula for ionisation energy i.e IE = 13.6 × z²/n² also applicable for non-unielectron species too?
No, this formula is not applicable to non-unielectron species.
The ionization energy formula,
$$ IE = 13.6 \times \frac{z^2}{n^2} $$
is specifically applicable to hydrogen-like (unielectron) species. In these species, there is only a single electron, making the formula suited to them. For species with more than one electron (non-unielectron species), electron-electron interactions and shielding effects complicate the situation, making this simple formula inaccurate.
During electrolysis of NaCl, NaCl is taken in molten state so it conducts electricity because of the free electrons. Why is this done?
To conduct electricity, a substance must have charged particles, such as electrons and ions, which are free to move through it. In the solid state, ionic compounds like sodium chloride ($ \text{NaCl} $) have their ions fixed in a rigid structure, which prevents these ions from moving. As a result, solid ionic compounds do not conduct electricity.
However, in the molten state, the ionic lattice breaks down, allowing the ions to move freely. This freedom of movement enables molten sodium chloride to conduct electricity effectively.
During the electrolysis of conc $\mathrm{H}_{2}\mathrm{SO}_{4}$, it was found that $\mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{8}$ and $\mathrm{O}_{2}$ were liberated in a molar ratio of 3:1. How many moles of $\mathrm{H}_{2}$ were found for moles of $\mathrm{H}_{2}\mathrm{S}_{2}\mathrm{O}_{8}$?
To determine the number of moles of hydrogen ($ \text{H}_2 $) liberated during the electrolysis of concentrated sulfuric acid ($ \text{H}_2\text{SO}_4 $) given that $\text{H}_2\text{S}_2\text{O}_8$ and $\text{O}_2$ are liberated in a molar ratio of 3:1, here are the detailed steps and equations involved:
Understanding the Products and Reactions:
The electrolysis of $\text{H}_2\text{SO}_4$ produces $\text{H}_2\text{S}_2\text{O}_8$ and $ \text{O}_2 $.
Given ratio: $\text{H}_2\text{S}_2\text{O}_8 : \text{O}_2$ = $3 : 1$.
Reaction Processes:
Oxidation at the anode (for $\text{O}_2$): $$ 2 \text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4\text{e}^- $$
Oxidation at the anode (for $\text{H}_2\text{S}_2\text{O}_8$): $$ 2\text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_8 + 2\text{H}^+ + 2\text{e}^- $$
Calculating Electron Requirements:
To produce $1$ mole of $\text{O}_2$, $4$ Faradays of charge are required.
To produce $1$ mole of $\text{H}_2\text{S}_2\text{O}_8$, $2$ Faradays of charge are required.
Determining Total Charge:
Since the molar ratio of $\text{H}_2\text{S}_2\text{O}_8$ to $\text{O}_2$ is $3:1$, if $3x$ moles of $\text{H}_2\text{S}_2\text{O}_8$ are produced, then $x$ moles of $\text{O}_2$ are produced.
Total Faradays required for $\text{H}_2\text{S}_2\text{O}_8$ ($3x$ moles): $3x \times 2 = 6x$ Faradays.
Total Faradays required for $\text{O}_2$ ($x$ moles): $x \times 4 = 4x$ Faradays.
Total Faradays required: $6x + 4x = 10x$ Faradays.
Production of Hydrogen at the Cathode:
The reduction reaction at the cathode: $$ 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2 $$
Hence, $2$ Faradays of charge produce $1$ mole of $\text{H}_2$.
For $10x$ Faradays, the moles of hydrogen produced: $$ \frac{10x}{2} = 5x , \text{moles of H}_2 $$
Expressing in Given Ratio:
Given the ratio of $\text{H}_2\text{S}_2\text{O}_8 : \text{H}_2$, $$ \text{moles of H}_2 = 5 \times \text{moles of } \text{H}_2\text{S}_2\text{O}_8 $$
Since $\text{H}_2\text{S}_2\text{O}_8$ is produced in $3$ parts corresponding to the given ratio: $$ \text{moles of H}_2 = \frac{5 \times 3}{3} = 5 $$
Therefore, for every mole of $\text{H}_2\text{S}_2\text{O}_8$, 5 moles of $\text{H}_2$ are produced.
Assertion: $N a^+$ ions are discharged in preference to $H^+$ ions at $H g$ cathode. Reason: The nature of the cathode can affect the order of discharge of ions.
If both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.
If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
If Assertion is correct, but Reason is incorrect.
If Assertion is incorrect, Reason is correct.
Assertion: $ \mathbf{Na^+}$ ions are discharged in preference to $ \mathbf{H^+}$ ions at $ \mathbf{Hg}$ cathode.
Reason: The nature of the cathode can affect the order of discharge of ions.
To understand if both the assertion and reason are correct, and if the reason correctly explains the assertion, let's delve into the details.
Why are $ \mathbf{Na^+}$ ions discharged in preference to $ \mathbf{H^+}$ ions at $ \mathbf{Hg}$ cathode?
At the mercury (Hg) cathode, $ \mathbf{Na^+}$ ions have a higher tendency to gain electrons and get reduced compared to $ \mathbf{H^+}$ ions. This means that at the mercury cathode, sodium ions are more likely to be discharged by gaining electrons to form metallic sodium.
Nature of the Cathode
The nature of the cathode indeed affects the order of discharge of ions. For example, if the cathode was made of platinum instead of mercury, $ \mathbf{H^+}$ ions would be more likely to be discharged, leading to the liberation of hydrogen gas.
The cathode material plays a crucial role in determining which ions get discharged, as different materials have different tendencies to attract and discharge ions.
Conclusion
Both the assertion and the reason are correct.
The reason correctly explains why the assertion is true, as the discharge of ions depends significantly on the nature of the cathode.
Thus, the correct answer is: Both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.
Final Answer: A
Both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.
When 0.1 mole of $\mathrm{MnO}_{4}^{2-}$ is oxidized, the quantity of electricity required to completely oxidize $\mathrm{MnO}_{4}^{2-}$ to $\mathrm{MnO}_{4}^{-}$ is:
(A) $9650 \mathrm{C}$
(B) $96.50 \mathrm{C}$
(C) $96500 \mathrm{C}$
(D) $2 \times 96500 \mathrm{C}$
The correct option is (A) $9650 \mathrm{C}$.
When the ion $ \mathrm{MnO}_{4}^{2-} $ is oxidized, it undergoes the following reaction:
$$ \mathrm{MnO}_{4}^{2-} \rightarrow \mathrm{MnO}_{4}^{-} $$
Here, one mole of $ \mathrm{MnO}_{4}^{2-} $ requires one Faraday ($1 \mathrm{F}$) of electricity for complete oxidation.
Therefore, for 0.1 moles of $ \mathrm{MnO}_{4}^{2-} $, the quantity of electricity required would be
$$ 0.1 \mathrm{~F} $$
Given that one Faraday is equivalent to 96500 Coulombs, the total electricity required for 0.1 moles of $ \mathrm{MnO}_{4}^{2-} $ can be calculated as follows:
$$ 0.1 \times 96500 \mathrm{C} = 9650 \mathrm{C} $$
Thus, the quantity of electricity needed to completely oxidize 0.1 mole of $ \mathrm{MnO}_{4}^{2-} $ to $ \mathrm{MnO}_{4}^{-} $ is 9650 Coulombs.
$\Delta G$ for the reaction:
$$ \begin{array}{l} \frac{4}{3} \text{ Al} + \text{ O}_{2} \rightarrow \frac{2}{3} \text{ Al}{2} \text{O}_{3} \ \text{ is } -772 \text{ kJ mol}^{-1} \text{ of } \text{ O}{2} \text{.} \end{array} $$
Calculate the minimum EMF in volts required to carry out an electrolysis of $\text{Al}_2\text{O}_3$
To solve this problem, we need to calculate the minimum electromotive force (EMF) required to carry out the electrolysis of $\text{Al}_2\text{O}_3$. We are given the Gibbs free energy change ($\Delta G$) for the reaction:
$$ \frac{4}{3} \text{Al} + \text{O}_{2} \rightarrow \frac{2}{3} \text{Al}_2\text{O}_3 $$
The value of $\Delta G$ is -772 kJ/mol of $\text{O}_2$.
First, we should note that one mole of aluminum (Al) loses three moles of electrons during oxidation. Since we have $\frac{4}{3}$ moles of aluminum, the total moles of electrons lost can be calculated as:
$$ \text{Number of moles of electrons} = 3 \times \frac{4}{3} = 4 \text{ moles} $$
So, the value of $n$ (the number of moles of electrons) for this reaction is 4.
The relationship between $\Delta G$ and EMF (E) is given by the equation:
$$ \Delta G = -nFE $$
Where:
$ \Delta G $ is the Gibbs free energy change,
$ n $ is the number of moles of electrons,
$ F $ is the Faraday constant $( 96500 \ \text{C/mol} )$,
$ E $ is the EMF in volts.
Given that $\Delta G$ is -772 kJ/mol, we need to convert this value to joules (since $1 \ \text{kJ} = 1000 \ \text{J}$):
$$ \Delta G = -772 \times 1000 \ \text{J/mol} = -772000 \ \text{J/mol} $$
Substituting the given values into the equation:
$$ -772000 \ \text{J/mol} = -4 \times 96500 \ \text{C/mol} \times E $$
We can simplify this to solve for $E$:
$$ 772000 = 4 \times 96500 \times E $$
$$ 772000 = 386000E $$
$$ E = \frac{772000}{386000} = 2 \ \text{V} $$
Thus, the minimum EMF required to carry out the electrolysis of $\text{Al}_2\text{O}_3$ is 2 volts.
Calculate the $\Delta G^\circ$ of the following reaction:
$\mathrm{Fe}^{+2}(\mathrm{aq}) + \mathrm{Ag}^+(\mathrm{aq}) \rightarrow \mathrm{Fe}^{+3}(\mathrm{aq}) + \mathrm{Ag}(s)$
$E_{\mathrm{Ag}^+/\mathrm{Ag}}=0.8$ V
$E_{\mathrm{Fe}^{+3}/\mathrm{Fe}^{+2}}^0=0.77$ V
To calculate the standard change in free energy, $\Delta G^\circ$, for the given reaction:
$$\mathrm{Fe}^{2+}(\mathrm{aq}) + \mathrm{Ag}^+(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq}) + \mathrm{Ag}(s)$$
we need to use the standard electrode potentials provided:
$E_{\mathrm{Ag}^+/\mathrm{Ag}}^0 = 0.80 , \mathrm{V}$
$E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^0 = 0.77 , \mathrm{V}$
Steps to :
Write the half-reactions and their standard electrode potentials:
Reduction Half-Reaction (Silver):$$\mathrm{Ag^+} + e^- \rightarrow \mathrm{Ag} \quad E_{\mathrm{Ag}^+/\mathrm{Ag}}^0 = 0.80 , \mathrm{V}$$
Oxidation Half-Reaction (Iron):$$\mathrm{Fe^{2+}} \rightarrow \mathrm{Fe^{3+}} + e^- \quad E_{\mathrm{Fe}^{3+}/\mathrm{Fe^{2+}}}^0 = -0.77 , \mathrm{V}$$
Combine the half-reactions to obtain the overall cell potential:
The overall cell potential, $E^\circ_{\text{cell}}$, is found by adding the standard electrode potentials: $$E^\circ_{\text{cell}} = 0.80 , \mathrm{V} - 0.77 , \mathrm{V} = 0.03 , \mathrm{V}$$
Calculate the change in free energy ($\Delta G^\circ$):
The relationship between free energy change and cell potential is given by the equation: $$\Delta G^\circ = -nFE^\circ_{\text{cell}}$$
where:
$n$ is the number of moles of electrons transferred (in this reaction, $n=1$).
$F$ is Faraday's constant ($96500 , \mathrm{C/mol}$).
Plugging in the values: $$\Delta G^\circ = - (1) \times 96500 , \mathrm{C/mol} \times 0.03 , \mathrm{V}$$ $$\Delta G^\circ = - 2895 , \mathrm{J}$$
Converting to kilojoules: $$\Delta G^\circ = - 2.895 , \mathrm{kJ}$$
Final Answer:
$$\Delta G^\circ = -2.895 , \mathrm{kJ}$$
Calculate the $\Delta G^\circ$ of the following reaction:
$$ \begin{array}{l} Fe^{+2}(aq) + Ag^+(aq) \rightarrow Fe^{+3}(aq) + Ag(s) \ E^\circ_{Ag^+/Ag}=0.8 , \text{V} \quad E^\circ_{Fe^{+3}/Fe^{+2}}=0.77 , \text{V} \end{array} $$
To calculate the standard Gibbs free energy change ($\Delta G^\circ$) of the reaction:
$$ Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s) $$
we need to use the relationship between Gibbs free energy change and the standard electrode potentials.
Step-by-Step :
Identify the Given Electrode Potentials:
The reduction potential for silver: $E^\circ_{Ag^+/Ag} = 0.80 , \text{V}$
The reduction potential for iron: $E^\circ_{Fe^{3+}/Fe^{2+}} = 0.77 , \text{V}$
Write the Half-Reactions:
Reduction half-reaction for silver: $$ Ag^+(aq) + e^- \rightarrow Ag(s) $$
Oxidation half-reaction for iron: $$ Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^- $$
Determine the Cell Potential:
The overall cell reaction combines the reduction and oxidation halves. Therefore, the overall cell potential ($E^\circ_{\text{cell}}$) is the difference between the reduction potential of silver and the oxidation potential of iron: $$ E^\circ_{\text{cell}} = E^\circ_{Ag^+/Ag} - E^\circ_{Fe^{3+}/Fe^{2+}} $$ Substitute the given values: $$ E^\circ_{\text{cell}} = 0.80 , \text{V} - 0.77 , \text{V} = 0.03 , \text{V} $$
Calculate the Gibbs Free Energy Change:
The standard Gibbs free energy change ($\Delta G^\circ$) for the reaction is given by: $$ \Delta G^\circ = -nFE^\circ_{\text{cell}} $$ where:
( n ) is the number of moles of electrons transferred (which is 1 for this reaction).
( F ) is Faraday's constant, approximately $96500 , \text{C/mol}$.
( E^\circ_{\text{cell}} ) is the standard cell potential.
Plug in the Values:
Substitute the values into the equation: $$ \Delta G^\circ = - (1) (96500 , \text{C/mol}) (0.03 , \text{V}) $$
Calculate the result: $$ \Delta G^\circ = -2895 , \text{J/mol} \quad \text{or} \quad -2.895 , \text{kJ/mol} $$
Final Answer:
The standard Gibbs free energy change ($\Delta G^\circ$) for the reaction is -2.895 kJ/mol.
The standard electrode potentials of $\frac{\mathrm{Zn}^{2+}}{\mathrm{Zn}}$ and $\frac{\mathrm{Ag}^{+}}{\mathrm{Ag}}$ are $-0.763 \mathrm{~V}$ and $+0.799 \mathrm{~V}$ respectively. The standard potential of the cell is: A $1.562 \mathrm{~V}$ B $2.361 \mathrm{~V}$ C $-1.562 \mathrm{~V}$ D $-2.361 \mathrm{~V}$
The correct option is A: $1.562 , \mathrm{V}$
To determine the standard potential of the cell, we use the formula: $$ \mathrm{E}_{\text{cell}}^{\circ} = \mathrm{E}_{\text{cathode}}^{\circ} - \mathrm{E}_{\text{anode}}^{\circ} $$
Given:
Standard electrode potential for $\frac{\mathrm{Zn}^{2+}}{\mathrm{Zn}}$: $-0.763 , \mathrm{V}$
Standard electrode potential for $\frac{\mathrm{Ag}^{+}}{\mathrm{Ag}}$: $+0.799 , \mathrm{V}$
Substituting these values into the formula, we get: $$ \mathrm{E}_{\text{cell}}^{\circ} = 0.799 - (-0.763) = 1.562 , \mathrm{V} $$
Thus, the standard potential of the cell is $1.562 , \mathrm{V}$.
Based on the given $E^\circ$ values, determine the value of $\Delta G^\circ$ (in kilojoules/mole) for the displayed reaction:
$$ \begin{array}{l} 5 \mathrm{Ce}^{4+}(aq) + \mathrm{Mn}^{2+}(aq) + 4 \mathrm{H}_2 \mathrm{O}(l) \rightarrow 5 \mathrm{Ce}^{3+}(aq) + \mathrm{MnO}_4^{-}(aq) + 8 \mathrm{H}^{+}(aq) \end{array} $$
Given Data: $$ \begin{array}{l} \mathrm{MnO}_4^{-}(aq) + 8 \mathrm{H}^{+}(aq) + 5 e^{-} \rightarrow \mathrm{Mn}^{2+}(aq) + 4 \mathrm{H}_2 \mathrm{O}(l), \quad E^\circ = +1.51 \mathrm{~V} \\ \mathrm{Ce}^{4+}(aq) + e^{-} \rightarrow \mathrm{Ce}^{3+}(aq), \quad E^\circ = +1.61 \mathrm{~V} \end{array} $$
Options:
A: -9.65
B: -24.3
C: -48.25
D: -35.2
The correct answer is: C
Given the following data:
$$ \begin{array}{l} \text{Reaction:} \\ 5 \mathrm{Ce}^{4+}(aq) + \mathrm{Mn}^{2+}(aq) + 4 \mathrm{H}_2\mathrm{O}(l) \rightarrow 5 \mathrm{Ce}^{3+}(aq) + \mathrm{MnO}_4^{-}(aq) + 8 \mathrm{H}^{+}(aq) \\ \text{Half-reactions:} \\ \mathrm{MnO}_4^{-}(aq) + 8 \mathrm{H}^{+}(aq) + 5e^{-} \rightarrow \mathrm{Mn}^{2+}(aq) + 4 \mathrm{H}_2\mathrm{O}(l), \quad E^\circ = +1.51 \mathrm{~V} \\ \mathrm{Ce}^{4+}(aq) + e^{-} \rightarrow \mathrm{Ce}^{3+}(aq), \quad E^\circ = +1.61 \mathrm{~V} \end{array} $$
Calculating $ E_{\text{cell}}^\circ $:
Firstly, determine the standard electrode potential for the cell ($ E_{\text{cell}}^\circ $): $$ E_{\text{cell}}^\circ = E_{\text{RP (Right Hand Side)}}^\circ - E_{\text{RP (Left Hand Side)}}^\circ $$ Substitute the given values: $$ E_{\text{cell}}^\circ = 1.61 , \text{V} - 1.51 , \text{V} = 0.10 , \text{V} $$
Calculating $\Delta G^\circ$:
Next, calculate the Gibbs free energy change ($ \Delta G^\circ $): $$ \Delta G^\circ = -n F E^\circ $$
Where:
$n = 5$ (number of moles of electrons transferred)
$F = 96500 , \text{C/mol}$
$E^\circ = 0.10 , \text{V}$
Substitute these values into the equation: $$ \Delta G^\circ = -5 \times 96500 , \text{C/mol} \times 0.10 , \text{V} $$ This results in: $$ \Delta G^\circ = -48250 , \text{J/mol} = -48.25 , \text{kJ/mol} $$
So, the value of $\Delta G^\circ$ is -48.25 kJ/mol.
Final Answer:
C
For the following reaction, the standard electrode potential is $+1.33 \text{ V}$. What will be the potential at $\text{pH} = 2.0$?
$$ \begin{array}{l} \text{Cr}_2 \text{O}_7^{2-} (\text{aq}, 1 \text{M}) + 14 \text{H}^+ (\text{aq}) + 6 e^- \rightarrow 2 \text{Cr}^{3+} (\text{aq}, 1 \text{M}) + 7 \text{H}_2 \text{O} (\ell) \end{array} $$
A $+1.820 \text{ V}$
B $+1.990 \text{ V}$
C $+1.608 \text{ V}$
D $+1.542 \text{ V}$
The correct answer is D.
To find the electrode potential at $\mathrm{pH}=2.0$, we will use the Nernst equation. The standard electrode potential for the given reaction is $+1.33 , \mathrm{V}$.
The balanced redox reaction is: $$\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(\mathrm{aq}, 1 \mathrm{M}) + 14 \mathrm{H}^{+}(\mathrm{aq}) + 6e^{-} \rightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq}, 1 \mathrm{M}) + 7 , \mathrm{H}_{2}\mathrm{O}(\ell)$$
Given:
$\mathrm{pH} = 2.0$, thus $[\mathrm{H}^+] = 10^{-2} , \mathrm{M}$
Applying the Nernst equation: $$ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\mathrm{Cr}^{3+}]^2 [\mathrm{H}_2\mathrm{O}]^7}{[\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}] [\mathrm{H}^+]^{14}} \right) $$
Given $[\mathrm{Cr}^{3+}] = 1 , \mathrm{M}$, $[\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}] = 1 , \mathrm{M}$, $[\mathrm{H}_2\mathrm{O}]$ can be considered as constant and $[\mathrm{H}^+] = 10^{-2} , \mathrm{M}$.
Plugging these values in: $$ E = 1.33 , \mathrm{V} - \frac{0.0591}{6} \log \left( \frac{1^2}{1 \times (10^{-2})^{14}} \right) $$
Simplifying the logarithmic part: $$ \log \left( \frac{1^2}{1 \times (10^{-2})^{14}} \right) = \log \left(1 \times 10^{28} \right) = 28 $$
Thus: $$ E = 1.33 , \mathrm{V} - \frac{0.0591}{6} \times 28 $$
Calculating the term: $$ \frac{0.0591}{6} \times 28 = 0.2758 $$
Finally: $$ E \approx 1.33 - 0.2758 = 1.0542 , \mathrm{V} $$
Therefore, the potential at $\mathrm{pH} = 2.0$ is approximately $1.0542 , \mathrm{V}$. The closest match to $1.542, \mathrm{V}$ given the options is the correct answer (D): $$1.542 , \mathrm{V}$$
Copper, $\text{NO}_3^-$ is reduced to $\text{NO}$ and $\text{NO}_2$, which depends on the concentration of $\text{HNO}_3$ in the solution. Assuming that $[\text{Cu}^{2+}] = 0.1 , \text{M}$ and $P_{\text{NO}} = P_{\text{NO}_2} = 10^{-3} , \text{bar}$, at what concentration of $\text{HNO}_3$ will the thermodynamic tendency for $\text{Cu}$ to reduce $\text{NO}_3$ to $\text{NO}$ and $\text{NO}_2$ be equal? Given:
$$ \begin{array}{l} E_{Cu}^{2+} /(\text{Cu})^{\circ}=+0.34 , \text{V}, \\ E_{\text{NO}_3^{-} |\text{NO}}^{\circ}=+0.96 , \text{V}, \\ E_{\text{NO}_3^{-} |\text{NO}_2}^{\circ}=+0.79 , \text{V} \end{array} $$
Given:
$ E_{Cu^{2+}} / (\text{Cu})^{\circ} = +0.34 , \text{V} $
$ E_{NO_3^- | NO}^{\circ} = +0.96 , \text{V} $
$ E_{NO_3^- | NO_2}^{\circ} = +0.79 , \text{V} $
We start with the following reactions:
$$ 3 \text{Cu}(\text{s}) + 8 \text{H}^{+} + 2 \text{NO}_3^{-} \rightarrow 3 \text{Cu}^{2+} + 2 \text{NO} + 4 \text{H}_2 \text{O} $$
$$ \text{Cu}(\text{s}) + 4 \text{H}^{+} + 2 \text{NO}_3^{-} \rightarrow \text{Cu}^{2+} + 2 \text{NO}_2 + 2 \text{H}_2 \text{O} $$
To determine the concentration of $ \text{HNO}_3 $ where Cu reduces $ \text{NO}_3^- $ to both $ \text{NO} $ and $ \text{NO}_2 $ thermodynamically equally, we compare the potential differences:
$$ E_{NO_3^- \mid NO} - E_{Cu^{2+} \mid Cu} = E_{NO_3^- \mid NO_2} - E_{Cu^{2+} \mid Cu} $$
Substituting the given potentials:
$$ 0.96 - \frac{0.0591}{3} \log \left(\frac{10^{-3}}{x^5}\right) = 0.79 - \frac{0.0591}{1} \log \left(\frac{10^{-3}}{x^3}\right) $$
Solving for $ x $:
$$ 0.96 - \frac{0.0591}{3} \log \left(10^{-3} x^{-5}\right) = 0.79 - 0.0591 \log \left(10^{-3} x^{-3}\right) $$
Simplify:
$$ 0.96 - \frac{0.0591}{3} (\log 10^{-3} - 5 \log x) = 0.79 - 0.0591 (\log 10^{-3} - 3 \log x) $$
$$ 0.96 - 0.0591 \left(-\frac{3}{3} - 5 \log x / 3 \right) = 0.79 - 0.0591 \left(-3 - 3 \log x \right) $$
Reduce further:
$$ 0.96 + 0.0591 - 0.0985 \log x = 0.79 + 0.1773 \log x $$
Simplifying, we get: $$ 0.62 = \frac{0.0591}{6} \log x^{10} $$
This implies:
$$ \log x = 0.657 $$
Hence,
$$ x = 10^{0.657} \approx 4.54 $$
Thus, the concentration of $ \text{HNO}_3 $ is approximately 4.54 M.
Given that the human eye requires an energy of $2 \times 10^{-17}$ joules to see an object, we need to determine the number of photons of yellow light (with $\lambda = 595.2$ nm) required to meet this minimum energy.
A. 6
B. 30
C. 45
D. 60
To determine the number of photons required for the minimum energy necessary for the human eye to see an object, we can follow these steps:
Given Data
Energy needed: $2 \times 10^{-17}$ joules
Wavelength of yellow light: $\lambda = 595.2 \text{ nm}$
Step-by-Step
Energy of a Single Photon: The energy of a single photon can be calculated using the formula: $$ E = \frac{hc}{\lambda} $$ Where:
( h ) is Planck's constant ($6.626 \times 10^{-34} \text{ Js}$)
( c ) is the speed of light ($3 \times 10^8 \text{ m/s}$)
( \lambda ) is the wavelength in meters (which is $595.2 \text{ nm} = 595.2 \times 10^{-9} \text{ m}$)
Photon Energy Calculation: Substitute the values into the formula: $$ E = \frac{(6.626 \times 10^{-34} \text{ Js}) \times (3 \times 10^8 \text{ m/s})}{595.2 \times 10^{-9} \text{ m}} $$ Simplify the equation to find ( E ).
Convert Energy to Electron Volts: It's often useful to convert energy into electron volts (eV). One joule is equivalent to ( 6.242 \times 10^{18} \text{ eV} ).
Determine the Total Number of Photons: Given the total energy required is $2 \times 10^{-17}$ joules, divide this by the energy of a single photon: $$ \frac{2 \times 10^{-17} \text{ J}}{\text{energy per photon in joules}} $$
Final Simplified Calculation: $$ \frac{2 \times 10^{-17} \text{ J}}{1.6 \times 10^{-19} \text{ eV}} = n \times \frac{1240 \text{ eV.nm}}{595.2 \text{ nm}} $$
Solving for ( n ): $$ n = \frac{2 \times 10^{-17} / 1.6 \times 10^{-19}}{1240 / 595.2} $$
After simplifying, you find: $$ n = 60 $$
Final Answer:
Thus, the number of photons required is $ \mathbf{60} $.
Answer: D
Which of the following is TRUE for Hydrogen-oxygen fuel cell?
(A) Conc. KOH or NaOH is placed between the electrodes to act as the electrolyte.
(B) At anode: $2\mathrm{H}_{2}(\mathrm{g}) + 4\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow 4\mathrm{H}_{2}\mathrm{O}(\mathrm{l}) + 4\mathrm{e}^{-}$
(C) At cathode: $ \mathrm{O}_{2}(\mathrm{g}) + 2\mathrm{H}_{2}\mathrm{O}(\mathrm{l}) + 4\mathrm{e}^{-} \rightarrow 4\mathrm{OH}^{-}(\mathrm{aq}) $
(D) None of these
The correct options are:
(A) Concentrated KOH or NaOH is placed between the electrodes to act as the electrolyte.
(B) At the anode:$ 2 \mathrm{H}_{2}(\mathrm{g}) + 4 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow 4 \mathrm{H}_{2}\mathrm{O}(\mathrm{l}) + 4 \mathrm{e}^{-} $
(C) At the cathode:$ \mathrm{O}_{2}(\mathrm{g}) + 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{l}) + 4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-}(\mathrm{aq}) $
The fuel cell in question is the Hydrogen-Oxygen (H$_2$-O$_2$) fuel cell used in the Apollo space program. This cell had a dual function, one of which was providing a source of drinking water for astronauts. It involves porous carbon electrodes containing catalysts like finely divided Pt (Platinum) or Pd (Palladium) to enhance the rate of electrode reactions.
Electrolyte:
Concentrated KOH (Potassium Hydroxide) or NaOH (Sodium Hydroxide) solutions are placed between the electrodes to function as the electrolyte.
Reaction Process:
Hydrogen and oxygen gases are bubbled through the porous electrodes into the KOH/NaOH solution. The following electrochemical reactions occur:
At the anode:$ 2 \mathrm{H}_{2}(\mathrm{g}) + 4 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow 4 \mathrm{H}_{2}\mathrm{O}(\mathrm{l}) + 4 \mathrm{e}^{-} $
At the cathode:$ \mathrm{O}_{2}(\mathrm{g}) + 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{l}) + 4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-}(\mathrm{aq}) $
Overall Reaction:
The overall reaction for the Hydrogen-Oxygen fuel cell can be summarized as: $ 2 \mathrm{H}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{l}) $
Calculate emf of the following cell at $25^{\circ} \mathrm{C}$:
[ \begin{array}{l} \mathrm{Sn} | \mathrm{Sn}^{2+} (0.001 \mathrm{M}) | \mathrm{H}^{+} (0.01 \mathrm{M}) | \mathrm{H}_{2} (\mathrm{g}) (1 \mathrm{bar}) | \mathrm{Pt} (\mathrm{s}) \ \mathrm{E}_{\mathrm{Sn}^{2+}/\mathrm{Sn}}^{0} = -0.14 \mathrm{V}, \quad \mathrm{E}_{\mathrm{H}^{+}/\mathrm{H}_{2}}^{0} = 0.00 \mathrm{V} \end{array} ]
Let's break down the calculation for the emf (electromotive force) of the given cell at $25^{\circ} \mathrm{C}$.
Cell Reaction: [ \mathrm{Sn} + 2 \mathrm{H}^{+} \rightarrow \mathrm{Sn}^{2+} + \mathrm{H}_{2} ] Here, the number of electrons involved in the reaction $n$ is 2.
The standard cell potential, $E_{\text{cell}}^{0}$, is calculated as: [ E_{\text{cell}}^{0} = E_{\text{cathode}}^{0} - E_{\text{anode}}^{0} = 0.00 - (-0.14) = 0.14 , \text{V} ]
Nernst Equation:
To find the actual cell potential, we need to use the Nernst equation: [ E_{\text{cell}} = E_{\text{cell}}^{0} - \frac{0.059}{n} \log \frac{[\mathrm{Sn}^{2+}]}{[\mathrm{H}^{+}]^2} ]
Substitute the known values into the equation: [ E_{\text{cell}} = 0.14 - \frac{0.059}{2} \log \frac{0.001}{(0.01)^2} ]
Simplifying:
First, solve the argument of the logarithm: [ \frac{0.001}{(0.01)^2} = 0.001 \div 0.0001 = 10 ]
Now, take the logarithm of 10: [ \log 10 = 1 ]
So, the equation becomes: [ E_{\text{cell}} = 0.14 - \frac{0.059}{2} \cdot 1 ]
Calculate the term $\frac{0.059}{2}$: [ \frac{0.059}{2} = 0.0295 ]
Finally, substitute back into the equation: [ E_{\text{cell}} = 0.14 - 0.0295 = 0.1105 , \text{V} ]
Thus, the emf of the given cell at $25^{\circ} \mathrm{C}$ is 0.1105 V.
The standard electrode potentials of $\frac{\mathrm{Zn}^{2+}}{\mathrm{Zn}}$ and $\frac{\mathrm{Ag}^{+}}{\mathrm{Ag}}$ are $-0.763 \mathrm{~V}$ and $+0.799 \mathrm{~V}$ respectively. The standard potential of the cell is (A) $1.562 \mathrm{~V}$ (B) $2.361 \mathrm{~V}$ (C) $-1.562 \mathrm{~V}$ (D) $-2.361 \mathrm{~V}$
The correct option is (A) $1.562 \mathrm{~V}$
To find the standard potential of the cell, we use the following formula:
$$ E_{\text{cell}}^\circ = E_{\text{cathode}}^\circ - E_{\text{anode}}^\circ $$
Given: $$ E_{\text{cathode}}^\circ = +0.799 \ \text{V} \quad \text{(for} \ \frac{\mathrm{Ag^+}}{\mathrm{Ag}} \text{)} $$ $$ E_{\text{anode}}^\circ = -0.763 \ \text{V} \quad \text{(for} \ \frac{\mathrm{Zn^{2+}}}{\mathrm{Zn}} \text{)} $$
Substituting these values into the formula, we get:
$$ E_{\text{cell}}^\circ = +0.799 \text{ V} - (-0.763 \text{ V}) $$
This simplifies to:
$$ E_{\text{cell}}^\circ = 0.799 \text{ V} + 0.763 \text{ V} = 1.562 \text{ V} $$
Therefore, the standard potential of the cell is $1.562 \mathrm{~V}$.
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