Redox Reaction - Class 11 Chemistry - Chapter 7 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Redox Reaction | NCERT | Chemistry | Class 11
The oxidation state of chromium in the final product formed by the reaction between $\mathrm{KI}$ and acidified potassium dichromate solution is:
A. +4
B. +6
C. +2
D. +3
The correct answer is D +3.
Here is how we arrive at this result:
The chemical reaction involving acidified potassium dichromate ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$) and potassium iodide ($\mathrm{KI}$) is given by: $$ \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + 6\mathrm{KI} + 7\mathrm{H}_2\mathrm{SO}_4 \rightarrow 4\mathrm{K}_2\mathrm{SO}_4 + \mathrm{Cr}_2(\mathrm{SO}_4)_3 + 7\mathrm{H}_2\mathrm{O} + 3\mathrm{I}_2 $$
In this equation, the dichromate ion ($\mathrm{Cr}_2\mathrm{O}_7^{2-}$) is reduced to chromium sulfate ($\mathrm{Cr}_2(\mathrm{SO}_4)_3$). In this compound, each chromium atom achieves an oxidation state of +3, as evident in the formula $\mathrm{Cr}_2(\mathrm{SO}_4)_3$ where $\mathrm{Cr}$ must balance the charge of $\mathrm{SO}_4^{2-}$ ions.
Silver jewelry becomes black on prolonged exposure to air. This is due to the formation of
(A) $\mathrm{Ag}_{3} \mathrm{N}$
B $\mathrm{Ag}_{2} \mathrm{O}$
C $\mathrm{Ag}{2} \mathrm{S}$ and $\mathrm{Ag}{3} \mathrm{N}$
D $\mathrm{Ag}_{2} \mathrm{S}$
Solution
Silver, when exposed to air for a prolonged duration, reacts with sulfur dioxide ($\mathrm{SO}_{2}$) present in the air. This reaction leads to the formation of silver sulfide ($\mathrm{Ag}_{2}\mathrm{S}$), which is responsible for the blackening of the silver jewelry. The reaction can be represented by the following equation:
$$ 2 \mathrm{Ag} + \mathrm{SO}_{2} \rightarrow \mathrm{Ag}_{2}\mathrm{S} + \mathrm{O}_{2} $$
Therefore, the correct answer is (D) $\mathrm{Ag}_{2} \mathrm{S}$.
The oxidation state of $\mathrm{S}$ in $\mathrm{H}{2} \mathrm{SO}{5}$ is
A) +8
B) +6
C) -2
D) +2
Solution
The correct answer is Option B: +6.
To determine the oxidation state of sulfur $(\mathrm{S})$ in $\mathrm{H}_2\mathrm{SO}_5$ (Peroxymonosulfuric acid or Caro's acid), consider the following:
The general oxidation state calculation formula is: $$ 2(+1) + x + 5(-2) = 0 $$ Here, $x$ represents the oxidation state of sulfur. Simplifying the equation: $$ 2 + x - 10 = 0 \quad \Rightarrow \quad x = +8 $$ However, this result appears incorrect because the maximum oxidation state sulfur can attain is +6, as it has only 6 valence electrons.
In $\mathrm{H}_2\mathrm{SO}_5$, two of the oxygen atoms form a peroxide linkage. This peroxide linkage implies that each of these oxygen atoms exhibit an oxidation state of -1 instead of the usual -2. Therefore, the calculation needs adjustment to: $$ 2(+1) + x + 3(-2) + 2(-1) = 0 $$ Simplifying, we have: $$ 2 + x - 6 - 2 = 0 \quad \Rightarrow \quad x = +6 $$
Therefore, the correct oxidation state of sulfur in $\mathrm{H}_2\mathrm{SO}_5$ is +6.
$\mathrm{H}{2} \mathrm{S}$ acts only as a reducing agent while $\mathrm{SO}{2}$ can act both as a reducing and oxidizing agent because
A. $\mathrm{S}$ in $\mathrm{H}{2} \mathrm{S}$ has -2 oxidation state. B. $\mathrm{S}$ in $\mathrm{SO}{2}$ has oxidation state +4. C. Hydrogen in $\mathrm{H}{2} \mathrm{S}$ is more positive than oxygen. D. Oxygen is more negative in $\mathrm{SO}{2}$.
Solution:
The correct options are A and B:
A: The sulfur ($\mathrm{S}$) in hydrogen sulfide ($\mathrm{H}_2\mathrm{S}$) has an oxidation state of -2.
B: The sulfur ($\mathrm{S}$) in sulfur dioxide ($\mathrm{SO}_2$) has an oxidation state of +4.
In hydrogen sulfide ($\mathrm{H}_2\mathrm{S}$), the sulfur atom is in a -2 oxidation state, which is the lowest possible for sulfur, indicating it can only increase in oxidation state, therefore, acting purely as a reducing agent. The highest oxidation state sulfur can achieve is +6, as seen in sulfuric acid ($\mathrm{H}_2\mathrm{SO}_4$).
Conversely, in sulfur dioxide ($\mathrm{SO}_2$), sulfur is in a +4 oxidation state. This position allows sulfur the flexibility to either increase to a higher state (up to +6, behaving as a reducing agent) or decrease to a lower state (down to -2, acting as an oxidizing agent). This dual capacity makes $\mathrm{SO}_2$ both an oxidizing and reducing agent.
The oxidation number of $\mathrm{Ba}$ in barium peroxide is:
A) +6
B) +2
C) +1
D) +4
The correct answer is B) +2.
In barium peroxide ($\mathrm{BaO_2}$), barium ($\mathrm{Ba}$) belongs to group 2 of the periodic table. Elements in this group typically exhibit an oxidation state of +2. Hence, barium in barium peroxide has an oxidation number of +2.
A ore can be reduced to its respective oxide ore by:
A) nitrate B) hydroxide C) roasting D) calcination
Solution The correct options are:
B) hydroxide
C) roasting
D) calcination
Calcination is a process where a concentrated ore is transformed into its oxide form by heating it below its melting point and in the absence of air. This method is typically applied to ores that are carbonates or hydroxides. For instance, the calcination process can be represented by the following chemical equations:
$$ 2 \mathrm{Al}(\mathrm{OH})_3(\mathrm{s}) \rightarrow \mathrm{Al}_2\mathrm{O}_3(\mathrm{s}) + 3 \mathrm{H}_2\mathrm{O}(\mathrm{l}) $$
$$ \mathrm{CaCO}_3(\mathrm{s}) \rightarrow \mathrm{CaO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g}) $$
These reactions show how aluminium hydroxide and calcium carbonate decompose into their respective oxides and other byproducts.
Consider the following statements:
A - Iron + Moisture + Oxygen = Iron Oxide.
B - The rust that develops in rocks with iron content cannot be removed, thereby, keeping the original structure of the rock intact.
Identify the correct statements.
A. A only
B. B only
C. Both A and B
D. Neither A nor B
The correct answer is A) A only.
Statement A is correct as it describes the basic chemical reaction for rusting, wherein iron ($\text{Fe}$) reacts with moisture ($\text{H}_2\text{O}$) and oxygen ($\text{O}_2$) to form iron oxide ($\text{Fe}_2\text{O}_3$), commonly known as rust:
$$ 4\text{Fe} + 3\text{O}_2 + 6\text{H}_2\text{O} \rightarrow 4\text{Fe(OH)}_3 $$
The hydrated iron(III) oxide forms the rust.
Statement B is incorrect because rust that develops in rocks with iron content can indeed change the structure of the rock. Instead of preserving the original structure, the rusting process often causes the rock to weaken and can lead to structural changes, causing the rock to break down more easily over time.
The standard hydrogen electrode has zero electrode potential because
A) Hydrogen is the easiest to oxidize.
B) This electrode potential is assumed to be zero.
C) Hydrogen atom has only one electron.
D) Hydrogen is the lightest element.
The correct answer is B) This electrode potential is assumed to be zero.
The rationale behind this choice is that the absolute value of electrode potential cannot be determined independently. Consequently, the electrical potential of the standard hydrogen electrode is conventionally assigned a value of zero. This zero value serves as a reference, allowing the electrode potentials of all other elements to be measured relative to it.
The oxidation number of '$\mathrm{S}$' in $\left(\mathrm{CH}{3}\right){2}\mathrm{SO}$ is:
A. 1
B. 2 (C). 0.0
D. 3
Solution:
The correct option is C. 0.0
Let's denote the oxidation number of sulfur ($\mathrm{S}$) as $a$. The oxidation number of methyl group $\mathrm{(CH_3)}$ is $+1$ and for oxygen ($\mathrm{O}$) it is $-2$.
Given that the molecule $\left(\mathrm{CH}_3\right)_2\mathrm{SO}$ is neutral, the sum of the oxidation numbers must also be zero. Therefore, we can set up the following equation:
$$ 2 \times (+1) + a + (-2) = 0 $$
Solving for $a$ gives:
$$ 2 + a - 2 = 0 \implies a = 0 $$
Thus, the oxidation number of sulfur ($\mathrm{S}$) in $\left(\mathrm{CH}{3}\right){2}\mathrm{SO}$ is 0.0.
Bleeding due to a cut can be stopped by applying ferric chloride solution in the laboratory. This is due to
A) Coagulation of negatively charged blood particles by $\mathrm{Fe}^{3+}$ ions.
B) Coagulation of positively charged blood particles by $\mathrm{Cl}^{-}$ ions.
C) Reaction taking place between ferric ions and the haemoglobin forming a complex.
D) Common element, iron, in both $\mathrm{FeCl}_{3}$ and haemoglobin.
Solution
The correct answer to the query is option A: Coagulation of negatively charged blood particles by $\mathrm{Fe}^{3+}$ ions.
This occurs because the $\mathrm{Fe}^{3+}$ ions, which are positively charged, effectively neutralize and coagulate the negatively charged components of the blood. This process helps in stopping the bleeding by forming a clot.
Which of the following reactions shows a redox reaction?
A Rusting of iron
B Burning of coal
C Both (Burning of coal and Rusting of iron)
D None of the given options
The correct answer is C (Both Burning of coal and Rusting of iron).
Both rusting of iron and burning of coal are examples of redox reactions.
In the rusting of iron:
Iron ($\mathrm{Fe}$) is oxidized to iron ions ($\mathrm{Fe}^{3+}$).
Oxygen ($\mathrm{O}_2$) from the air is reduced to oxide ions ($\mathrm{O}^{2-}$).
In the burning of coal:
Carbon ($\mathrm{C}$) in coal is oxidized to carbon dioxide ($\mathrm{CO}_2$, where carbon is present as $\mathrm{C}^{4+}$).
Oxygen ($\mathrm{O}_2$) from the air is reduced to become part of $\mathrm{CO}_2$ as $\mathrm{O}^{2-}$ ions.
Both these processes involve the transfer of electrons, with one reactant being oxidized and another being reduced, fitting the criteria for redox reactions.
One mole of $\mathrm{PbS}$ is oxidised to $\mathrm{PbSO}_{4}$ by $\mathrm{H}_{2} \mathrm{O}_{2}$. In this process, the number of electrons involved are $\left[N_{0}=\text{Avogadro number}\right]$
A) $2 \text{ }N_{0}$
B) $3 \text{ }N_{0}$ C) $6 \text{ }N_{0}$ D) $8 \text{ }N_{0}$
The correct answer is D: $8 N_0$
To determine the number of electrons involved in the oxidation of $\mathrm{PbS}$ to $\mathrm{PbSO}_4$, we first identify the changes in oxidation states.
In $\mathrm{PbS}$ (lead sulfide), lead (Pb) is in the +2 oxidation state and sulfur (S) is in the -2 oxidation state.
In $\mathrm{PbSO}_4$ (lead sulfate), lead (Pb) remains in the +2 oxidation state, but sulfur (S) changes to the +6 oxidation state in $\mathrm{SO}_4^{2-}$.
Key steps:
Sulfur changes its oxidation state from -2 in sulfide $\mathrm{S}^{2-}$ to +6 in sulfate $\mathrm{SO}_4^{2-}$, a change of 8 units.
Since this is a per atom calculation and there is one sulfur atom per molecule of $\mathrm{PbS}$, each molecule involves the gain or loss of 8 electrons.
For one mole of $\mathrm{PbS}$, which contains $N_0$ (Avogadro's number) molecules, the total number of electrons transferred in converting all sulfur atoms from -2 to +6 oxidation state is $8 N_0$.
Hence, option D is correct.
"What is rusting?"
Rusting is the common name for the corrosion of iron and its alloys, such as steel. While similar processes of corrosion can occur in other metals, the oxides they form are typically not referred to as rust. In the presence of oxygen and water, any iron object will ultimately deteriorate completely into rust over time.
Rust specifically refers to various forms of iron oxides, especially red oxides, that are formed from the reaction of iron with oxygen, facilitated by water or moisture in the air. Rust is notably composed of hydrated iron(III) oxides $ \text{Fe}_2\text{O}_3 \cdot n\text{H}_2\text{O} $ and iron(III) oxide-hydroxide (either $ \text{FeO(OH)} $ or $ \text{Fe(OH)}_3 $). These forms of rust can be visually and spectroscopically distinct and develop under different environmental conditions.
The mass of oxygen gas obtained by the decomposition of $231.73 \mathrm{~g}$ of silver oxide is: (Molar mass of $\mathrm{Ag}_{2}\mathrm{O}=231.735 \mathrm{~g} \mathrm{~mol}^{-1}$)
A. $16 \mathrm{~g}$
B. $32 \mathrm{~g}$
C. $48 \mathrm{~g}$
D. $64 \mathrm{~g}$
Solution The correct option is A $$ 16 \mathrm{~g} $$
The relevant decomposition reaction for silver oxide ($\mathrm{Ag}_2\mathrm{O}$) is: $$ \mathrm{Ag}_2\mathrm{O} \rightarrow 2\mathrm{Ag} + \mathrm{O}_2 $$
Analysis of moles:
- 1 mole of $\mathrm{Ag}_2\mathrm{O}$ decomposes to yield 1 mole of $\mathrm{O}_2$.
- The molar mass of $\mathrm{Ag}_2\mathrm{O}$ is 231.735 g/mol.
Calculations:
- First, calculate the moles of $\mathrm{Ag}_2\mathrm{O}$ decomposed: $$ \text{Moles of } \mathrm{Ag}_2\mathrm{O} = \frac{231.73 \mathrm{~g}}{231.735 \mathrm{~g/mol}} \approx 1 \mathrm{~mol} $$
- Based on the reaction stoichiometry, the moles of $\mathrm{O}_2$ produced: $$ \text{Moles of } \mathrm{O}_2 = 1 \mathrm{~mol} $$
- Convert this to mass using the molar mass of $\mathrm{O}_2$ (32 g/mol): $$ \text{Mass of } \mathrm{O}_2 = 1 \mathrm{~mol} \times 32 \mathrm{~g/mol} = 32 \mathrm{~g} $$
Conclusion: The mass of oxygen gas obtained by the decomposition of 231.73 g of silver oxide is 16 g.
For a spontaneous reaction, $\Delta G$, equilibrium constant $(K)$, and $E_{\text{cell}}^{\circ}$ will be respectively:
A -ve, > 1, +ve
B +ve, > 1, -ve
C -ve, < 1, -ve D -ve, > 1, -ve
The correct option for a spontaneous reaction regarding $\Delta G$, the equilibrium constant $K$, and $E_{\text{cell}}^{\circ}$ is Option A:
$\Delta G$ is negative ($\textbf{-ve}$)
$K$ is greater than 1 ($\textbf{> 1}$)
$E_{\text{cell}}^{\circ}$ is positive ($\textbf{+ve}$)
For a reaction to be spontaneous, $\Delta G$ must be negative. This indicates that the reaction releases free energy. Similarly, when the equilibrium constant $K > 1$, it implies that the reaction heavily favors the formation of products over reactants. Furthermore, when $E_{\text{cell}}^{\circ}$ is positive, it indicates that the electrochemical cell can perform work spontaneously. This is consistent with the criteria for a spontaneous chemical reaction. Therefore, Option A perfectly aligns with the properties of a spontaneous reaction: $$ \underline{\text{A -ve, > 1, +ve}} $$
Equal weights $(1.00 \mathrm{~g})$ of iron and sulfur are heated together and react to form $\mathrm{FeS}$. What percentage of the original weight is left unreacted?
$$
\left(\mathrm{Fe}=56 \mathrm{gmol}^{-1}, \mathrm{S}=32 \mathrm{gmol}^{-1}\right) %
$$
The chemical reaction between iron (Fe) and sulfur (S) to form iron sulfide (FeS) can be expressed as: $$ \mathrm{Fe} + \mathrm{S} \rightarrow \mathrm{FeS} $$
To find the number of moles for each, we use their respective molar masses:
Moles of Fe: $$ \frac{1.00 \text{ g}}{56 \text{ g/mol}} = 0.01786 \text{ moles} $$
Moles of S: $$ \frac{1.00 \text{ g}}{32 \text{ g/mol}} = 0.03125 \text{ moles} $$
From the stoichiometry of the reaction, Fe and S react in a 1:1 ratio. The lesser amount of moles determines the limiting reagent. Here, Fe is the limiting reagent since we have fewer moles of Fe compared to S.
The excess moles of S not used in the reaction are: $$ 0.03125 - 0.01786 = 0.01339 \text{ moles} $$
The mass of unreacted sulfur is: $$ 0.01339 \text{ moles} \times 32 \text{ g/mol} = 0.42848 \text{ g} $$
The percentage of the original total weight ($2.00 \text{ g}$) that remains unreacted is computed as: $$ \left( \frac{0.42848 \text{ g}}{2.00 \text{ g}} \right) \times 100% = 21.42% $$
Thus, 21.42% of the original weight remains unreacted after the formation of FeS.
"A solution containing $2.68 \times 10^{-3}$ mol of $\mathrm{A}^{n+}$ ions requires $1.608 \times 10^{-3}$ mol of $\mathrm{MnO}{4}^{-}$ for the complete oxidation of $\mathrm{A}^{n+}$ to $\mathrm{AO}{3}^{-}$ in acidic medium. What is the value of $n$?"
Here is a rewritten explanation of the solution using Markdown formatting and highlighting important phrases:
The reduction product of $\mathrm{MnO}_4^-$ in an acidic medium is $\mathrm{Mn}^{2+}$. Thus, the change in oxidation state for $\mathrm{MnO}_4^-$ is from +7 in $\mathrm{MnO}_4^-$ to +2 in $\mathrm{Mn}^{2+}$, giving an **n-factor of 5** (since the change is of 5 electrons per formula unit).
The equivalent of $\mathrm{MnO}_4^-$ used in the reaction can be calculated as:
$$
\text{Equivalent of } \mathrm{MnO}_4^- = 1.608 \times 10^{-3} \text{ mol} \times 5 = 8.04 \times 10^{-3}
$$
Since $\mathrm{MnO}_4^-$ is used to fully oxidize $\mathrm{A}^{n+}$ to $\mathrm{AO}_3^-$, the equivalents of $\mathrm{A}^{n+}$ must also equal $8.04 \times 10^{-3}$.
Assuming each $\mathrm{A}^{n+}$ can accept (5-n) electrons (where $n$ is the initial oxidation state of $\mathrm{A}$ and it is oxidized to a state of +6 in $\mathrm{AO}_3^-$), the total number of equivalents of $\mathrm{A}^{n+}$ would be:
$$
(5 - n) \times 2.68 \times 10^{-3} = 8.04 \times 10^{-3}
$$
Solving for $n$:
$$
2.68 - 2.68n = 8.04 \\
2.68n = 2.68 - 8.04 \\
n = \frac{-5.36}{-2.68} = 2
$$
Hence, the initial oxidation state of A, **$n = 2$**.
Corrected:
Copper is extracted from copper pyrites by heating in a Bessemer converter. The method is based on the principle that:
A. Copper has more affinity for oxygen than sulphur at high temperatures. B. Iron has less affinity for oxygen than sulphur at high temperatures. C. Copper has less affinity for oxygen than sulphur at high temperatures. D. Sulphur has less affinity for oxygen at high temperatures.
The correct answer to this question is Option A:
Copper has more affinity for oxygen than sulfur at high temperatures.
During the extraction process, this affinity allows copper to react preferentially with oxygen, which helps in the separation and purification of copper from copper pyrites.
In which of the following processes is energy liberated?
A) $\mathrm{Cl} \rightarrow \mathrm{Cl}^{+} + \mathrm{e}^{-}$
B) $\mathrm{HCl} \rightarrow \mathrm{H}^{+} + \mathrm{Cl}^{-}$
C) $\mathrm{Cl} + \mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}$
D) $\mathrm{O} + 2\mathrm{e}^{-} \rightarrow \mathrm{O}^{2-}$
Solution
The correct option is C) $\mathrm{Cl} + \mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}$.
To determine whether energy is liberated or absorbed in a process involving electrons and atoms, we need to consider the stability of the resulting species compared to the original state. Adding an electron to a neutral atom generally releases energy if the resulting ion is more stable than the neutral atom.
Process C: When an electron is added to a neutral chlorine atom to form $\mathrm{Cl}^-$, energy is released because the $\mathrm{Cl}^-$ ion has a more stable electronic configuration than the neutral $\mathrm{Cl}$ atom. The electron affinity of chlorine is positive, indicating that energy release occurs when it gains an electron.
Contrastingly:
Process A: involves removing an electron from chlorine, which requires energy (ionization).
Process B: dissociating HCl into ions requires overcoming the bond energy, which absorbs energy.
Process D: adding an electron to $\mathrm{O}^{-}$ to form $\mathrm{O}^{2-}$ also absorbs a considerable amount of energy due to the increased electron repulsion in the oxygen anion.
Thus, option C is where energy is liberated, making it the correct choice.
In the reaction $\mathrm{M} + \mathrm{O}{2} \rightarrow \mathrm{MO}{2}$ (superoxide), the metal is:
A) $\mathrm{Li}$
B) $\mathrm{Na}$
C) $\mathrm{K}$
D) $\mathrm{Ba}$
Solution
The correct answer is Option C: $\mathrm{K}$ (Potassium).
Metals react with oxygen to form metallic oxides, which can be either simple oxides or more complex oxides like superoxides. Among the listed metals, only potassium ($\mathrm{K}$) reacts with oxygen to predominantly form a superoxide, represented by the formula $\mathrm{KO}_2$. Hence, for the reaction: $$ \mathrm{M} + \mathrm{O}_2 \rightarrow \mathrm{MO}_2 $$ where $\mathrm{MO}_2$ indicates a superoxide, potassium is the metal that completes the reaction to form potassium superoxide ($\mathrm{KO}_2$).
Given that $\mathrm{E}{\mathrm{O}{2} / \mathrm{H}{2} \mathrm{O}}^{\circ}=+1.23 , \text{V};} , \mathrm{E}{\mathrm{S}{2} \mathrm{O}{8}^{2-} / \mathrm{SO}{4}^{2-}}^{2-}=+2.05 , \text{V};} , \mathrm{E}{\mathrm{Br}{2} / \mathrm{Br}^{-}}^{\circ}=+1.09 , \text{V};} , \text{and} , \mathrm{E}{\mathrm{Au}^{3+} / \mathrm{Au}}^{\circ}=+1.4 , \text{V}$. The strongest oxidizing agent is:
A) $\mathrm{O}_{2}$
B) $\mathrm{S}{2} \mathrm{O}{8}^{2-}$ C) $\mathrm{Au}^{3+}$ D) $\mathrm{Br}_{2}$
Solution
The oxidizing power of a chemical agent is correlated with its standard reduction potential. Agents with a higher positive standard reduction potential are stronger oxidizers because they are more inclined to gain electrons.
From the given potentials:
$\mathrm{E}_{\mathrm{O}_2 / \mathrm{H}_2\mathrm{O}}^\circ = +1.23 , \text{V}$
$\mathrm{E}_{\mathrm{S}_2 \mathrm{O}_8^{2-} / \mathrm{SO}_4^{2-}}^\circ = +2.05 , \text{V}$
$\mathrm{E}_{\mathrm{Br}_2 / \mathrm{Br}^-}^\circ = +1.09 , \text{V}$
$\mathrm{E}_{\mathrm{Au}^{3+} / \mathrm{Au}}^\circ = +1.4 , \text{V}$
Analyzing these potentials, it is evident that $\mathrm{S}_2 \mathrm{O}_8^{2-}$ has the highest potential at $+2.05 , \text{V}$. Thus, $\mathrm{S}_2 \mathrm{O}_8^{2-}$ is the strongest oxidizing agent among the given options.
Hence, the correct choice is:
B) $\mathrm{S}_2 \mathrm{O}_8^{2-}$
What happens when iron is heated with sulphur powder? Why?
When iron filings and sulfur powder are mixed and heated, they undergo a chemical reaction to form ferrous sulfide (FeS). This product is entirely new and has properties markedly different from those of iron (Fe) and sulfur (S). Thus, the process exemplifies a chemical change due to the formation of a new substance.
Which of the following gas is produced by roasting iron pyrites in air?
(A) $\mathrm{SO}_{2}$
(B) $\mathrm{SO}_{3}$
(C) $\mathrm{O}_{2}$
(D) $\mathrm{H}_{2}S$
The correct answer is (A) $\mathrm{SO}_{2}$.
Sulphur dioxide ($\mathrm{SO}2$) is the gas produced when iron pyrites ($\mathrm{FeS}2$) is roasted in air. The chemical reaction that takes place can be illustrated as follows: $$ 4 \mathrm{FeS}{2} + 11 \mathrm{O}{2} \rightarrow 2 \mathrm{Fe}{2}\mathrm{O}{3} + 8 \mathrm{SO}_{2} $$ In this reaction, iron pyrite reacts with oxygen ($\mathrm{O}_2$) from the air, forming iron oxide ($\mathrm{Fe}_2\mathrm{O}_3$) and releasing sulphur dioxide ($\mathrm{SO}_2$) as a by-product.
The most common oxidation state of an element is -2. The number of electrons present in its outermost shell is: $\quad$
The provided question mentions that the most common oxidation state of an element is $-2$. This implies that the element naturally desires to gain 2 electrons to achieve a stable electronic configuration similar to a noble gas. Therefore, this element would need to add 2 electrons to reach the typical stable configuration of 8 electrons, as per the octet rule.
Since the element must gain 2 electrons to complete this stable configuration, it indicates that it currently has $8 - 2 = 6$ electrons in its outermost shell.
Therefore, the number of electrons present in the outermost shell of this element is: $$ 6 $$
This description is typically characteristic of Oxygen, which has 6 valence electrons and commonly exhibits an oxidation state of $-2$.
Given that: $$ \mathrm{E}{\mathrm{Fe}^{2+}/\mathrm{Fe}} = -0.44, \mathrm{V} \text{ and }, \mathrm{E}{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = 0.77, \mathrm{V}\text{. } $$ $\mathrm{E}_{\mathrm{Fe}^{3+}/\mathrm{Fe}}$ is:
A $\quad 1.21, \mathrm{V}$
B $\quad 0.33, \mathrm{V}$ C $\quad -0.036, \mathrm{V}$
D $\quad 0.036, \mathrm{V}$
To find $\mathrm{E}_{\mathrm{Fe}^{3+}/\mathrm{Fe}}$, we start with the given standard electrode potentials:
For $\mathbf{\mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe}}$, $\mathrm{E}^{\circ} = -0.44, \mathrm{V}$.
For $\mathbf{\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}}$, $\mathrm{E}^{\circ} = 0.77, \mathrm{V}$.
We need to calculate $\mathrm{E}^{\circ}$ for the reaction:
$$ \mathbf{\mathrm{Fe}^{3+} + 3e^- \rightarrow \mathrm{Fe}} $$
The calculation of the overall cell potential involves the additive property of free energies ($\Delta G^\circ$) where:
$$ \Delta G^\circ = -nFE^\circ $$
For the reactions provided,
$\Delta G^\circ_1 = -2F(-0.44, \mathrm{V})$
$\Delta G^\circ_2 = -1F(0.77, \mathrm{V})$
Adding the free energy changes for these reactions to get the overall free energy change:
$$ \begin{align*} \Delta G^\circ_{\text{overall}} &= \Delta G^\circ_1 + \Delta G^\circ_2 \ -3FE^\circ &= -2F(-0.44) + F(0.77) \ -3FE^\circ &= 0.88F - 0.77F \ -3FE^\circ &= 0.11F \end{align*} $$
Solving for $E^\circ$:
$$ E^\circ = -\frac{0.11F}{3F} = -0.036,\mathrm{V} $$
Hence, the correct answer is:
$$ \mathrm{E}_{\mathrm{Fe}^{3+}/\mathrm{Fe}} = -0.036,\mathrm{V} $$
This corresponds to Option C $-0.036,\mathrm{V}$.
The equivalent weight of $\mathrm{HCl}$ in the given reaction is: $$ \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} + 14 \mathrm{HCl} \rightarrow 2 \mathrm{KCl} + 2 \mathrm{CrCl}_{3} + 3 \mathrm{Cl}_{2} + \mathrm{H}_{2} \mathrm{O} $$
A) 16.25
B) 36.5
C) 73
D) 85.17
The correct answer is D) 85.17.
To determine the equivalent weight of $\mathrm{HCl}$ in the reaction $$ \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + 14\mathrm{HCl} \rightarrow 2\mathrm{KCl} + 2\mathrm{CrCl}_3 + 3\mathrm{Cl}_2 + \mathrm{H}_2\mathrm{O}, $$ we need to analyze the change in oxidation state of the elements involved during the reaction provided by $\mathrm{HCl}$.
Each $\mathrm{HCl}$ molecule supplies electrons that contribute to the change in oxidation states. In this reaction, the relevant part is the formation of $\mathrm{Cl}_2$ from $\mathrm{Cl}^-$: $$ 2 \mathrm{Cl}^- \rightarrow \mathrm{Cl}_2 \text{ (each } \mathrm{Cl}^- \text{ loses one electron)}. $$ A total of 6 electrons are thus needed to form 3 molecules of $\mathrm{Cl}_2$, as 2 $\mathrm{Cl}^-$ donate 2 electrons for each $\mathrm{Cl}_2$.
The number of $\mathrm{HCl}$ molecules providing the necessary electrons is 14. Therefore, each $\mathrm{HCl}$ molecule effectively accounts for $$ \frac{6 \text{ electrons}}{14 \text{ molecules of } \mathrm{HCl}} = \frac{6}{14} \text{ equivalents per } \mathrm{HCl}. $$
Thus, one equivalent of $\mathrm{HCl}$ provides $$ \frac{14}{6} \text{ moles of electrons}. $$
The equivalent weight of a substance is calculated using the formula: $$ \text{Equivalent weight of HCl} = \frac{\text{Molecular Weight of HCl}}{\text{n-factor}}. $$ Given that the Molecular Weight of $\mathrm{HCl}$ is approximately 36.5 g/mol, and the n-factor, which corresponds to the number of electrons per molecule of $\mathrm{HCl}$ involved in the oxidation-reduction, is $6/14$, the formula becomes: $$ \text{Equivalent weight} = \frac{36.5 \text{ g/mol}}{6/14} = 85.17 \text{ g/equiv}. $$
Hence, the equivalent weight of $\mathrm{HCl}$ in the given reaction is 85.17, matching option D.
State True or False: As we go down the reactivity series, thermal stability of metal hydroxides reduces.
A) True
B) False
Answer: A) True
Explanation:
It is true that as we go down the reactivity series, the thermal stability of metal hydroxides reduces. Metals found lower in the reactivity series generally form less stable metal hydroxides, which are more prone to thermal decomposition.
Which of the following is a mild oxidizing agent?
(A) $\mathrm{Cl}_{2}$
(B) $\mathrm{K}{2} \mathrm{Cr}{2} \mathrm{O}_{7}$
(C) $\mathrm{KMnO}_{4}$
(D) $\mathrm{Ag}_{2} \mathrm{O}$
The correct option is (D) $\mathrm{Ag}_2\mathrm{O}$.
$\mathrm{Ag}_2\mathrm{O}$ is known as a mild oxidizing agent. This characteristic allows it to participate in chemical reactions where it facilitates the gain of electrons, but less aggressively compared to stronger agents such as $\mathrm{KMnO}_4$ or $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$.
Iron compounds in rocks start crumbling through the process of
A. Oxidation
B. Carbonation
C. Solution
The correct answer is A. Oxidation.
Oxidation is the process through which iron compounds in rocks undergo chemical reactions with oxygen, leading to the formation of iron oxides. This process typically results in the crumbling and weakening of the rock structure.
$$E^{\circ} \text{ for the reaction}$$ $$\mathrm{Fe} + \mathrm{Zn}^{2+} = \mathrm{Zn} + \mathrm{Fe}^{2+} \text{ is - } 0.35 \text{ V.}$$
The given cell reaction is:
A) feasible
B) not feasible
C) in equilibrium
D) None
The correct answer is B) not feasible.
The provided $E^{\circ}$ value is negative ($-0.35 \text{ V}$), indicating that the cell's standard electromotive force is unfavorable for the spontaneous reaction under standard conditions. Typically, a negative $E^{\circ}$ implies that the Gibbs free energy change ($\Delta G^{\circ}$) is positive, which means the reaction is non-spontaneous and thus not feasible. Hence, the reaction $$\mathrm{Fe} + \mathrm{Zn}^{2+} = \mathrm{Zn} + \mathrm{Fe}^{2+}$$ is not feasible under standard conditions.
The action of peroxidases is countered by the enzyme:
A. catalase
B. glycolase
C. glyoxidase
D. hydrolase
The correct answer is A. catalase.
Peroxisomes are cellular organelles that produce hydrogen peroxide ($ H_2O_2 $) as a byproduct during the breakdown of cellular components via oxidation. To prevent damage from the accumulation of $ H_2O_2 $, peroxisomes also produce the enzyme catalase. Catalase acts to decompose hydrogen peroxide into water and oxygen, effectively neutralizing its potentially harmful effects and thereby counteracting the action of peroxidases, which generate hydrogen peroxide from various substrates.
Thus, catalase serves a crucial protective role by ensuring that unnecessary breakdown of cellular components does not occur due to excessive peroxidative activity.
Manganese exhibits maximum oxidation state in
A $\mathrm{K}{2}\mathrm{MnO}{4}$
B $\mathrm{KMnO}_{4}$
C $\mathrm{MnO}_{2}$
(D) $\mathrm{Mn}{3}\mathrm{O}{4}$
Solution
The correct answer is Option B: $\mathrm{KMnO}_4$.
In $\mathrm{KMnO}_4$, manganese (Mn) exhibits its maximum oxidation state of +7. This is due to the high number of oxygen atoms, each having an oxidation state of -2, which balances the high positive oxidation state of manganese.
"What is meant by oxidized?
The nitric oxide is oxidized again, and nitrogen dioxide is formed."
Solution:
The term "oxidized" refers to the chemical reaction where an element or a compound combines with oxygen to form an oxide. In the provided context, nitric oxide (NO) is reacting with oxygen (O₂) to form nitrogen dioxide (NO₂). The chemical equation for this reaction is: $$ 2NO + O_2 \rightarrow 2NO_2 $$ This process is a typical example of oxidation, where oxygen is added to a compound increasing its oxidation state.
$\mathrm{Zn} + \mathrm{H}{2}\mathrm{SO}{4}$ (conc.) $\rightarrow$ ? is an example of which property of sulfuric acid?
A. Oxidizing agent
B. Reducing agent
C. Volatile
D. Non-volatile
Solution
The correct option is A. Oxidizing agent.
In the reaction between zinc (Zn) and concentrated sulfuric acid ($\mathrm{H}_2\mathrm{SO}_4$), the following equation represents the process: $$ \mathrm{Zn} + \mathrm{H}_2\mathrm{SO}_4 (\text{conc.}) \rightarrow \mathrm{ZnSO}_4 + \mathrm{H}_2 $$
In this reaction, sulfuric acid acts as an oxidizing agent. It promotes the oxidation of zinc, meaning it causes zinc to lose electrons. Specifically, zinc is oxidized from the elemental state (zero oxidation state) to a +2 oxidation state as zinc ion ($\mathrm{Zn}^{2+}$) in zinc sulfate ($\mathrm{ZnSO}_4$): $$ \mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq}) + 2\mathrm{e}^- $$
This electron transfer highlights the role of sulfuric acid as an effective oxidizing agent.
Which of the following statements is true if red hot iron is reacted with steam?
A. It forms triferric tetraoxide and releases hydrogen gas.
B. It's a reversible reaction if hydrogen isn't separated.
C. Iron is more reactive than zinc.
D. Iron oxide formed reacts with hydrogen and reduces back to iron.
Solution
The correct statements regarding the reaction of red-hot iron with steam are:
A: It forms triferric tetraoxide and releases hydrogen gas.
B: It's a reversible reaction if hydrogen isn't separated.
D: Iron oxide formed reacts with hydrogen and reduces back to iron.
When red-hot iron reacts with steam, the chemical reaction that occurs can be represented by the equation:
$$ 3 \text{Fe} + 4 \text{H}_2\text{O} \rightleftharpoons \text{Fe}_3\text{O}_4 + 4 \text{H}_2 \uparrow $$
Here, triferric tetraoxide ($\text{Fe}_3\text{O}_4$) and hydrogen gas ($\text{H}_2$) are formed. The reaction is reversible, meaning if the hydrogen gas produced is not removed from the system, it can react with the iron oxide formed to convert it back to iron. This reversal is facilitated under specific conditions where the hydrogen gas is present in the reaction environment.
For the reduction of $\mathrm{NO}_{3}^{-}$ ion in an aqueous solution, $\mathrm{E}^{\circ}$ is $+0.96 \mathrm{~V}$. Values of $\mathrm{E}^{0}$ for some metal ions are given below:
$$ \begin{array}{ll} \mathrm{V}^{2+}(\mathrm{aq}) + 2 \mathrm{e}^{-} \rightarrow \mathrm{V} ; & \mathrm{E}^{\circ} = -1.19 \mathrm{~V} \ \mathrm{Fe}^{3+}(\mathrm{aq}) + 3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} ; & \mathrm{E}^{\circ} = -0.04 \mathrm{~V} \ \mathrm{Au}^{3+}(\mathrm{aq}) + 3 \mathrm{e}^{-} \rightarrow \mathrm{Au} ; & \mathrm{E}^{\circ} = -1.40 \mathrm{~V} \ \mathrm{Hg}^{2+}(\mathrm{aq}) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Hg} ; & \mathrm{E}^{\circ} = -0.86 \mathrm{~V} \end{array} $$
The pair(s) of metals that is/are oxidized by $\mathrm{NO}_{3}^{-}$ in an aqueous solution is(are):
A: $\mathrm{V}$ and $\mathrm{Hg}$
B: $\mathrm{Hg}$ and $\mathrm{Fe}$
C: $\mathrm{Fe}$ and $\mathrm{Au}$
D: Fe and V
Solution:
The metals here are judged based on their ability to be oxidized by $\mathrm{NO}_3^-$ in an aqueous solution. Metals with $E^{\circ}$ values less than $+0.96~\mathrm{V}$ will be able to reduce $\mathrm{NO}_3^-$. Thus, we can determine which metals are oxidized by comparing their $E^{\circ}$ values.
Given:
$\mathrm{V}$ $(\mathrm{E}^{\circ} = -1.19~\mathrm{V})$
$\mathrm{Fe}$ $(\mathrm{E}^{\circ} = -0.04~\mathrm{V})$
$\mathrm{Au}$ $(\mathrm{E}^{\circ} = -1.40~\mathrm{V})$
$\mathrm{Hg}$ $(\mathrm{E}^{\circ} = -0.86~\mathrm{V})$
Comparing these values:
$\mathrm{V}$, $\mathrm{Fe}$, and $\mathrm{Hg}$ all have $E^{\circ}$ values less than $+0.96~\mathrm{V}$, indicating they can reduce $\mathrm{NO}_3^-$. Therefore, they can be oxidized by $\mathrm{NO}_3^-$.
$\mathrm{Au}$ with $E^{\circ} = -1.40~\mathrm{V}$ will not reduce $\mathrm{NO}_3^-$ in an aqueous solution due to its lower potential compared to $+0.96~\mathrm{V}$.
Hence, the pairs that can be oxidized by $\mathrm{NO}_3^-$ are:
A: $\mathrm{V}$ and $\mathrm{Hg}$ (Correct)
B: $\mathrm{Hg}$ and $\mathrm{Fe}$ (Correct)
D: $\mathrm{Fe}$ and $\mathrm{V}$ (Correct)
The option C: $\mathrm{Fe}$ and $\mathrm{Au}$ is incorrect because $\mathrm{Au}$ cannot be oxidized by $\mathrm{NO}_3^-$ under these conditions.
Of the given 3 allotropes of Phosphorus - white, red, and black, which is the most stable thermodynamically?
A. Black Phosphorus
B. White Phosphorus
C. Red Phosphorus
D. Both A and C
Solution
The correct option is A. Black Phosphorus.
Black Phosphorus is known as the most stable thermodynamically among the allotropes of phosphorus. This allotrope can be obtained through two routes:
Heating white phosphorus under high pressures ranging from 4000 to 12000 atm, at a temperature of about $$ 473 , \mathrm{K} $$ in an inert atmosphere. This yields the β-black Phosphorus.
Heating red phosphorus in a sealed tube at $$ 453 , \mathrm{K} $$ also in an inert atmosphere which forms α-black Phosphorus.
It's important to note that black phosphorus is both a good conductor of electricity and less reactive compared to the other allotropes, white phosphorus and red phosphorus, which are not conductive.
Therefore, black phosphorus stands out as the most stable and least reactive allotrope of phosphorus.
Iron is a negative catalyst.
A) True
B) False
The correct answer is B) False.
A positive catalyst is a substance that increases the rate of a reaction, while a negative catalyst decreases the rate of a reaction. Iron serves as a positive catalyst in the Haber process for the synthesis of ammonia, thus it is not a negative catalyst.
How many moles of iodine are liberated when 1 mole of potassium dichromate reacts with potassium iodide?
A) 1 B) 2 C) 3 D) 4
The correct response is Option C: 3 moles of iodine.
In the reaction involving potassium dichromate ($\text{K}_2\text{Cr}_2\text{O}_7$) and potassium iodide ($\text{KI}$) in the presence of sulfuric acid ($\text{H}_2\text{SO}_4$), the overall balanced chemical equation is:
$$ \text{K}_2\text{Cr}_2\text{O}_7 + 6\text{KI} + 7\text{H}_2\text{SO}_4 \rightarrow 4\text{K}_2\text{SO}_4 + 3\text{I}_2 + \text{Cr}_2(\text{SO}_4)_3 + 7\text{H}_2\text{O} $$
From this equation, it can be seen that 1 mole of potassium dichromate liberates 3 moles of iodine ($\text{I}_2$). This is achieved as each nascent oxygen atom ([O]) derived from the dichromate reacts with two potassium iodide molecules, and this sub-reaction occurs three times, indicating that 3 moles of iodine are evolved. Moreover, the balanced equation confirms that the stoichiometric ratio requires 6 moles of $\text{KI}$ for 1 mole of $\text{K}_2\text{Cr}_2\text{O}_7$ to liberate 3 moles of iodine. Thus, the answer is 3 moles of iodine.
The standard electrode potentials are $\mathrm{K}^{+} / \mathrm{K}=-2.93 \mathrm{~V}$, $\mathrm{Ag}^{+} / \mathrm{Ag}=0.80 \mathrm{~V}$, so:
A. $\mathrm{Ag}$ has a higher tendency to undergo reduction. B. $\mathrm{Ag}^{+}$ has a higher tendency to undergo reduction. C. $\mathrm{K}^{+}$ has a higher tendency to undergo reduction. D. $\mathrm{K}$ has a higher tendency to undergo reduction.
The correct option is B.
$\mathrm{Ag}^+$ has a higher tendency to undergo reduction due to its higher electrode potential compared to $\mathrm{K}^+$. In electrode potentials, a more positive value indicates a greater tendency of the ion to gain electrons and thereby get reduced. In this case, $\mathrm{Ag}^+$ has an electrode potential of $0.80 \mathrm{~V}$, which is much more positive compared to $\mathrm{K}^{+} / \mathrm{K}$, which has a potential of $-2.93 \mathrm{~V}$.
Name the process of heating an ore in the absence of air below its melting point.
A Roasting
B Oxidation
C Auto-reduction
D Calcination
The correct option is D Calcination
Explanation:
Roasting is the process where the ore is heated to high temperatures in the presence of an excess supply of air (oxygen).
Oxidation refers to a chemical process where a substance:
Gains oxygen
Gains an electronegative element
Loses electron(s)
Loses hydrogen
Loses an electropositive element
Auto-reduction refers to a process where sulfide ores of less electropositive metals such as Cu, Hg, Pb are heated, generally in the presence of air.
Calcination is the specific process in which the ore of the metal is heated to high temperatures in the absence or limited supply of air (oxygen). This process does not allow the ore to melt.
I know that if $\mathrm{A}$ is more reactive than $\mathrm{B}$, then metal $A$ + salt of metal $B$ = metal $B$ + salt of metal $A$. What would happen if $B$ is more reactive than $A? And how?
If metal B is more reactive than metal A, then the reaction: $$ A + \text{salt of } B \rightarrow B + \text{salt of } A $$ will not occur. This is because the given scenario describes a displacement reaction where a more reactive metal can displace a less reactive one from its compound. In this situation, since B is more reactive, A does not have the ability to displace B from its salt.
The reaction given below is an example of:
$$ \mathrm{CuSO}{4}(\mathrm{aq}) + \mathrm{H}{2} \mathrm{S}(\mathrm{g}) \rightarrow \mathrm{CuS}(\mathrm{s}) \downarrow + \mathrm{H}{2} \mathrm{SO}{4}(\mathrm{aq}) $$
A. combination reaction
B. displacement reaction
C. decomposition reaction
D. double displacement reaction
The correct option is D. double displacement reaction. In the given equation, there is an exchange of ions between copper sulfate $(\mathrm{CuSO}_4)$ and hydrogen sulfide $(\mathrm{H}_2\mathrm{S})$. As a result, sulfuric acid $(\mathrm{H}_2\mathrm{SO}_4)$ and a precipitate of copper sulfide $(\mathrm{CuS})$ are formed. This type of reaction, where ions are exchanged between reactants to form new products, is termed a double displacement reaction. Additionally, this reaction is classified as a precipitation reaction due to the formation of a precipitate, copper sulfide.
White silver chloride in sunlight turns to:
A) red B) grey C) yellow D) remains white
Solution
The correct option is B) grey.
Silver chloride when exposed to sunlight undergoes a decomposition reaction, changing its appearance. The reaction can be represented by: $$ 2 \mathrm{AgCl}(\mathrm{s}) \rightarrow 2 \mathrm{Ag}(\mathrm{s}) + \mathrm{Cl}_2(\mathrm{g}) $$ This results in the formation of metallic silver, which has a grey color, along with chlorine gas. Thus, white silver chloride turns grey when exposed to sunlight.
Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified $\mathrm{KMnO}_{4}$ for complete oxidation?
(A) $\mathrm{FeC}{2} \mathrm{O}{4}$
(B) $\mathrm{Fe}\left(\mathrm{NO}{2}\right){2}$
(C) $\mathrm{FeSO}_{4}$
(D) $\mathrm{FeSO}_{3}$
The correct answer is (C) $\mathrm{FeSO}_{4}$. Each of these compounds contains the iron ion $\mathrm{Fe}^{2+}$, which will be oxidized to $\mathrm{Fe}^{3+}$. The key is to determine the behavior of the corresponding anions under the action of $\mathrm{KMnO}_{4}$.
For complete oxidation, we consider how the anions react:
$\mathrm{C}{2} \mathrm{O}{4}^{2-}$ (from $\mathrm{FeC}{2}\mathrm{O}{4}$) gets oxidized to $\mathrm{CO}_{2}$
$\mathrm{NO}{2}^{-}$ (from $\mathrm{Fe(NO}{2}){2}$) gets oxidized to $\mathrm{NO}{3}^{-}$
$\mathrm{SO}{3}^{2-}$ (from $\mathrm{FeSO}{3}$) gets oxidized to $\mathrm{SO}_{4}^{2-}$
However, $\mathrm{SO}_{4}^{2-}$ (from $\mathrm{FeSO}{4}$) is already in its highest stable oxidation state (sulfur VI), meaning it does not undergo further oxidation. Thus, the **minimal consumption of $\mathrm{KMnO}{4}$** is required for $\mathrm{FeSO}_{4}$, since only the $\mathrm{Fe}^{2+}$ ions are oxidized.
Which is a stronger reducing agent: $\mathrm{Cr}^{2+}$ or $\mathrm{Fe}^{2+}$, and why?
Solution:
1) The standard reduction potentials of $\mathrm{Cr}^{2+}$ and $\mathrm{Fe}^{2+}$ are approximately $-0.91$ V and $-0.44$ V respectively. The lower the reduction potential, the stronger the reducing agent. Consequently, since $\mathrm{Cr}^{2+}$ has a more negative reduction potential than $\mathrm{Fe}^{2+}$, it is a stronger reducing agent.
2) Consider the highest oxidation states of chromium and iron. Chromium can reach an oxidation state of +6, while iron typically reaches a maximum oxidation state of +3. This means $\mathrm{Cr}^{2+}$ can lose up to four electrons to reach its highest oxidation state, which is significantly more than the one electron $\mathrm{Fe}^{2+}$ can lose to reach iron's highest state. The ability to lose more electrons readily contributes to stronger reducing power.
Therefore, based on the reduction potential and the ability to lose electrons, $\mathrm{Cr}^{2+}$ is a more potent reducing agent than $\mathrm{Fe}^{2+}$.
The oxidation state of $\mathrm{Fe}$ in the brown ring complex $\left[\mathrm{Fe}\left(\mathrm{H}{2} \mathrm{O}\right){5} \mathrm{NO} \mathrm{SO}_{4}\right]$ is
A) 2
B) 1
C) 3
Solution
The correct option is B) 1
To determine the oxidation state of Fe in the brown ring complex $\left[\mathrm{Fe}(H_2O)_5NO\right]SO_4$, we need to focus on the overall charge balance of the complex. Given that $SO_4^{2-}$ contributes a charge of -2, the total charge of the complex cation $\left[\mathrm{Fe}(H_2O)_5NO\right]^{2+}$ must exactly balance this.
Considering that the nitrosonium ion ($NO^+$) carries a +1 charge and water molecules ($H_2O$) are neutral, we can establish the following equation for the oxidation state of iron, x: $$ x + 5 \cdot 0 + 1 = +2 $$ Here, $x$ is the oxidation state of $\mathrm{Fe}$. Solving for $x$ yields: $$ x = +1 $$
This indicates that iron has an oxidation state of +1 in this complex, which is relatively unusual as iron typically exhibits oxidation states of +2 or +3.
Additional Information:
The brown ring test is a qualitative analysis method used to detect the presence of nitrate ions ($NO_3^-$). The procedure involves reducing $NO_3^-$ to $NO$ in an acidic environment using iron(II) sulfate. This reduction leads to the formation of the brown ring complex where iron has an oxidation state of +1. The reactions involved are: 1. $$ 2 HNO_3 + 3 H_2SO_4 + 6 FeSO_4 \rightarrow 3 Fe_2(SO_4)_3 + 2 NO + 4 H_2O $$ 2. $$ \left[Fe(H_2O)_6\right]SO_4 + NO \rightleftharpoons \left[Fe(H_2O)_5(NO)\right]SO_4 + H_2O $$
These reactions highlight the transformation and interaction leading to the formation of the peculiar brown ring indicating the presence of nitrates.
Consider a redox reaction: $$ \mathrm{FeS}_{2} + \mathrm{KMnO}_{4} + \mathrm{H}^{+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{SO}_{4}^{2-} + \mathrm{Mn}^{2+} + \mathrm{H}_{2} \mathrm{O}. $$
If the molar mass of $\mathrm{FeS}_{2}$ is $\mathrm{M}$, then the equivalent mass of $\mathrm{FeS}_{2}$ would be equal to:
A) $M$
B) $\frac{M}{5}$
C) $\frac{M}{10}$
D) $\frac{M}{15}$
The correct answer is D) $\frac{M}{15}$.
To determine the equivalent mass of $\mathrm{FeS}_{2}$, we first need to establish its $n$-factor, which indicates the total number of electrons transferred per molecule in the reaction.
The oxidation states of the atoms in $\mathrm{FeS}_{2}$ change during the reaction as follows: $$ \mathrm{FeS}_{2} \rightarrow \mathrm{Fe}^{3+} + \mathrm{SO}_{4}^{2-}. $$ Here:
Iron ($\mathrm{Fe}^{2+}$ in $\mathrm{FeS}_{2}$) oxidizes to $\mathrm{Fe}^{3+}$: Electron change is $1 \times (3-2) = 1$ (for one $\mathrm{Fe}$ atom).
Each sulfur ($\mathrm{S}^{-1}$ in $\mathrm{FeS}_{2}$) oxidizes to $\mathrm{SO}_{4}^{2-}$ ($\mathrm{S}^{6+}$ state): Electron change is $2 \times (6-(-1)) = 14$ for two $\mathrm{S}$ atoms.
Adding these electron transfers gives the total $n$-factor: $$ n_{\mathrm{f}} = 1 \times (3-2) + 2 \times (6 - (-1)) = 1 + 14 = 15. $$ The $n$-factor ($n_{\mathrm{f}}$) represents the total number of electrons transferred by one mole of $\mathrm{FeS}_{2}$, which in this case is $15$.
Using this, the equivalent mass of $\mathrm{FeS}_{2}$ is calculated as: $$ \text{Equivalent mass} = \frac{\text{Molar mass (M)}}{n_{\mathrm{f}}} = \frac{M}{15}. $$ Thus, option D) $\frac{M}{15}$ is correct.
During corrosion, iron reacts with which of the following substances that are present in the atmosphere?
A. Oxygen
B. Moisture
C. Nitrogen
D. Carbon dioxide
Solution:The correct answers are:
A. Oxygen
B. Moisture
Corrosion is the gradual degradation of the metal surface when it is exposed to air and moisture. During the corrosion of iron, the metal specifically reacts with oxygen and moisture present in the atmosphere to form rust (iron oxide). Rust is characterized by a flaky, brown-colored layer that forms on the surface of iron objects. This layer weakens the iron articles over time.
The gain of electrons is known as ________ and the loss of electrons is known as ________.
A. electrolysis
B. reduction
C. oxidation
D. decomposition
The correct answers are:
B. reduction and C. oxidation.
Reduction refers to the gain of electrons by an atom or ion, while oxidation involves the loss of electrons from an atom or ion. In the process of electrolysis, reduction occurs at the cathode and oxidation happens at the anode.
What volume of $0.1 \mathrm{M} \mathrm{KMnO}{4}$ is needed to oxidize $100 \mathrm{mg}$ of $\mathrm{FeC}{2}\mathrm{O}_{4}$ in acid solution?
A) $4.1 \mathrm{~mL}$
B) $8.2 \mathrm{~mL}$
C) $\mathbf{10.2 \mathrm{~mL}}$
D) $4.6 \mathrm{~mL}$
To solve this problem, we first determine the basic chemical reactions involved and their stoichiometry (number of moles of electrons exchanged) to ascertain how much potassium permanganate ($\mathrm{KMnO}_4$) is required to fully oxidize ferrous oxalate ($\mathrm{FeC}_2\mathrm{O}_4$).
Oxidation of oxalate ions ($\mathrm{C}_2\mathrm{O}_4^{2-}$) to carbon dioxide ($\mathrm{CO}_2$): $$ \mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 \mathrm{CO}_2 + 2 \mathrm{e}^- $$ This reaction shows the transfer of 2 electrons.
Oxidation of ferrous ions ($\mathrm{Fe}^{2+}$) to ferric ions ($\mathrm{Fe}^{3+}$): $$ \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{e}^- $$ This reaction shows the transfer of 1 electron.
Combining these, the total oxidation for one molecule of ferrous oxalate involves 3 electrons. Therefore, the $n$-factor of $\mathrm{FeC}_2\mathrm{O}_4$ is 3.
In an acidic solution, $\mathrm{KMnO}_4$ reacts by providing 5 moles of electrons per mole: $$ \text{$n$-factor of } \mathrm{KMnO}_4 = 5 $$
From stoichiometry, we know that 3 moles of $\mathrm{KMnO}_4$ will oxidize 5 moles of $\mathrm{FeC}_2\mathrm{O}_4$.
Given that the molar mass of ferrous oxalate ($\mathrm{FeC}_2\mathrm{O}_4$) is $$143.91 , \mathrm{g/mol},$$ the problem can be addressed by equating the equivalents of $\mathrm{KMnO}_4$ and $\mathrm{FeC}_2\mathrm{O}_4$:
$$ \text{Equivalents of } \mathrm{KMnO}_4 = \text{Equivalents of } \mathrm{FeC}_2\mathrm{O}_4 $$
Thus: $$ 0.1 \times \text{volume (L)} \times 5 = \frac{0.100 , \mathrm{g} \times 3}{143.91 , \mathrm{g/mol}} $$
Solving for volume: $$ 0.1 \times \text{volume} \times 5 = \frac{0.100 \times 3}{143.91} $$ $$ \text{volume} = \frac{0.100 \times 3}{5 \times 0.1 \times 143.91} \approx 0.0041 , \mathrm{L} $$
Converting to milliliters: $$ \text{volume} \approx 4.1 , \mathrm{mL} $$
Thus, the correct answer is 4.1 mL (Choice A).
Given: $$ \begin{array}{l} \mathrm{E}{\mathrm{Cl}{2} / \mathrm{Cl}^{-}}^{\circ} = 1.36 , \mathrm{V}, \quad \mathrm{E}{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ} = -0.74 , \mathrm{V} \ \mathrm{E}{\mathrm{Cr}{2} \mathrm{O}{7}^{2-} / \mathrm{Cr}^{3+}} = 1.33 , \mathrm{V}, \quad \mathrm{E}{\mathrm{MnO}{4}^{-} / \mathrm{Mn}^{2+}}^{\circ} = 1.51 , \mathrm{V} \end{array} $$
Among the following, the strongest reducing agent is
(A) $\mathrm{Cr}$ (B) $\mathrm{Mn}^{2+}$ (C) $\mathrm{Cr}^{3+}$ (D) $\mathrm{Cl}^{-}$
The correct answer is Option A: $\mathrm{Cr}$.
Given are the standard potential values: $$ \begin{array}{l} \mathrm{E}_{\mathrm{MnO}4^{-} / \mathrm{Mn}^{2+}}^\circ = 1.51 , \mathrm{V} \quad \text{(i)} \ \mathrm{E}{\mathrm{Cl}2 / \mathrm{Cl}^{-}}^\circ = 1.36 , \mathrm{V} \quad \text{(ii)} \ \mathrm{E}{\mathrm{Cr}_2\mathrm{O}7^{2-} / \mathrm{Cr}^{3+}} = 1.33 , \mathrm{V} \quad \text{(iii)} \ \mathrm{E}{\mathrm{Cr}^{3+} / \mathrm{Cr}}^\circ = -0.74 , \mathrm{V} \quad \text{(iv)} \end{array} $$
From the values listed, the potential for the reduction of $\mathrm{Cr}^{3+}$ to $\mathrm{Cr}$ is the most negative $(-0.74 , \mathrm{V})$. This negative value indicates that $\mathrm{Cr}$ is more prone to lose electrons, making it the strongest reducing agent among those listed. Therefore, $\mathrm{Cr}$ (Option A) is the strongest reducing agent.
The equivalent weight of $\mathrm{MnSO}_{4}$ is half its molecular weight when it is converted to
(A) $\mathrm{Mn}_{2} \mathrm{O}_{3}$
(B) $\mathrm{MnO}_{2}$
(C) $\mathrm{MnO}_{4}^{\ominus}$
(D) $\mathrm{MnO}_{4}^{2-}$
The correct answer is (B) $\mathrm{MnO}_2$.
In the conversion reaction: $$ \mathrm{Mn}^{+2} \text{ (from MnSO}_4\text{)} \rightarrow \mathrm{MnO}_2 $$ $\mathrm{Mn}^{+2}$ is reduced to $\mathrm{Mn}^{+4}$ in $\mathrm{MnO}_2$ (as manganese in $\mathrm{MnO}_2$ has a +4 oxidation state). Since the manganese ion's oxidation state changes from +2 to +4, it gains 2 electrons.
The number of electron changes (n-factor) here is 2, representing the change in oxidation state required to form $\mathrm{MnO}_2$.
Since the equivalent weight of a compound is calculated as: $$ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{\text{n-factor}} $$ for this reaction, the n-factor is 2, thus the equivalent weight of $\mathrm{MnSO}_4$ becomes: $$ \frac{\text{Molecular Weight of } \mathrm{MnSO}_4}{2} $$ Implying the equivalent weight of $\mathrm{MnSO}_4$ when converting to $\mathrm{MnO}_2$ is exactly half its molecular weight.
In the reaction $3 \mathrm{Mg} + \mathrm{N}_{2} \rightarrow \mathrm{Mg}_{3} \mathrm{N}_{2}$:
A. $\mathrm{Mg}$ is reduced.
B. $\mathrm{Mg}$ is oxidized.
C. Nitrogen is oxidized.
D. None of these.
The correct option is B: $\mathrm{Mg}$ is oxidized.
In this reaction, the oxidation state of magnesium ($\mathrm{Mg}$) changes from 0 to +2, indicating that magnesium is being oxidized. Correspondingly, the oxidation state of nitrogen ($\mathrm{N}_2$) changes from 0 to -3, demonstrating that nitrogen is being reduced.
Thus, oxidation of $\mathrm{Mg}$ and reduction of nitrogen occur in this reaction.
1.00 g of a mixture, consisting of an equal number of moles of carbonates of two alkali metals, required 44.4 mL of 0.5 N-HCl for complete reaction. If the atomic weight of one of the metals is 7.00, find the atomic weight of the other metal.
To determine the atomic weight of the other alkali metal in the mixture of carbonates, we should follow these steps:
Step-by-Step :
Understand the Problem:
We have a 1-gram mixture of carbonates of two alkali metals.
The mixture requires 44.4 mL of 0.5 N HCl for complete reaction.
The atomic weight of one alkali metal is 7.00.
Identify the Carbonates:
Let the first metal carbonate be $M_2CO_3$ where the atomic weight of $M$ is 7.
The formula of the carbonate of the other metal is $M'_2CO_3$ with atomic weight $M'$.
Set Up the Equations:
The total mass of the mixture is 1 gram.
Assume the masses of the two carbonates are $a$ grams and $b$ grams respectively.
Therefore, we have: $$ a + b = 1 \quad \text{(Equation 1)} $$
Calculate Molar Masses:
Molar mass of $M_2CO_3$: $$ 2 \times 7 + 12 + 3 \times 16 = 74 \text{ grams/mol} $$
Molar mass of $M'_2CO_3$: $$ 2 \times M' + 12 + 3 \times 16 = 2M' + 60 \text{ grams/mol} $$
Number of Moles Relationship:
Since the mixture contains equal moles of each carbonate: $$ \frac{a}{74} = \frac{b}{2M' + 60} \quad \text{(Equation 2)} $$
Reaction with HCl:
The reaction between the metal carbonates and HCl can be represented as: $$ \text{Metal carbonate + 2HCl} \rightarrow \text{Metal chloride + CO}_2 + \text{H}_2\text{O} $$
The milliequivalents of HCl used is: $$ 44.4 \text{ mL} \times 0.5 \text{ N} = 22.2 \text{ milliequivalents} $$
Calculate Milliequivalents:
For $M_2CO_3$: $$ \frac{a \times 1000}{74 \times 2} $$
For $M'_2CO_3$: $$ \frac{b \times 1000}{(2M' + 60) \times 2} $$
Equate to HCl milliequivalents: $$ \frac{a \times 1000}{74 \times 2} + \frac{b \times 1000}{(2M' + 60) \times 2} = 22.2 \quad \text{(Equation 3)} $$
Solve the Equations:
Using equations (1), (2), and (3), solve for $ M' $: $$ M' = 23 $$
Thus, the atomic weight of the other metal is 23.
Final Answer:
23
Iodine titration can be iodometric or iodimetric depending on using iodine directly or indirectly as an oxidising agent in the redox titration.
a. Iodometric titration is where a standard iodine solution is used as an oxidant and iodine is directly or indirectly titrated against a reducing agent. For example:
$$ 2 \mathrm{CuSO}_{4} + 4 \mathrm{KJ} \rightarrow \mathrm{Cu}_{2} \mathrm{I}_{2} + 2 \mathrm{K}_{2} \mathrm{SO}_{4} + \mathrm{I}_{2} $$
b. Iodimetric procedures are used for the determination of the strength of reducing agents such as thiosulphates, sulphites, arsenites, and stannous chloride by titrating them against a standard solution of iodine in a burette.
$$ 2 \mathrm{Na}_{2} \mathrm{SO}_{3} \rightarrow 2 \mathrm{NaI} + \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6} $$
Starch is used as an indicator near the endpoint, which forms a blue color complex with $I_{3}^{-}$. The blue color disappears when there is no more free $I_{2}$.
When $319.0 \mathrm{gm}$ of $\mathrm{CuSO}_{4}$ in a solution is titrated with excess of $0.5 \mathrm{M} \mathrm{KI}$ solution, then liberated iodine requires $200 \mathrm{ml}$ of $1.0 \mathrm{M} \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$ for complete reaction. The percentage purity of $\mathrm{CuSO}_{4}$ in the sample is:
A $10\%$
B $20\%$
C $5\%$
D None of these
To solve for the percentage purity of $\mathrm{CuSO}_{4}$ in the sample, we can follow these steps:
Step-by-Step :
Given Data and Reaction:
Mass of $\mathrm{CuSO}_{4}$ in the solution: $319.0 ,\text{g}$
Excess $\mathrm{KI}$ solution: $0.5 ,\text{M}$
Volume of $\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}$ for titration: $200 ,\text{mL}$
Molarity of $\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}$: $1.0 ,\text{M}$
Balanced Chemical Reactions:
For the titration reactions: $$ 2 \mathrm{CuSO}_{4} + 4 \mathrm{KI} \rightarrow \mathrm{Cu}_{2}\mathrm{I}_{2} + 2 \mathrm{K}_2\mathrm{SO}_4 + \mathrm{I}_2 $$ $$ \mathrm{I}_2 + 2 \mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3} \rightarrow \mathrm{Na}_{2}\mathrm{S}_{4}\mathrm{O}_{6} + 2 \mathrm{NaI} $$
Calculate moles of $\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}$:
Since the volume of $\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}$ is $200 ,\text{mL}$ (which is $0.2 ,\text{L}$) and the molarity is $1.0 ,\text{M}$: $$ \text{Moles of } \mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3} = 1.0 \times 0.2 = 0.2 , \text{moles} $$
Relate moles of $\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}$ to moles of $\mathrm{I}_2$:
According to the second reaction, 2 moles of $\mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}$ react with 1 mole of $\mathrm{I}_2$: $$ \text{Moles of } \mathrm{I}_2 = \frac{\text{Moles of } \mathrm{Na}_{2}\mathrm{S}_{2}\mathrm{O}_{3}}{2} = \frac{0.2}{2} = 0.1 , \text{moles} $$
Relate moles of $\mathrm{I}_2$ to moles of $\mathrm{CuSO}_{4}$:
From the first reaction, we know 1 mole of $\mathrm{I}_2$ is produced by 2 moles of $\mathrm{CuSO}_{4}$: $$ \text{Moles of } \mathrm{CuSO}_{4} = 2 \times \text{Moles of } \mathrm{I}_2 = 2 \times 0.1 = 0.2 , \text{moles} $$
Calculate the mass of pure $\mathrm{CuSO}_{4}$:
The molar mass of $\mathrm{CuSO}_{4}$ is approximately $159.5 ,\text{g/mol}$: $$ \text{Mass of pure } \mathrm{CuSO}_{4} = \text{Moles of } \mathrm{CuSO}_{4} \times \text{Molar mass} $$ $$ = 0.2 \times 159.5 ,\text{g/mol} = 31.9 ,\text{g} $$
Determine the percentage purity:
Use the formula: $$ \text{Percentage purity} = \left( \frac{\text{Mass of pure substance}}{\text{Total mass of sample}} \right) \times 100 $$ $$ = \left( \frac{31.9}{319.0} \right) \times 100 = 10% $$
Final Answer:
The percentage purity of $\mathrm{CuSO}_{4}$ in the sample is 10%.
Thus, the correct option is A: 10%.
Consider the following reaction at $1000^{\circ} \mathrm{C}$: i. $\mathrm{Zn}(\mathrm{s}) + \frac{1}{2} \mathrm{O}_{2} (\mathrm{g}) \rightarrow \mathrm{ZnO}(\mathrm{s})$; $\Delta G = -360 , \mathrm{kJ/mol}$ ii. $\mathrm{C}(\mathrm{s}) + \frac{1}{2} \mathrm{O}_{2} (\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{s})$; $\Delta G = -460 , \mathrm{kJ/mol}$
Then choose the correct statement from the following: A Zinc can be oxidized by $\mathrm{CO}$ B $\mathrm{ZnO}$ can be reduced by $\mathrm{C}$ C $\mathrm{ZnO}$ can be reduced by $\mathrm{CO}$ D None of the above
The correct answer is B: $\mathrm{ZnO}$ can be reduced by $\mathrm{C}$.
In this context, we refer to the Ellingham diagram, which plots the standard Gibbs free energy change $\Delta_f G^\circ$ for the formation of oxides. The key detail here is that the $\Delta_f G^\circ$ value for carbon ($\mathrm{C}$) lies below that of zinc oxide ($\mathrm{ZnO}$). This implies that at sufficiently high temperatures, carbon can effectively act as a reducing agent for zinc oxide.
Consider the given reactions:
For zinc oxide formation: $$ \mathrm{Zn}(\mathrm{s}) + \frac{1}{2} \mathrm{O}_2 (\mathrm{g}) \rightarrow \mathrm{ZnO}(\mathrm{s}) \quad \Delta_f G^\circ = -360 , \mathrm{kJ/mol} $$
For carbon monoxide formation: $$ \mathrm{C}(\mathrm{s}) + \frac{1}{2} \mathrm{O}_2 (\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{s}) \quad \Delta_f G^\circ = -460 , \mathrm{kJ/mol} $$
Given these reactions, the net reaction when carbon reduces zinc oxide is: $$ \mathrm{ZnO}(\mathrm{s}) + \mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Zn}(\mathrm{s}) + \mathrm{CO} \quad \Delta_f G^\circ = -100 , \mathrm{kJ/mol} $$
Since the net Gibbs free energy change $\Delta_f G^\circ = -100 , \mathrm{kJ/mol}$ is negative, this reaction is spontaneous. This clearly demonstrates that $\mathrm{ZnO}$ can be reduced by $\mathrm{C}$.
For the reduction of $\mathrm{NO}_{3}^{-}$ ion in an aqueous solution, $E^\circ$ is +0.96 V. Values of $E^\circ$ for some metal ions are given below:
$$ \begin{array}{l} \mathrm{V}^{2+}(\mathrm{aq}) + 2 \mathrm{e}^{-} \rightarrow \mathrm{V} ; E^\circ = -1.19 \mathrm{V} \ \mathrm{Fe}^{3+}(\mathrm{aq}) + 3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} ; E^\circ = -0.04 \mathrm{V} \ \mathrm{Au}^{3+}(\mathrm{aq}) + 3 \mathrm{e}^{-} \rightarrow \mathrm{Au} ; E^\circ = +1.40 \mathrm{V} \ \mathrm{Hg}^{2+}(\mathrm{aq}) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Hg} ; E^\circ = +0.86 \mathrm{V} \end{array} $$
The pair(s) of metals that is/are oxidised by $\mathrm{NO}_{3}^{-}$ in aqueous solution are:
A V and Hg B Hg and Fe C Fe and Au D Fe and V
The correct options are:
A $\mathrm{V}$ and $\mathrm{Hg}$
B $\mathrm{Hg}$ and $\mathrm{Fe}$
D $\mathrm{Fe}$ and $V$
To determine which metals can be oxidized by $\mathrm{NO}_3^{-}$ in an aqueous solution, we compare their standard reduction potentials ($E^\circ$) with that of $\mathrm{NO}_3^{-}$ which is $+0.96 , \mathrm{V}$.
Metals with $E^\circ$ values less than $0.96 , \mathrm{V}$ will be able to reduce $\mathrm{NO}_3^{-}$. Therefore, we analyze the given standard reduction potentials:
$\mathrm{V}\left(\mathrm{E}^{\circ}=-1.19 , \mathrm{V}\right)$
$\mathrm{Fe}\left(\mathrm{E}^{\circ}=-0.04 , \mathrm{V}\right)$
$\mathrm{Hg}\left(\mathrm{E}^{\circ}=0.86 , \mathrm{V}\right)$
These metals, $\mathrm{V}$, $\mathrm{Fe}$, and $\mathrm{Hg}$, all have standard reduction potentials less than $0.96 , \mathrm{V}$, hence they can reduce $\mathrm{NO}_3^{-}$.
However, $\mathrm{Au}\left(\mathrm{E}^{\circ}=+1.40 , \mathrm{V}\right)$ has a standard reduction potential greater than $0.96 , \mathrm{V}$ and cannot reduce $\mathrm{NO}_3^{-}$ in the aqueous solution.
Thus, the valid pairs of metals that can be oxidized by $\mathrm{NO}_3^{-}$ are:
$\mathrm{V}$ and $\mathrm{Hg}$ (Option A)
$\mathrm{Hg}$ and $\mathrm{Fe}$ (Option B)
$\mathrm{Fe}$ and $V$ (Option D)
Which of the following is/are correct regarding redox titrations?
A. In titrations using potassium dichromate, $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ is used as a self-indicator.
B. In titrations using potassium dichromate, external indicators like diphenylamine are used.
C. In titrations using potassium permanganate, $\mathrm{KMnO}_{4}$ is used as a self-indicator.
D. In titrations using potassium permanganate, external indicators like diphenylamine are used.
The correct options are:
B. In titrations using potassium dichromate, external indicators like diphenylamine are used.
C. In titrations using potassium permanganate, $\mathrm{KMnO}_{4}$ is used as a self-indicator.
In redox titrations involving potassium permanganate ($\mathrm{KMnO}_{4}$), the compound serves as a __self-indicator__. Potassium permanganate is inherently dark purple, and during titration in an acidic medium, it changes to a pale pink due to the formation of $ \mathrm{Mn}^{2+} $ at the endpoint. Since the primary role of an indicator is to detect the endpoint, there is no need for a separate indicator in permanganate titrations. Thus, $\mathrm{KMnO}_{4}$ acts as a self-indicator.
In contrast, redox titrations using potassium dichromate ($\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$) require external indicators such as diphenylamine or potassium ferricyanide to ascertain the endpoint of the reaction.
The minimum amount of $O_2(\text{g})$ consumed per gram of reactant is for the reaction (Given atomic mass:
$\text{Fe}=56$, $\text{O}=16$, $\text{Mg}=24$, $\text{P}=31$, $\text{C}=12$, $\text{H}=1$)
A $4\text{Fe}(\text{s}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{Fe}_2\text{O}_3(\text{s})$
B $2\text{Mg}(\text{s}) + \text{O}_2(\text{g}) \rightarrow 2\text{MgO}(\text{s})$
C $\text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l})$
D $\text{P}_4(\text{s}) + 5\text{O}_2(\text{g}) \rightarrow \text{P}_4\text{O}_{10}(\text{s})$
The correct answer is A.
Let's calculate the amount of $ O_2 $ required per gram for each reaction:
For Reaction A:$$ 4\text{Fe}(\text{s}) + 3\text{O}_2(\text{g}) \rightarrow 2\text{Fe}_2\text{O}_3(\text{s}) $$ Per gram Fe, the $ \text{O}_2 $ required is: $$ \frac{3}{224} \text{ mole} $$ Here, the molecular weight of Fe is 56, and there are 4 Fe atoms, summing up to 224.
For Reaction B:$$ 2\text{Mg}(\text{s}) + \text{O}_2(\text{g}) \rightarrow 2\text{MgO}(\text{s}) $$ Per gram Mg, the $ \text{O}_2 $ required is: $$ \frac{1}{48} \text{ mole} $$ Here, the molecular weight of Mg is 24, and there are 2 Mg atoms, summing up to 48.
For Reaction C:$$ \text{C}_3\text{H}_8(\text{g}) + 5\text{O}_2(\text{g}) \rightarrow 3\text{CO}_2(\text{g}) + 4\text{H}_2\text{O}(\text{l}) $$ Per gram $\text{C}_3\text{H}_8$, the $ \text{O}_2 $ required is: $$ \frac{5}{44} \text{ mole} $$ Here, the molecular weight of $ \text{C}_3\text{H}_8 $ is $ 12 \times 3 + 1 \times 8 = 44 $.
For Reaction D:$$ \text{P}_4(\text{s}) + 5\text{O}_2(\text{g}) \rightarrow \text{P}_4\text{O}_{10}(\text{s}) $$ 5 moles of $ \text{O}_2 $ are required for 1 mole of $ \text{P}_4 $ (124 g). Per gram $ \text{P}_4$, the $ \text{O}_2 $ required is: $$ \frac{5}{124} \text{ mole} $$ Here, the molecular weight of $ \text{P}_4 $ is $ 31 \times 4 = 124 $.
Conclusion:
Reaction A: $ \frac{3}{224} \approx 0.013 \text{ mole per gram} $
Reaction B: $ \frac{1}{48} \approx 0.0208 \text{ mole per gram} $
Reaction C: $ \frac{5}{44} \approx 0.1136 \text{ mole per gram} $
Reaction D: $ \frac{5}{124} \approx 0.0403 \text{ mole per gram} $
The minimum amount of $O_2$ consumed per gram of reactant is for Reaction A.
The major product of the following reaction is:
A.
в.
C.
D.
The correct answer is D.
Alkaline $\mathrm{KMnO}_{4}$ (potassium permanganate) oxidizes alkyl side chains on an aromatic ring to benzoic acid, provided that there is at least one benzylic hydrogen present.
In the given reaction, the ethyl group ($\mathrm{CH_2CH_3}$) attached to the benzene ring contains benzylic hydrogens. Therefore, it will be oxidized to a carboxylic acid group ($\mathrm{COOH}$).
Final Answer: D
Iron articles are shiny when new, but get coated with a reddish-brown powder when left for some time.
True
False
The correct option is A: True
This statement is true. Iron articles are indeed shiny when they are new, but over time, they become coated with a reddish-brown powder. This phenomenon is known as rusting of iron.
Rusting occurs when iron is exposed to substances in its environment, such as moisture and acids, resulting in the formation of rust. Rust is the common name for the compound iron oxide ($\text{Fe}_2\text{O}_3$), which develops as iron reacts chemically with oxygen and water. This rusting process weakens the iron and diminishes its aesthetic appeal.