Question

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?

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Answer

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: $\mathbf{0.439 \text{ g}}$


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