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Q2.16:
Three electrolytic cells A,B,C containing solutions of (ZnSO_{4}), (AgNO_{3}) and (CuSO_{4}), respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Solution :

Step 1: Identifying Variables and Theoretical Constants

Given data for the electrolytic cells connected in series:

  • Current ($I$) = $1.5 \text{ A}$
  • Mass of silver deposited ($m_{\text{Ag}}$) = $1.45 \text{ g}$
  • Molar mass of silver ($M_{\text{Ag}}$) = $107.87 \text{ g/mol}$
  • Faraday’s constant ($F$) = $96485 \text{ C/mol}$

Step 2: Calculating the Charge Passed ($Q$)

The reduction reaction for silver is: $\text{Ag}^+ + e^- \rightarrow \text{Ag}(s)$. [Per Faraday's First Law of Electrolysis, $m = \frac{M \cdot I \cdot t}{n \cdot F}$].

First, calculate the number of moles of silver deposited ($n_{\text{Ag}}$):

$n_{\text{Ag}} = \frac{1.45 \text{ g}}{107.87 \text{ g/mol}} \approx 0.01344 \text{ mol}$

Since the stoichiometry is $1:1$, the amount of charge required ($Q$) is:

$Q = n_{\text{Ag}} \times F = 0.01344 \text{ mol} \times 96485 \text{ C/mol} \approx 1296.76 \text{ C}$

Step 3: Calculating Time ($t$)

Using the relationship $Q = I \times t$:

$t = \frac{Q}{I} = \frac{1296.76 \text{ C}}{1.5 \text{ A}} \approx 864.5 \text{ seconds}$

Converting to minutes: $t \approx 14.41 \text{ minutes}$.

Step 4: Calculating Mass of Copper and Zinc Deposited

Because the cells are in series, the same quantity of electricity ($1296.76 \text{ C}$) passes through each cell. We apply the formula $m = \frac{M \cdot Q}{n \cdot F}$.

Metal Half-reaction $n$ (electrons) Molar Mass ($M$)
Copper $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ $2$ $63.55 \text{ g/mol}$
Zinc $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$ $2$ $65.38 \text{ g/mol}$

Mass of Copper ($m_{\text{Cu}}$):

$m_{\text{Cu}} = \frac{63.55 \text{ g/mol} \times 1296.76 \text{ C}}{2 \times 96485 \text{ C/mol}} \approx 0.427 \text{ g}$

Mass of Zinc ($m_{\text{Zn}}$):

$m_{\text{Zn}} = \frac{65.38 \text{ g/mol} \times 1296.76 \text{ C}}{2 \times 96485 \text{ C/mol}} \approx 0.439 \text{ g}$

Step 5: Visual Representation of Series Circuit

ZnSO4 (A) AgNO3 (B) CuSO4 (C)

Final Solution: The current flowed for $864.5 \text{ seconds}$ (approx. $14.41 \text{ minutes}$). The mass of copper deposited is $0.427 \text{ g}$ and the mass of zinc deposited is $0.439 \text{ g}$.


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