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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: Establishing the Theoretical Framework

According to [Faraday’s First Law of Electrolysis], the mass ($m$) of a substance deposited at an electrode is directly proportional to the quantity of electricity ($Q$) passed through the electrolyte: $m = ZIt = \frac{MIt}{nF}$, where $M$ is the molar mass, $I$ is current in Amperes, $t$ is time in seconds, $n$ is the number of moles of electrons involved, and $F$ is Faraday’s constant ($\approx 96487 \, C \, mol^{-1}$).

Step 2: Calculating Time ($t$) from Silver Deposition

For cell B ($AgNO_{3}$), the cathodic reaction is: $Ag^+ + e^- \rightarrow Ag(s)$. Here, $n = 1$.

  • Mass of Silver ($m_{Ag}$) = $1.45 \, g$
  • Molar Mass of Silver ($M_{Ag}$) = $107.87 \, g \, mol^{-1}$
  • Current ($I$) = $1.5 \, A$

Rearranging the formula for time ($t$):

$t = \frac{m_{Ag} \times n \times F}{M_{Ag} \times I} = \frac{1.45 \, g \times 1 \times 96487 \, C \, mol^{-1}}{107.87 \, g \, mol^{-1} \times 1.5 \, A}$

$t = \frac{139906.15}{161.805} \approx 864.66 \, s$

Step 3: Calculating Mass of Copper Deposited

For cell C ($CuSO_{4}$), the cathodic reaction is: $Cu^{2+} + 2e^- \rightarrow Cu(s)$. Here, $n = 2$.

  • Molar Mass of Copper ($M_{Cu}$) = $63.55 \, g \, mol^{-1}$

$m_{Cu} = \frac{M_{Cu} \times I \times t}{n \times F} = \frac{63.55 \times 1.5 \times 864.66}{2 \times 96487}$

$m_{Cu} = \frac{82424.3}{192974} \approx 0.427 \, g$

Step 4: Calculating Mass of Zinc Deposited

For cell A ($ZnSO_{4}$), the cathodic reaction is: $Zn^{2+} + 2e^- \rightarrow Zn(s)$. Here, $n = 2$.

  • Molar Mass of Zinc ($M_{Zn}$) = $65.38 \, g \, mol^{-1}$

$m_{Zn} = \frac{M_{Zn} \times I \times t}{n \times F} = \frac{65.38 \times 1.5 \times 864.66}{2 \times 96487}$

$m_{Zn} = \frac{84795.1}{192974} \approx 0.439 \, g$

Electrochemical Cell Schematic

Cell A: Zn Cell B: Ag Cell C: Cu
SubstanceMass Deposited (g)
Silver (Ag)1.45
Copper (Cu)0.427
Zinc (Zn)0.439

Final Solution: The current flowed for 864.66 seconds. The mass of copper deposited is 0.427 g, and the mass of zinc deposited is 0.439 g.


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