$10 \, A$ current is passed through $CuSO_4$ solution for $193 \, \text{minutes}$ at $1 \, \text{bar}$ pressure and $300 \, K$ temperature. If the cell has $80 \%$ efficiency,find the weight of $Cu$ deposited and the volume of $O_2$ evolved. $[Cu = 63.5 \, u, R = 0.08314 \, \text{L bar K}^{-1} \text{mol}^{-1}]$
(A) Total charge passed $Q = I \times t = 10 \, A \times 193 \times 60 \, s = 115800 \, C$.
Effective charge considering $80 \%$ efficiency $= 115800 \times 0.8 = 92640 \, C$.
In terms of Faraday: $Q = \frac{92640}{96500} \approx 0.96 \, F$.
Cathode reaction: $Cu^{2+} + 2e^- \rightarrow Cu_{(s)}$.
Anode reaction: $H_2O \rightarrow \frac{1}{2} O_2 + 2H^+ + 2e^-$.
Weight of $Cu$ deposited $= \frac{0.96 \, F}{2 \, F} \times 63.5 \, g \approx 30.48 \, g$.
Moles of $O_2$ produced $= \frac{0.96 \, F}{4 \, F} = 0.24 \, \text{mol}$.
Using $PV = nRT$,$V = \frac{nRT}{P} = \frac{0.24 \times 0.08314 \times 300}{1} \approx 5.986 \, L$.
Effective charge considering $80 \%$ efficiency $= 115800 \times 0.8 = 92640 \, C$.
In terms of Faraday: $Q = \frac{92640}{96500} \approx 0.96 \, F$.
Cathode reaction: $Cu^{2+} + 2e^- \rightarrow Cu_{(s)}$.
Anode reaction: $H_2O \rightarrow \frac{1}{2} O_2 + 2H^+ + 2e^-$.
Weight of $Cu$ deposited $= \frac{0.96 \, F}{2 \, F} \times 63.5 \, g \approx 30.48 \, g$.
Moles of $O_2$ produced $= \frac{0.96 \, F}{4 \, F} = 0.24 \, \text{mol}$.
Using $PV = nRT$,$V = \frac{nRT}{P} = \frac{0.24 \times 0.08314 \times 300}{1} \approx 5.986 \, L$.
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