Question

100 ml $CuS{O}_4\left(aq\right)$ was electrolyzed using inert electrodes by passing 0.965 A till the pH of the resulting solution was 1. The solution after electrolysis was neutralized, treated with excess $KI$ and titrated with 0.04 M $N{a}_2{S}_2{O}_3$. Volume of $N{a}_2{S}_2{O}_3$ required was 35 ml. Assuming no volume change during electrolysis, calculate the duration of electrolysis (in sec) if current efficiency is 80%.

Answer 1

$I_2+N{a}_2{S}_2{O}_3\to I^{-}+N{a}_2{S}_4{O}_6$

moles of $N{a}_2{S}_2{O}_3=0.04\times 35\times {10}^{-3}$

moles of $I_2=\frac{0.04\times 35\times {10}^{-3}}{2}moles$

moles of $KI=2\times \frac{0.04\times 35\times {10}^{-3}}{2}moles$

$KI$ and $CuS{O}_4$ reaction can be written as:

$4KI+2CuS{O}_4\to 2CuI+I_2+2K_{2}S{O}_4$

Moles of $CuS{O}_4$ reacted with $KI=\frac{0.04\times 35\times {10}^{-3}}{2}=0.7\times {10}^{-3}moles$

Moles of $CuS{O}_4$ electrolyzed$=(6.4-0.7)\times {10}^{-3}=5.7\times {10}^{-3}moles$

$1F$ charge deposits $0.5$ moles of $Cu$

$5.7\times {10}^{-3}$ moles will be deposited by $\frac{1\times 5.7\times {10}^{-3}}{0.5}F$ charge

Considering $80\%$ efficiency:

Time taken$=\frac{2\times 5.7\times {10}^{-3}}{0.965\times {10}^{-5}}\times 1.25=1476seconds$