Question

Electrolysis of $NaCl$ solution with inert electrodes for a certain period of time gave 600 $c{m}^3$ of 1.0 M $NaOH$ in the electrolytic cell. During the same period, 31.80 g of copper was deposited in a copper voltameter in series with the electrolytic cell. What is the percent current efficiency in the electrolytic cell? (At. Wt. of Cu = 63.6)___

Answer 1

Since both processes happen in series, the same current flows through. Using this idea, we can get the number of equivalents of copper desposited by that current, compare it with the amount of $NaOH$ produced to get the efficiency.
$NaC{l}_{\left(aq\right)}$ (cathode)
$2H_{2}O+2e^{-}\to H_{2\left(g\right)}+2O{H}^{-}\left(aq\right)$
$CuS{O}_{4\left(aq\right)}$ (cathode):$C{u}_{\left(aq\right)}^{2+}+2e^{-}\to C{u}_{\left(s\right)}$
Equivalent of $O{H}^{-}$ = mole of $O{H}^{-}$ formed
$=\frac{600\times 1}{1000}=0.6$
Equivalent of $Cu$ deposited = $\frac{31.8}{\frac{63.6}{2}}=1.0$
Current efficiency $=\frac{0.6\times 100}{1}$% =60%