Question

The volume of a solution of KCl in an electrolytic cell (1) after electrolyzing it for a certain time was found to be 0.6 L. The molarity of solution was calculated to be 1.0 M. During the same time 32.7 g of Zn was deposited in another electrolytic cell(II) which was joined in series with it. The current efficiency in cell (II) is 100%. The current efficiency (in %) in cell (I) is __________.

Answer 1

Equivalent of Zn deposited = $\frac{32.7}{63.5/2}=1Eq$
Number of Faradays produced = Number of Faradays consumed
= 1 Eq
= $0.6L\times 1M1\left(nfactor\right)$
= 0.6 F.
$\Rightarrow$ Current efficiency $=\frac{0.6}{1}\times 100=60\%$