Question

A current of 0.1A was passed for 4hr through a solution of cuprocyanide and 0.3745 g of copper was deposited on the cathode. Calculate the current efficiency for the copper deposition:

[Cu: 63.5]

• A. 79%
• B. 39.5%
• C. 63.25%
• D. 63.5%

Answer 1

The correct option is B 39.5%

Cuprocyanide is $CuCN$

Reduction reaction:

$C{u}^{+}+e^{-}\to Cu$

Since $1F$ charge is required to get $63.5gCu$

amount of $Cu$ obtained from $1C=\frac{63.5}{96500}g$

charge supplied:$Q=It$

$=0.1\times 4\times 3600C$

maximum mass from charge supplied should be:

$=\frac{63.5}{96500}\times (0.1\times 4\times 3600)g$

$=0.9475g$

The mass obtained is $0.3745g$

percent current efficiency $\eta =\frac{\text{mass obtained}}{\text{maximum mass}}\times 100$

$\eta =\frac{0.3745}{0.9475}\times 100=39.5$%

Hence, the correct option is $\text{B}$