Question

A current of 19.3 mA is passed for 9375 seconds through 500 ml of 2 mM $ZnS{O}_4$ solution. If the final molarity of $Z{n}^{2+}$ is 0.5 mM, then the current efficiency of the source is (assume volume remains constant during this process) :

• A. 80%
• B. 75%
• C. 72%
• D. 65%

Answer 1

The correct option is A 80%

A current of 19.3 mA is passed for 9375 sec.

Let the current efficiency be x %.

Charge passed $=\frac{x}{100}\times \frac{19.3}{1000}\times 9375$

Moles of $Z{n}^{2+}$ initially $=\frac{1}{2}\times \frac{2}{1000}={10}^{-3}$ mol

Moles of $Z{n}^{2+}$ finally $=\frac{1}{2}\times \frac{1}{2000}=0.25\times {10}^{-3}$ mol

Moles of Zn deposited $=0.75\times {10}^{-3}=\frac{3}{4}\times {10}^{-3}$ mol

Moles of $e^{-}$ involved $=\frac{3}{4}\times {10}^{-3}\times 2$

$\Rightarrow \frac{3}{4}\times {10}^{-3}\times 2\times 96500=\frac{x}{100}\times \frac{19.3}{1000}\times 9375.$

$\Rightarrow x=80$.