Question

A current of $965$ A is passed for $1s$ through $1L$ solution of $0.02N$ $NiS{O}_4$ using Ni electrodes. What is the new concentration of $NiS{O}_4$?

• A. $0.01N$
• B. $0.01M$
• C. $0.002M$
• D. $0.02M$

Answer 1

The correct option is B $0.01M$

$\frac{W}{E_w}=\frac{965\times 1}{96500}=0.01$
Equivalent of $NiS{O}_4$ present initially = $1\times 0.02=0.02$. If $Ni$ electrodes are used no change in concentration of $NiS{O}_4$, i.e., $0.02N$ or $0.01M$.
If $Pt$ electrodes are used, then equivalent of $NiS{O}_4$ left $=0.01$