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Extraction of Copper

A current of ...

Question

A current of $965$ A is passed for $1 s$ through $1 L$ solution of $0.02 N$ $N i S O_{4}$ using Ni electrodes. What is the new concentration of $N i S O_{4}$ ?

0.01 N

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0.01 M

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0.002 M

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0.02 M

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Solution

The correct option is B $0.01 M$
$\frac{W}{E_{w}} = \frac{965 \times 1}{96500} = 0.01$
Equivalent of $N i S O_{4}$ present initially = $1 \times 0.02 = 0.02$ . If $N i$ electrodes are used no change in concentration of $N i S O_{4}$ , i.e., $0.02 N$ or $0.01 M$ .
If $P t$ electrodes are used, then equivalent of $N i S O_{4}$ left $= 0.01$

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