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Faraday's Second Law

Passage of on...

Question

Passage of one ampere current through $0.1 M$ $N i (N O_{3})_{2}$ solution using $N i$ electrodes brings in the concentration of solution to ________ in $60$ seconds.

0.1 M

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0.015 M

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0.2 M

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0.025 M

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Solution

The correct option is B $0.1 M$
Cell reaction:
$N i^{2 +} + 2 e^{-} \to N i$
for depositing 1 mol $N i : 2 \times 96500 C$ charge required
Charge passed, $Q = I t = 1 \times 60 C$

Mass of Nickel deposited, $w = Z I t = \frac{E}{96500} \times I \times t$
$= \frac{(58.69 / 2)}{96500} \times 1 \times 60 = 0.0182 g$ [ $∵ E = \frac{A t . w t .}{x - f a c t o r}$ ]
number of moles deposited $= 0.0182 / 58.69 = 3.1 \times 10^{- 4} m o l$

moles of $N i^{2 +}$ remaining in 1 L solution: $0.1 - 3.1 \times 10^{- 4} = 0.09969 m o l$

Therefore molarity $\approx 0.1 M$

Option A is correct.

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