The standard oxidation potential of $N i / N i^{2 +} (N i^{2 +} = 1 M)$ electrode is $0.236 V$ . If this is combined with a hydrogen electrode $(P_{H_{2}} = 1 a t m)$ in acid solution, the $p H$ at which the measured e.m.f. of the solution will be zero at $25$ $^{o}$ C is:

Open in App

Solution

The standard reduction potential of hydrogen electrode will be 0.236. This will make the cell potential zero.
$E_{e l} = E_{e l}^{0} - \frac{0.0592}{n} l o g \frac{P_{H_{2}}}{[H^{+}]^{2}}$
$0.236 = 0 - \frac{0.0592}{2} l o g \frac{1}{[H^{+}]^{2}}$
$l o g \frac{1}{[H^{+}]^{2}} = 7.9729$
$[H^{+}]^{2} = 1.06 \times 10^{- 8}$
$[H^{+}] = 1.03 \times 10^{- 4}$
$p H = - l o g [H^{+}] = - l o g 1.03 \times 10^{- 4} = 4$

Suggest Corrections

Similar questions

The standard oxidation potential of $N i / N i^{2 +}$ electrode is 0.236 V. If this is combined with a hydrogen electrode in acid solution, at what pH of the solution will the measured emf be zero at $25^{0} C$ ? assume $[N i^{2 +}] = 1 M$ and $P_{H 2} = 1 a t m$