First slide

Nernst Equation

Question

If the hydrogen electrode is dipped in a solution of pH = 2 at 25° C, then the reduction potential of the electrode would be

Easy

A

0.118 V
B

0.084 V
C

0.059 V
D

-0.118 V

Solution

$H^{+} + e^{-} \to \frac{1}{2} H_{2} (g) ∴ E_{H^{+} / H_{2}} = {E^{0}}_{H^{+} / H_{2}} - \frac{0 .0591}{1} \log \frac{1}{[H^{+}]} = 0 - \frac{0 .0591}{1} \log [10^{- 2}]$
= – 0.0591 × 2 = – 0.118V

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