What is reduction potential of hydrogen gas electrode when pure hydrogen gas is at 1 atmospheric pressure and platinum electrode is in contact with HCl solution of PH 1 at 298 K ?

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The reduction half-reaction for the hydrogen electrode is:

$H^{+} + e^{-} \to \frac{1}{2} H_{2} (g)$

The Nernst equation for the reduction potential is given by:

$E = E^{\circ} - \frac{0.0592}{n} lo g_{10} \frac{P _{H_{2}}^{1/2}}{[ H ^{+} ]}$

For the standard hydrogen electrode, $E^{\circ} = 0$ V. Here, $n = 1$ , $P_{H_{2}} = 1$ atm, and $p H = 1$ .

Since $p H = - lo g_{10} [H^{+}]$ , we have $[H^{+}] = 1 0^{- 1}$ M.

Substituting the values into the Nernst equation:

$E = 0 - \frac{0.0592}{1} lo g_{10} \frac{1 ^{1/2}}{1 0 ^{- 1}}$

$E = - 0.0592 lo g_{10} (1 0^{1})$

$E = - 0.0592 \times 1 = - 0.0592$ V

Answer: -0.0592 V

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