Question

the reduction potential of H-electrode in pure water at 298K and 1atm pressure
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Answer 1

Answer:

Reduction potential of H-electrode is -0.4137eV in pure water.

Explanation:

Cell reaction is:    2H⁺ (aq.) + 2e⁻ → H₂(g)

The concentration of [H⁺] in pure water = 10⁻⁷

Pressure of H₂ gas takes 1atm, T= 298K

Standard reduction potential of H-electrode E⁰ = 0 Volt

Nernst's equation :-

                    $E=E^{o} - \frac{0.0591}{n} log\frac{P_H_{2}}{[H^{+} ]^{2} }$

                      $E=0 - \frac{0.0591}{2} log\frac{{1}}{[10^{-7} ]^{2} }$

                     $E=(-0.02955)log10^{14} =14(-0.02955)log10$       [ ∵㏒10=1]

         ∴           $E=-0.4137 Volt$