Question

The reduction potential of hydrogen electrode ($P_{H_2}$ = 1 atms; [H+] = 0.1 M) at 25°C will be -

• A. 0.00 V
• B. -0.059 V
• C. 0.118 V
• D. 0.059 V

Answer 1

Answer: (B)

-0.059 V

Answer 2

The reduction potential is calculated using the Nernst equation: $E = E^0 - \frac{0.059}{n} \log Q$. For the hydrogen electrode, the half-reaction is $H^+_{(aq)} + e^- \rightarrow \frac{1}{2}H_{2(g)}$. The standard reduction potential $E^0$ is 0 V. The reaction quotient $Q = \frac{P_{H_2}^{1/2}}{[H^+]}$. Given $P_{H_2} = 1$ atm and $[H^+] = 0.1$ M, and $n=1$. $E = 0 - \frac{0.059}{1} \log \frac{(1)^{1/2}}{0.1} = -0.059 \log \frac{1}{0.1} = -0.059 \log 10 = -0.059$ V.