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NEET CHEMISTRYMedium

The pressure of H2\text{H}_2 required to make the potential of H2\text{H}_2-electrode zero in pure water at 298 K298 \text{ K} is

A

10−12 atm10^{-12} \text{ atm}

B

10−10 atm10^{-10} \text{ atm}

C

10−4 atm10^{-4} \text{ atm}

D

10−14 atm10^{-14} \text{ atm}

Step-by-Step Solution

For a hydrogen electrode, the reduction half-reaction is: 2H+(aq)+2e−→H2(g)2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g) According to the Nernst equation at 298 K298 \text{ K}: E=E∘−0.05912log⁡PH2[H+]2E = E^\circ - \frac{0.0591}{2} \log \frac{P_{\text{H}_2}}{[\text{H}^+]^2} We are given that the potential E=0E = 0, and the standard potential of the hydrogen electrode E∘=0E^\circ = 0. In pure water at 298 K298 \text{ K}, the concentration of hydrogen ions is [H+]=10−7 M[\text{H}^+] = 10^{-7} \text{ M}. Substituting these values into the Nernst equation: 0=0−0.05912log⁡PH2(10−7)20 = 0 - \frac{0.0591}{2} \log \frac{P_{\text{H}_2}}{(10^{-7})^2} 0=log⁡PH210−140 = \log \frac{P_{\text{H}_2}}{10^{-14}} PH210−14=100=1\frac{P_{\text{H}_2}}{10^{-14}} = 10^0 = 1 PH2=10−14 atmP_{\text{H}_2} = 10^{-14} \text{ atm}

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