Question

The pressure of $H_2$ required to make the potential of $H_2$-electrode zero in pure water at 298 K is

• A. $10^{-12} $ atm
• B. $10^{-10} $ atm
• C. $10^{-4} $ atm
• D. $10^{-14} $ atm

Answer 1

The Correct Option is D

Electrode Potential Calculation for Hydrogen 

The given reaction for the hydrogen electrode is:

\(2H^{+} + 2e^{-} \rightarrow H_2(g)\)

The Nernst equation for this half-reaction is:

\(E_{H^{+}/H_2} = - \frac{0.0591}{2} \log \frac{P_{H_2}}{[H^{+}]^2}\)

If the logarithmic term equals zero, we get:

\(\log \frac{P_{H_2}}{[H^{+}]^2} = 0\)

This leads to:

\(\frac{P_{H_2}}{[H^{+}]^2} = 10^0 = 1\)

So, we have:

\(P_{H_2} = [H^{+}]^2\)

For Pure Water:

In pure water, the concentration of \( H^{+} \) is \( 10^{-7} \, \text{M} \). Therefore:

\(P_{H_2} = (10^{-7})^2 = 10^{-14} \, \text{atm}\)

Conclusion:

The partial pressure of hydrogen gas (\( P_{H_2} \)) in pure water is \( 10^{-14} \, \text{atm} \).