The pressure of H2 required to make the potential of H2 electrode zero in pure water at 298K is (Given E°cell = 0.0 V)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(10^{-14} atm\)
For the hydrogen electrode the reduction half-reaction and its Nernst equation at 298 K are:
\( \ce{2H+(aq) + 2e- -> H2(g)} \)
\( \displaystyle E = E^\circ - \frac{0.059}{2}\log\frac{p_{H_2}}{[H^+]^2} \)
E° of the hydrogen electrode is 0 and the potential is to be zero, so
\( \displaystyle 0 = 0 - \frac{0.059}{2}\log\frac{p_{H_2}}{[H^+]^2} \;\Rightarrow\; p_{H_2} = [H^+]^2 \)
In pure water at 298 K, \( \ce{[H^{+}]} \) = 10−7 mol L−1:
\( p_{H_2} = (10^{-7})^2 = 10^{-14}\ \text{atm} \)