How much will the potential of a hydrogen electrode change when its solution initially at pH = 0 is neutralized to pH = 7?
Answer & explanation
Correct answer: option 4
The correct answer is option 4. decrease by 0.413 V.
From the question, the reaction is
\(H^+ + e^−\) ⇌ \(\frac{1}{2}H_2(g)\)
We know, from Nertnst equation,
\(E = E^0 − \frac{0.0591}{1}logQ\)
\(E = 0.0 − \frac{0.0591}{1}log\frac{H_2^{\frac{1}{2}}}{[H^+]}\)
\(E = −\frac{0.0591}{1}log\frac{1}{10^{−7}}\)
\(E = −0.0591 × 7 × log10\)
\(E = − 0.413V\)