How much will the potential of a hydrogen electrode change when its solution initially at pH = 0 is neutralized to pH = 7?

decrease by 0.413 V

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,

\(E = E^0 − \frac{0.0591}{1}logQ\)

\(E = 0.0 − \frac{0.0591}{1}log\frac{H_2^{\frac{1}{2}}}{[H^+]}\)

\(E = −\frac{0.0591}{1}log\frac{1}{10^{−7}}\)

\(E = −0.0591 × 7 × log10\)

\(E = − 0.413V\)