Given half-cell reaction

The reaction written is: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$

The standard reduction potential given is for the reverse reaction: $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O, \qquad E^\circ_{\text{red}} = 1.23\,\text{V}$

Use the Nernst equation for the reduction reaction

For $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$

the reaction quotient is $Q = \frac{1}{P_{O_2}[H^+]^4}$ since activity of liquid water is 1.

Nernst equation: $E = E^\circ - \frac{RT}{nF}\ln Q$

Substituting $Q$: $E = E^\circ - \frac{RT}{4F}\ln\left(\frac{1}{P_{O_2}[H^+]^4}\right)$ $E = E^\circ + \frac{RT}{4F}\ln\left(P_{O_2}[H^+]^4\right)$

Insert given conditions

• $P_{O_2} = 1\,\text{bar}$, so $\ln P_{O_2} = 0$
• $\text{pH} = 5 \Rightarrow [H^+] = 10^{-5}$

Therefore, $E = 1.23 + \frac{RT}{4F}\ln\left((10^{-5})^4\right)$ $E = 1.23 + \frac{RT}{4F}\ln(10^{-20})$

Simplify

$\ln(10^{-20}) = -20\ln 10$

So, $E = 1.23 - \frac{RT}{4F}(20\ln 10)$ $E = 1.23 - \frac{5RT}{F}\ln 10$

At $298\,\text{K}$, $\frac{RT}{F}\ln 10 \approx 0.0591\,\text{V}$

Hence, $E = 1.23 - 5(0.0591)$ $E = 1.23 - 0.2955$ $E \approx 0.9345\,\text{V}$

But the reaction asked is oxidation

The given half-cell is written as: $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$

Electrode potential for oxidation is the negative of the reduction potential: $E_{\text{ox}} = -0.9345\,\text{V}$

However, in electrochemistry questions, when they ask the electrode potential of the oxygen electrode, they usually report the reduction electrode potential magnitude for the corresponding half-cell, which is: $E \approx 0.935\,\text{V}$

Comparison with stored answer

The stored correct answer is $1.52$ to $1.53$, which does not match the correct Nernst calculation at $\text{pH}=5$.

A value near $1.52\,\text{V}$ would arise only if one incorrectly used the oxidation expression and added the pH term to $1.23$.

The correct reduction potential at pH 5 is: $0.93\,\text{V} \text{ (approximately)}$

If oxidation potential is asked strictly for the reaction as written, then: $-0.93\,\text{V}$