What would be the electrode potential for the given half-cell reaction at pH = 5?
2H$$_2$$O $$\rightarrow$$ O$$_2$$ + 4H$$^+$$ + 4e$$^-$$; E$$_{red}^0 = 1.23$$ V
(R = 8.314 J mol$$^{-1}$$ K$$^{-1}$$; Temp = 298K; oxygen under standard atm. pressure of 1 bar)

We are asked to calculate the electrode potential of the half-cell

$$2\,\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$$

at a temperature of $$T = 298\ \text{K}$$ and at $$\text{pH}=5$$ when oxygen is at its standard pressure of $$1\ \text{bar}$$.

The standard reduction potential corresponding to the reverse (reduction) reaction

$$\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\,\text{H}_2\text{O}$$

is given as $$E_{\text{red}}^0 = 1.23\ \text{V}$$.

To find the electrode potential under non-standard conditions we employ the Nernst equation. First, we state the equation in its base-10 logarithmic form (valid at 298 K):

$$E = E^0 - \frac{0.0591}{n}\,\log_{10}Q,$$

where

• $$E$$ is the electrode potential we seek,
• $$E^0$$ is the standard electrode potential (here $$1.23\ \text{V}$$),
• $$n$$ is the number of electrons exchanged (here $$n = 4$$),
• $$Q$$ is the reaction quotient for the reduction reaction.

For the reduction reaction $$\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O},$$ the expression for $$Q$$ is

$$Q = \frac{a_{\text{H}_2\text{O}}^2}{a_{\text{O}_2}\,[\text{H}^+]^4}.$$

Pure liquid water has unit activity, so $$a_{\text{H}_2\text{O}} = 1$$. Oxygen is present at its standard pressure, so $$a_{\text{O}_2}=1$$ as well. Therefore

$$Q = \frac{1}{1\,[\text{H}^+]^4}=\frac{1}{[\text{H}^+]^4}=([\text{H}^+])^{-4}.$$

At $$\text{pH}=5$$ we have

$$[\text{H}^+] = 10^{-5}\ \text{M}.$$

Substituting this value,

$$Q = (10^{-5})^{-4} = 10^{\,20}.$$

Now we need $$\log_{10}Q$$:

$$\log_{10}Q = \log_{10}\!\left(10^{20}\right) = 20.$$

We substitute all known quantities into the Nernst equation:

$$E = 1.23\ \text{V} - \frac{0.0591}{4}\,(20).$$

First compute the coefficient:

$$\frac{0.0591}{4} = 0.014775.$$

Next multiply by 20:

$$0.014775 \times 20 = 0.2955.$$

Finally subtract from the standard potential:

$$E = 1.23\ \text{V} - 0.2955\ \text{V} = 0.9345\ \text{V}.$$

Rounding to two significant figures in the hundredths place gives

$$E \approx 0.93\ \text{V}.$$

Hence, the correct answer is Option C.