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)

To calculate the electrode potential at pH = 5, we can use the Nernst equation. The Nernst equation is given by:

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

Where:

• $$E^{\circ} = 1.23\text{ V}$$ (standard reduction potential)
• $$R = 8.314\text{ J mol}^{-1}\text{ K}^{-1}$$ (universal gas constant)
• $$T = 298\text{ K}$$ (temperature)
• $$n = 4$$ (number of electrons transferred)
• $$F = 96485\text{ C mol}^{-1}$$ (Faraday's constant)
• $$Q$$ is the reaction quotient.

At pH = 5, the concentration of $$\text{H}^{+}$$ ions is:

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

The reaction quotient $$Q$$ for the reduction half-cell reaction is given by:

$$Q = \frac{1}{P_{\text{O}_2} \cdot [\text{H}^{+}]^4}$$

Assuming standard atmospheric pressure for $$\text{O}_2$$, $$P_{\text{O}_2} = 1\text{ bar}$$, we have:

$$Q = \frac{1}{1 \cdot (10^{-5})^4} = \frac{1}{10^{-20}} = 10^{20}$$

Now substituting into the Nernst equation:

$$E = -1.23 - \frac{(8.314)(298)}{(4)(96485)} \ln(10^{20})$$

Calculating the logarithm:

$$\ln(10^{20}) = 20 \ln(10) \approx 20 \times 2.303 = 46.06$$

Now substituting this value back into the equation:

$$E = -1.23 - \frac{(8.314)(298)}{(4)(96485)} (46.06)$$

Calculating the prefactor:

$$\frac{(8.314)(298)}{(4)(96485)} \approx 0.00641$$

Thus:

$$E = -1.23 + (0.00641 \times 46.06)$$

Calculating the final potential:

$$E \approx -1.23 + 0.0917 \approx -0.93\text{ V}$$

E = -$$E_{ox}$$ = $$\approx$$0.93

Therefore, the electrode potential at pH = 5 is approximately  0.93V.