At what pH, given half cell MnO$$_4^-$$(0.1M) | Mn$$^{2+}$$(0.001 M) will have electrode potential of 1.282 V? (Nearest Integer)
Given E$$^0_{MnO_4^-/Mn^{2+}}$$ = 1.54 V, $$\frac{2.303RT}{F}$$ = 0.059 V

We have $$MnO_4^-(0.1M) | Mn^{2+}(0.001M)$$, electrode potential = 1.282 V, $$E^0 = 1.54$$ V, and $$\frac{2.303RT}{F} = 0.059$$ V.

The half-cell reaction is:

$$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

Using the Nernst equation with $$n = 5$$:

$$E = E^0 - \frac{0.059}{n}\log\frac{[Mn^{2+}]}{[MnO_4^-][H^+]^8}$$

Substituting the values:

$$1.282 = 1.54 - \frac{0.059}{5}\log\frac{0.001}{0.1 \times [H^+]^8}$$

$$-0.258 = -0.0118\left(-2 + 8\,\text{pH}\right)$$

$$\frac{0.258}{0.0118} = -2 + 8\,\text{pH}$$

$$21.864 = -2 + 8\,\text{pH}$$

$$8\,\text{pH} = 23.864$$

$$\text{pH} = 2.98 \approx 3$$

Hence, the pH is approximately $$3$$.