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

We are given the half-cell $$MnO_4^-(0.1 \text{ M}) \mid Mn^{2+}(0.001 \text{ M})$$ with electrode potential $$E = 1.282$$ V, and we need to find the pH.

We start by writing the half-cell reaction: $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

Next, we apply the Nernst equation: $$E = E^\circ - \frac{0.059}{n} \log \frac{[Mn^{2+}]}{[MnO_4^-][H^+]^8}$$ Here $$n = 5$$, $$E^\circ = 1.54$$ V, $$[MnO_4^-] = 0.1$$ M, and $$[Mn^{2+}] = 0.001$$ M.

Substituting the values gives $$1.282 = 1.54 - \frac{0.059}{5} \log \frac{0.001}{0.1 \times [H^+]^8}$$ which simplifies to $$1.282 - 1.54 = -0.0118 \log \frac{10^{-3}}{10^{-1} \times [H^+]^8}$$ and then $$-0.258 = -0.0118 \log \frac{10^{-2}}{[H^+]^8}.$$

This leads to $$\frac{0.258}{0.0118} = \log \frac{10^{-2}}{[H^+]^8}$$ and hence $$21.86 = \log(10^{-2}) - \log([H^+]^8) = -2 - 8\log[H^+].$$

Since $$\text{pH} = -\log[H^+]$$, we have $$21.86 = -2 + 8 \times \text{pH}$$, so $$8 \times \text{pH} = 23.86$$ and therefore $$\text{pH} = \frac{23.86}{8} \approx 2.98 \approx 3.$$

The pH of the solution is 3.