NVAMediumJEE Main 2023 · 1 February, Shift 1Nernst Equation

Chemistry Question from JEE Main 2023 · 1 February, Shift 1

At what pH, given half cell MnO4−\mathrm{MnO_4^-} (0.1 M0.1 \, \text{M}) | Mn2+\mathrm{Mn^{2+}} (0.001 M0.001 \, \text{M}) will have electrode potential of 1.282 V1.282 \, \text{V}? ____\ (Nearest Integer)

Given EMnO4−/Mn2+∘=1.54 VE^\circ_{\mathrm{MnO_4^-}/\mathrm{Mn^{2+}}} = 1.54 \, \text{V}, 2.303RTF=0.059V\frac{2.303RT}{F} = 0.059\text{V}

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