The reaction which occur in the Galvanic cell is
$\begin{matrix} {MnO}_{4}^{-} + 8 H^{+} + 5 {Fe}^{2 +} ⟶ {Mn}^{2 +} + 5 {Fe}^{3 +} + 4 H_{2} O \\ E^{\circ} ({MnO}_{4}^{-}, {Mn}^{2 +}, H^{+} ∣ Pt) = 1 .51 V and \\ E^{\circ} ({Fe}^{3 +}, {Fe}^{2 +} ∣ Pt) = 0 .77 V \end{matrix}$
(Q). If $[H_{3} O^{+}] = 2 .0 M, [{Fe}^{3 +}] = 2 .0 M, [{Fe}^{2 +}] = 0 .5 M$ $[{Mn}^{2 +}] = 0 .5 M, [{MnO}_{4}^{-}] = 1 .0 M$ Calculate the emf of the cell.

Question

Infinity Learn · Accepted Answer

Ecell $$=0.74V$$−$$RT5Fln$$⁡$$(0.5)(2.0)5(1.0)(2.0)8(0.5)5$$         $$=0.74V$$−$$0.0595log$$⁡$$1(2.0)3(0.5)4$$         $$=0.74V$$−$$0.0595log$$⁡$$2=0.74V$$−$$0.059$$×$$0.3015=0.736V$$

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The reaction which occur in the Galvanic cell is MnO4−+8H++5Fe2+⟶Mn2++5Fe3++4H2OE∘MnO4−,Mn2+,H+∣Pt=1.51V and E∘Fe3+,Fe2+∣Pt=0.77V(Q). If H3O+=2.0M,Fe3+=2.0M,Fe2+=0.5M Mn2+=0.5M,MnO4−=1.0MCalculate the emf of the cell.

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