The reaction which occur in the Galvanic cell is $$MnO4$$−$$+8H++5Fe2+$$⟶$$Mn2++5Fe3++4H2OE$$∘$$MnO4$$−,$$Mn2+$$,$$H+$$∣$$Pt=1.51V$$ and E∘$$Fe3+$$,$$Fe2+$$∣$$Pt=0.77V(Q)$$ How would the emf of the cell be increased above the standard emf ?

Correct option is 1By decreasing pH

Questions

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) How would the emf of the cell be increased above the standard emf ?

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By decreasing pH

By increasing Fe3+

By decreasing MnO4-

By increasing Mn2+

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detailed solution

Correct option is A

E=E∘−RT5Fln⁡Mn2+Fe3+5MnO4−H+8Fe2+5Lowering pH means increasing H+and it will make term less negative and Ecellbecomes positive.

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Similar Questions

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.