The reaction which occur in the Galvanic cell is $$MnO4$$−$$+8H++5Fe2+$$⟶$$Mn2++5Fe3++4H2OE$$∘$$MnO4$$−,$$Mn2+$$,$$H+$$∣$$Pt=1.51V$$ and E∘$$Fe3+$$,$$Fe2+$$∣$$Pt=0.77V(Q)$$. The other concentrations are kept at unity except reducing $$Mn2+to$$ $$0.50$$ M, the emf of the cell is increased by

Correct option is 23.55×10−3V

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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). The other concentrations are kept at unity except reducing $[{Mn}^{2 +}]$ to 0.50 M, the emf of the cell is increased by

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0.012 V

3.55×10−3V

0.059 V

0.018 V

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detailed solution

Correct option is B

ΔE=−0.0595log⁡(0.5)=+3.55×10−3V

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	Column I		Column II
A.	${Fe}^{3 +} and I^{-}$	p.	$E_{cell}^{\circ} = + ve$
B.	${Fe}^{3 +} and {Br}^{-}$	q.	${ΔG}^{\circ} = + ve$
C.	$Zn and H^{+}$	r.	$E_{cell}^{\circ} = - E_{electrode}^{\circ}$
D.	$Br and {Fe}^{2 +}$	s.	${ΔG}^{\circ} = - ve$

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). The other concentrations are kept at unity except reducing Mn2+to 0.50 M, the emf of the cell is increased by

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