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Question

E° value for various ions are as follows :

E°(MnO4-) = +1.51 V

E°(Ag/Ag+) = +0.7996 V

E°(Au/Au+) = +1.692 V

E°(Zn/Zn2+) = -0.761 V

Based on this data, permaganate can be used to oxidize :

This question was previously asked in
MH SET 2025 Paper 2 Hindi Question Paper (15-Jun-2025)
The correct answer is

Zn, Ag

An oxidising agent can oxidise a metal only if the overall cell potential is positive. Working with reduction potentials, permanganate will oxidise a metal whenever

\(E^{\circ}_{\mathrm{MnO_4^-}} > E^{\circ}_{\mathrm{M^{n+}/M}}\).

Permanganate sits at +1.51 V, so compare each metal against that figure.

Zinc, -0.761 V. Far below +1.51 V, so the difference is large and positive: \(1.51 - (-0.761) = +2.27\) V. Zinc is easily oxidised — it is a reactive, electropositive metal that gives up electrons readily.

Silver, +0.7996 V. Still below +1.51 V, giving \(1.51 - 0.80 = +0.71\) V. So silver is oxidised, though less vigorously than zinc.

Gold, +1.692 V. This lies above permanganate, so the cell potential would be \(1.51 - 1.692 = -0.18\) V — negative, and the reaction is not spontaneous. Gold is not oxidised by permanganate.

That result matches chemical experience: gold's nobility is precisely its resistance to oxidation, and dissolving it requires aqua regia, where the real work is done not by a stronger oxidant alone but by chloride complexation — forming [AuCl4]- pulls the effective potential down far enough to make the reaction go.

So permanganate oxidises zinc and silver but not gold, which rules out every option containing Au.

Hence permanganate can be used to oxidize Zn and Ag.

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