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Question

For the following sequential reaction Eo values in volts at 298 K are given:
MnO40.564 VMnO242.26 VMnO20.95 VMn+31.51 VMn+2
Calculate Eo for MnO4Mn+2.

A
-1.81 volt
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B
-1.18 volt
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C
1.51 volt
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D
1.18 volt
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Solution

The correct option is C 1.51 volt
Given:
ΔG=nFEcell
where n No. of electrons
F Faraday's constant
E Reduction potential
ΔG Gibbs free energy
Gibb's free energy is additive in nature,so
ΔGMnO4Mn2+=ΔGMnO4MnO24+ΔGMnO24MnO2+ΔGMnO2Mn3++ΔGMn3+Mn2+5FEcell=1×F×0.5642×F×2.261×F×0.951×F×1.515Ecell=0.5644.520.951.51Ecell=7.5445Ecell=1.51V

1120222_877570_ans_90caefdf2bee44fba1ac38be75e1ecb2.png

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