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Question

Given standard electrode potentials
Fe+++2eFe; E=0.440 V
Fe++++3eFe; E=0.036 V
The standard electrode potential (E) for Fe++++eFe++ is


A

- 0.476 V

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B

- 0.404 V

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C

+ 0.404 V

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D

+ 0.772 V

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Solution

The correct option is D

+ 0.772 V


ΔG=nEF
Fe2++2eFe .......(i)
ΔG=2×F×(0.440 V)=0.880 F
Fe3++3eFe ......(ii)
ΔG=3×F×(0.036)=0.108 F
On substracting equation (i) from (ii)
Fe3++eFe2+
ΔG=0.108 F0.880 F= 0.772F
E for the reaction =ΔGnF=(0.772 F)1×F=+0.772V


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