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Question

Given standard electrode potential Fe2++2eFe;E0=0.440V Fe3++3eFe;E0=0.036V. Calculate the electrode potential of cell.

A
+0.772V
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B
0.772V
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C
+0.569V
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D
0.569V
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Solution

The correct option is D +0.772V
ΔGo=nFEo
For the reaction,
Fe2++2eFe
ΔGo=2×F×(0.440V)=0.880F ....(I)
For the reaction,
Fe3++3eFe
ΔGo=3×F×(0.036)=0.108F ....(II)
On subtracting equation (I) from equation (II)
Fe3++eFe2+
ΔGo=0.108F0.880F=0.772F
Eo=ΔGonF=0.772F1×F
=+0.772V.

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