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Question

What will be the E0 value for the given half cell reaction?
Fe3+(aq)+3eFe(s)
Given:
E0 value for the half cell reaction,
Fe3+(aq)+eFe2+(aq); E0=0.77 V
Fe2+(aq)+2eFe(s); E0=0.44 V

A
0.33 V
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B
1.21 V
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C
0.036 V
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D
0.605 V
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Solution

The correct option is C 0.036 V
ΔG0=nFE0

Fe3+(aq)+eFe2+(aq)....(1); ΔG01=1F×0.77
Fe2+(aq)+2eFe(s)......(2); ΔG02=+2F×0.44

Adding equation (1) and (2), we get,
Fe3+(aq)+3eFe(s).......(3); ΔG03=3FE0

Thus,
ΔG03=ΔG01+ΔG02
3FE0=1F×0.77+2F×0.44
3E0=1×0.77+2×0.44
E0=0.770.883
E0=0.036 V




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