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Question

What will be the E0 value for the given half cell reaction?
Sn4+(aq)+4eSn(s)
Given:
E0 value for the half cell reaction,
Sn4+(aq)+2eSn2+(aq); E0=0.15 V
Sn2+(aq)+2eSn(s); E0=0.14 V

A
0.001 V
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B
0.005 V
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C
+0.001 V
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D
+0.005 V
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Solution

The correct option is D +0.005 V
ΔG0=nFE0

Sn4+(aq)+2eSn2+(aq)....(1); ΔG01=2F×0.15
Sn2+(aq)+2eSn(s)......(2); ΔG02=+2F×0.14

Adding equation (1) and (2), we get,
Sn4+(aq)+4eSn(s).......(3); ΔG03=4FE0

Thus,
ΔG03=ΔG01+ΔG02
4FE0=2F×0.15+2F×0.14
4E0=2×0.15+2×0.14
E0=+0.005 V




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