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Question

Given the half-cell reactions
i) Fe2+(aq)+2eFe(s); E0=0.44 V
ii) 2H+(aq)+12O2(g)+2eH2O(l); E0=+1.23 V
E0 for the reaction
Fe(s)+2H+(aq)+12O2(g)Fe2+(aq)+H2O(l) is

A
1.67 V
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B
+1.67 V
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C
0.77 V
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D
+0.77 V
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Solution

The correct option is B +1.67 V
Subtracting the first reaction from second, we get the third equation and using the free-energy concept, we have
ΔG02ΔG01=ΔG03 (n= no. of electrons involved)
n2E02F(n1E01F)=n3E03F
E03=n2E02n1E01n3
E03=2×1.232×(0.44)2
E03=+1.67 V

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