Given the half-cell reactions
i) Fe2+(aq)+2e−→Fe(s);E0=−0.44V
ii) 2H+(aq)+12O2(g)+2e−→H2O(l);E0=+1.23V E0 for the reaction Fe(s)+2H+(aq)+12O2(g)→Fe2+(aq)+H2O(l) is
A
−1.67V
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B
+1.67V
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C
−0.77V
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D
+0.77V
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Solution
The correct option is B+1.67V Subtracting the first reaction from second, we get the third equation and using the free-energy concept, we have ΔG02−ΔG01=ΔG03 (n= no. of electrons involved) −n2E02F−(−n1E01F)=−n3E03F E03=n2E02−n1E01n3 E03=2×1.23−2×(−0.44)2 E03=+1.67V