Eo values for the half cell reactions are given: Cu2++e−→Cu+; Eo=0.15V Cu2++2e−→Cu; Eo=0.34V What will be the Eo of the half-cell: Cu++e−→Cu?
+0.30V
+0.49V
+0.19V
+0.53V
A
+0.53V
B
+0.49V
C
+0.19V
D
+0.30V
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Solution
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Given,
Cu2++e−→Cu+;E01=0.15V,ΔG01,n1=1
Cu2++2e−→Cu;E02=0.34V,ΔG02,n2=2
Cu++e−→Cu;E03=?V,ΔG03,n3=1
As ΔG is an extensive property,
ΔG03=ΔG02−ΔG01
⟹−n3FE03=−n2FE02+n1FE01
−E03=−2×0.34+1×0.15
E03=0.68−0.15=0.53V
Hence, option C is correct.
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