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Question

The standard oxidation potentials, E, for the half reactions are as,
ZnZn2++2e;E=+0.76 volt
FeFe2++2e;E=+0.41 volt
The emf of the cell, Fe2++ZnZn2++Fe is:

A
+0.35 volt
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B
0.35 volt
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C
+1.17 volt
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D
1.17 volt
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Solution

The correct option is A +0.35 volt
Ecell or EMF= EocathodeEoanode
since, oxidation potentials are given so for converting them into reduction potential just reverse the sign.
EoZn2+/Zn=0.762
EoFe2+/Fe=0.41
EMF=EocathodeEoanode
EMF=EoFe/Fe2+EoZn/Zn+2
=0.41(0.762)
=0.41+0.76
EMF=+0.35V

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