The standard oxidation potentials, E∘, for the half reactions are as, Zn→Zn2++2e−;E∘=+0.76volt Fe→Fe2++2e−;E∘=+0.41volt The emf of the cell, Fe2++Zn→Zn2++Fe is:
A
+0.35volt
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B
−0.35volt
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C
+1.17volt
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D
−1.17volt
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Solution
The correct option is A+0.35volt Ecell or EMF= Eocathode−Eoanode
since, oxidation potentials are given so for converting them into reduction potential just reverse the sign.