Reduction potential for the following half-cell reactions are {\Zn}={\Zn}^{2+}+2 e^{-} ; \E^\circ=+0.76 {\V} {\Fe}={\Fe}^{2+}+2 e^{-} ; \E^\circ=+0.44 {\V} The EMF for the cell reaction {\Fe}^{2+}+{\Zn} \to{\Zn}^{2+}+{\Fe} will be (1996)

Class 12 NEET Chemistry Electrochemistry Questions

Question : 70 of 74

Marks: +1, -0

Reduction potential for the following half-cell reactions are
Zn=Zn2++2e−;E°=+0.76V
Fe=Fe2++2e−;E°=+0.44V
The EMF for the cell reaction
Fe2++Zn→Zn2++Fe will be

(1996)

−0.32V
+1.20V
−1.20V
+0.32V

Validate

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