Question

The standard reduction potentials $E^{\circ}$, for the half reactions are as
$ Zn = Zn ^{2+}+2 e^{-}, E^{\circ}=+0.76 \,V $
$ Fe = Fe ^{2+}+2 e^{-}, E^{\circ}=0.41 \,V $
The emf for the cell reaction,
$Fe ^{2+}+ Zn \rightarrow Zn ^{2+}+ Fe$ is

• A. - 0.35 V
• B. + 0.35 V
• C. + 1.17 V
• D. - 1.17 V

Answer 1

Answer: (B)

$Fe ^{2+}+ Zn \rightarrow Zn ^{2+}+ Fe$
$Zn \rightarrow Zn ^{2+}+2 e ^{-} ; E ^{\ominus}=+0.76 V$
$Fe \rightarrow Fe ^{2+}+2 e ^{-} ; E ^{\ominus}=+0.41 V$
These are oxidation potentials.
Reduction potentials are equal and opposite.
Fe forms cathode and $Zn$ forms anode.
$ \begin{array}{l} E _{\text {cell }}^{\ominus}=\left( E _{\text {red }}^{\ominus}\right)_{ c }+\left( E _{\text {oxid }}^{\ominus}\right)_{ a } \\ =(-0.41+0.76) V \\ =0.35 V \end{array} $