Question

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

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

Answer 1

Answer: (B)

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