Question

Calculate \(E_{\text {cell }}^{\circ}\) for a iron-cadmium cell in which the cell reaction is

$$ 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cd}^{2+}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Cd}(\mathrm{s}) $$

$$ \begin{array}{ll} \mathrm{Fe}^{3+}+e^{-} \rightarrow \mathrm{Fe}^{2+} & E^{\mathrm{O}}=0.77 \mathrm{~V} \\ \mathrm{Cd}^{2+}+2 e^{-} \rightarrow \mathrm{Cd} & E^{\mathrm{O}}=-0.40 \mathrm{~V} \end{array} $$

is oxidized in the reaction: \(2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cd}^{2+}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Cd}(\mathrm{s})\)

Select one:

A. -0.37 V
B. 0.37 V
C. -1.17 V
D. 1.17 V
E. none of these

Answer 1

2Fe2+(aq) + Cd2+(aq) → 2Fe3+(aq) + Cd(s)
Oxidation is the removal of an electron, It takes place at the anode. There
will be an increase in oxidation number during oxidation.

At anode

Fe2+ ==> Fe3+ + 1e-

reduction happens at the cathode where the electron is added to the element

at cathode

Cd+ 2e - --> Cd2+

E0cell = Eocathode- Eoanode = -0.40 V - 0.77 V = -1.17 V

The answer is c. -1.17 V; Fe2+