Question

Write the Nearest equation and calculate the e.m.f of the following cell at 298K.
Fe(s) | Fe$^{2+$(0.001M) || H$^+$(1M) H$_2$(1atm) | Pt.
Given E$^\circ_{Fe^{+2}/Fe} = -0.44$ V.}

Hint:

Pay attention to the signs. $\log(10^{-3})$ is $-3$, so the term adds to the standard potential. Lower product concentration increases voltage.

Answer 1

Cell Reaction: Fe(s) + 2H$^+$ $\to$ Fe$^{2+}$ + H$_2$(g)
Here, $n=2$ electrons transferred.
E$^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$
Cathode is SHE (H$^+$/H$_2$), so $E^\circ = 0.00$ V.
Anode is Fe/Fe$^{2+}$, so $E^\circ = -0.44$ V.
E$^\circ_{\text{cell}} = 0 - (-0.44) = +0.44$ V.
Nernst Equation:
$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]}$
$E_{\text{cell}} = 0.44 - \frac{0.059}{2} \log \frac{[\text{Fe}^{2+}]}{[\text{H}^+]^2}$
Substitute values ($[\text{Fe}^{2+}]=10^{-3}$, $[\text{H}^+]=1$):
$E_{\text{cell}} = 0.44 - 0.0295 \log \frac{10^{-3}}{1^2}$
$E_{\text{cell}} = 0.44 - 0.0295 \times (-3)$
$E_{\text{cell}} = 0.44 + 0.0885$
$E_{\text{cell}} = 0.5285$ V.