Electrochemistry 2 Question 18
17. Consider the following cell reaction,
$$ \begin{array}{r} 2 \mathrm{Fe}(s)+\mathrm{O}{2}(g)+4 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}{2} \mathrm{O}(l), \ E^{\circ}=1.67 \mathrm{~V} \end{array} $$
At $\left[\mathrm{Fe}^{2+}\right]=10^{-3} \mathrm{M}, \mathrm{P}\left(\mathrm{O}_{2}\right)=0.1 \mathrm{~atm}$ and $\mathrm{pH}=3$, the cell potential at $25^{\circ} \mathrm{C}$ is
(2011)
(a) $1.47 \mathrm{~V}$
(b) $1.77 \mathrm{~V}$
(c) $1.87 \mathrm{~V}$
(d) $1.57 \mathrm{~V}$
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Solution:
- The half reactions are $\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 e^{-} \times 2$
$$ \begin{gathered} \mathrm{O}{2}(g)+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}{2} \mathrm{O} \ 2 \mathrm{Fe}(s)+\mathrm{O}{2}(g)+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}{2} \mathrm{O}(l) \ E=E^{\mathrm{o}}-\frac{0.059}{4} \log \frac{\left(10^{-3}\right)^{2}}{\left(10^{-3}\right)^{4}(0.1)}=1.57 \mathrm{~V} \end{gathered} $$
