Electrochemistry Result Question 17
17. Consider the following cell reaction,
$2 \mathrm{Fe}(s)+\mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l), E^{\circ}=1.67 \mathrm{V}$
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 V
(b) 1.77 V
(c) 1.87 V
(d) 1.57 V
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Answer:
Correct Answer: 17. ( d )
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(\mathrm{~g})+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) ; \\ E=E^{\circ}-\frac{0.059}{4} \log \frac{\left(10^{-3}\right)^2}{\left(10^{-3}\right)^4(0.1)}=1.57 \mathrm{~V} \end{gathered} $
BETA
