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Consider the following cell reaction,
$$
\begin{array}{cc}
& 2 \mathrm{Fe}(\mathrm{s})+\mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q) \longrightarrow \\
& 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l), E^{\circ}=1.67 \mathrm{~V} \\
\text { At } & {\left[\mathrm{Fe}^{2+}\right]=10^{-3} \mathrm{M}, \mathrm{P}\left(\mathrm{O}_2\right)=0.1 \mathrm{~atm}}
\end{array}
$$
and $\mathrm{pH}=3$, the cell potential at $25^{\circ} \mathrm{C}$ is
Options:
$$
\begin{array}{cc}
& 2 \mathrm{Fe}(\mathrm{s})+\mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q) \longrightarrow \\
& 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l), E^{\circ}=1.67 \mathrm{~V} \\
\text { At } & {\left[\mathrm{Fe}^{2+}\right]=10^{-3} \mathrm{M}, \mathrm{P}\left(\mathrm{O}_2\right)=0.1 \mathrm{~atm}}
\end{array}
$$
and $\mathrm{pH}=3$, the cell potential at $25^{\circ} \mathrm{C}$ is
Solution:
1281 Upvotes
Verified Answer
The correct answer is:
$1.57 \mathrm{~V}$
$1.57 \mathrm{~V}$
The half reactions are $\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 e^{-} \times 2$
$$
\begin{array}{r}
\mathrm{O}_2(g)+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{Fe}(s) \\
+\mathrm{O}_2(g)+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}{ }_2 \mathrm{O}(l) \\
E=E^{\circ}-\frac{0.059}{4} \log \frac{\left(10^{-3}\right)^2}{\left(10^{-3}\right)^4(0.1)}=1.57 \mathrm{~V}
\end{array}
$$
$$
\begin{array}{r}
\mathrm{O}_2(g)+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{Fe}(s) \\
+\mathrm{O}_2(g)+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}{ }_2 \mathrm{O}(l) \\
E=E^{\circ}-\frac{0.059}{4} \log \frac{\left(10^{-3}\right)^2}{\left(10^{-3}\right)^4(0.1)}=1.57 \mathrm{~V}
\end{array}
$$
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