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JEE Advanced · Chemistry · 17. Electrochemistry

Consider the following cell reaction,
\(2 \mathrm{Fe}(\mathrm{s})+\mathrm{O}_2(g)+4 \mathrm{H}^{+}(a q) \longrightarrow \)
\( 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l), E^{\circ}=1.67 \mathrm{~V} \)
\( \text {At } {\left[\mathrm{Fe}^{2+}\right]=10^{-3} \mathrm{M}, \mathrm{P}\left(\mathrm{O}_2\right)=0.1 \mathrm{~atm}}\)
and \(\mathrm{pH}=3\), the cell potential at \(25^{\circ} \mathrm{C}\) is

  1. A \(1.47 \mathrm{~V}\)
  2. B \(1.77 \mathrm{~V}\)
  3. C \(1.87 \mathrm{~V}\)
  4. D \(1.57 \mathrm{~V}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(1.57 \mathrm{~V}\)

Step-by-step Solution

Detailed explanation

The half reactions are \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(\mathrm{aq})+2 e^{-} \times 2\)
\(\mathrm{O}_2(g)+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{Fe}(s) \)
\( +\mathrm{O}_2(g)+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}{ }_2 \mathrm{O}(l) \)
\( E=E^{\circ}-\frac{0.059}{4} \log \frac{\left(10^{-3}\right)^2}{\left(10^{-3}\right)^4(0.1)}=1.57 \mathrm{~V}\)
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