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JEE Advanced · Chemistry · 9. Redox Reactions

Paragraph:
Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential \(\left(E^{\circ}\right)\) of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their \(E^{\circ}(V\) with respect to normal hydrogen electrode) values. Using this data obtain the correct explanations to Questions 14-19.
\(\mathrm{I}_2+2 e^{-} \longrightarrow 2 \mathrm{I}^{-} E^{\circ}=0.54 \)
\( \mathrm{Cl}_2+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-} E^{\circ}=1.36 \)
\( \mathrm{Mn}^{3+}+e^{-} \longrightarrow \mathrm{Mn}^{2+} E^{\circ}=1.50 \)
\( \mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} E^{\circ}=0.77 \)
\( \mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O} E^{\circ}=1.23\)
Question:
While \(\mathrm{Fe}^{3+}\) is stable, \(\mathrm{Mn}^{3+}\) is not stable in acid solution because

  1. A \(\mathrm{O}_2\) oxidises \(\mathrm{Mn}^{2+}\) to \(\mathrm{Mn}^{3+}\)
  2. B \(\mathrm{O}_2\) oxidises both \(\mathrm{Mn}^{2+}\) and \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\)
  3. C \(\mathrm{Fe}^{3+}\) oxidises \(\mathrm{H}_2 \mathrm{O}\) to \(\mathrm{O}_2\)
  4. D \(\mathrm{Mn}^{3+}\) oxidises \(\mathrm{H}_2 \mathrm{O}\) to \(\mathrm{O}_2\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\mathrm{Mn}^{3+}\) oxidises \(\mathrm{H}_2 \mathrm{O}\) to \(\mathrm{O}_2\)

Step-by-step Solution

Detailed explanation

\({\left[\mathrm{Mn}^{3+}+e^{-}\right.}\left.\longrightarrow \mathrm{Mn}^{2+}\right] \times 4 \)
\( 2 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{H}^{+}+\mathrm{O}_2+4 e^{-} \)
\( 4 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{Mn}^{2+}+4 \mathrm{H}^{+}+\mathrm{O}_2 \)
\( E_{\text {cell }}^{\circ}= 1.50-1.23=0.27 \mathrm{~V} \)
\( E_{\text {cell }}^{\circ} \text { is + ve }\)
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