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NEET · Chemistry · STD 12 - 2. Electrochemistry
Consider the half-cell reduction reaction \(Mn^{2+} + 2e^- \rightarrow Mn,\, \)\(E^o = - 1.18\, V\) \(Mn^{2+} \rightarrow Mn^{3+} + e^-,\) \( E^o = - 1.51 \,V\) The \(E^o\) for the reaction \(3Mn^{2+} \rightarrow Mn^o + 2Mn^{3+},\) and possibility of the forward reaction are respectively
- A \(- 4.18\, V\) and yes
- B \(+ 0.33\, V\) and yes
- C \(+ 2.69\, V\) and no
- D \(- 2.69\, V\) and no
Answer & Solution
Correct Answer
(D) \(- 2.69\, V\) and no
Step-by-step Solution
Detailed explanation
\(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}, \mathrm{E}^{\circ}=-1.18 \mathrm{V}\) .... \((i)\) \(2\left(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}\right), E^{\circ}=+1.51 \mathrm{V}\).... \((ii)\)
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