Electrochemistry Result Question 30
Passage
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 Daniell 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.
(2007, 4 $\times$ 3M = 12M)
$ \mathrm{I}_2+2 e^{-} \rightarrow 2 \mathrm{I}^{-} $$\quad \quad E^{\circ}=0.54 $
$ \mathrm{Cl}_2+2 e^{-} \rightarrow 2 \mathrm{Cl}^{-} $$\quad \quad E^{\circ}=1.36 $
$ \mathrm{Mn}^{3+}+e^{-} \rightarrow \mathrm{Mn}^{2+} $$\quad \quad E^{\circ}=1.50 $
$ \mathrm{Fe}^{3+}+e^{-} \rightarrow \mathrm{Fe}^{2+} $$\quad \quad\quad E^{\circ}=0.77 $
$ \mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-} \rightarrow 2 \mathrm{H}_2 \mathrm{O} $$\quad \quad E^{\circ}=1.23$
30. While $\mathrm{Fe}^{3+}$ is stable, $\mathrm{Mn}^{3+}$ is not stable in acid solution because
(a) $\mathrm{O}_2$ oxidises $\mathrm{Mn}^{2+}$ to $\mathrm{Mn}^{3+}$
(b) $\mathrm{O}_2$ oxidises both $\mathrm{Mn}^{2+}$ to $\mathrm{Mn}^{3+}$ and $\mathrm{Fe}^{2+}$ to $\mathrm{Fe}^{3 +}$
(c) $\mathrm{Fe}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$
(d) $\mathrm{Mn}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$
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Answer:
Correct Answer: 30. ( d )
Solution:
- For the reaction:
(i) $4 \mathrm{Fe}^{3+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{Fe}^{2+}+4 \mathrm{H}^{+}+\mathrm{O}_2 ; E^{\circ}=-0.46 \mathrm{~V}$
(ii) $4 \mathrm{Mn}^{3+}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{Mn}^{2+}+4 \mathrm{H}^{+}+\mathrm{O}_2 ; E^{\circ}=+0.27 \mathrm{~V}$
As evidenced above, reaction (i) is non-spontaneous, therefore, $\mathrm{Fe}^{3+}$ is stable in acid solution. However, reaction (ii) is spontaneous $\mathrm{Mn}^{3+}$ oxidises $\mathrm{H}_2 \mathrm{O}$ to $\mathrm{O}_2$ and itself reduced to $\mathrm{Mn}^{2+}$ in acidic medium.
BETA
