Question

Redox reactions 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.
$I_2+2e^{-}\to 2I^{-}{E}^{\circ }=0.54$
$C{l}_2+2e^{-}\to 2C{l}^{-}{E}^{\circ }=1.36$
$M{n}^{3+}+e^{-}\to M{n}^{2+}{E}^{\circ }=1.50$
$F{e}^{3+}+e^{-}\to F{e}^{2+}{E}^{\circ }=0.77$
$O_2+4H^{+}+4e^{-}\to 2H_{2}O{E}^{\circ }=1.23$
While $F{e}^{3+}$ is stable, $M{n}^{3+}$ is not stable in acid solution because

• A. $O_2$ oxidises $M{n}^{2+}$ to $M{n}^{3+}$.
• B. $O_2$ oxidises both $M{n}^{2+}$ to $M{n}^{3+}$ and $F{e}^{2+}$ to $F{e}^{3+}$.
• C. $F{e}^{3+}$ oxidises $H_{2}O$ to $O_2$.
• D. $M{n}^{3+}$ oxidises $H_{2}O$ to $O_2$.

Answer 1

Answer: (D)

Reaction of $M{n}^{3+}$ with $H_{2}O$ is spontaneous.
At the anode:
$O_2+4H^{+}+4e^{-}\to 2H_{2}O{E}^{\circ }=1.23V$
At the cathode:
$M{n}^{3+}+e^{-}\to M{n}^{2+}{E}^{\circ }=1.50V$
$E_{\text{cell }}=E_{\text{cathode }}-E_{\text{anode }}$
$=1.50-1.23=0.27V=0.27V$
Since $E$ cell is positive, the reaction is spontaneous because Gibbs free energy $\left(=-NF{E}_{\text{cell }}\right)$ is negative.