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}+2 e^{-} \rightarrow 2 I ^{-} E ^{\circ}=0.54 $
$Cl _{2}+2 e^{-} \rightarrow 2 Cl ^{-} E ^{\circ}=1.36$
$ Mn ^{3+}+e^{-} \rightarrow Mn ^{2+} E ^{\circ}=1.50$
$Fe ^{3+}+e^{-} \rightarrow Fe ^{2+} E ^{\circ}=0.77$
$ O _{2}+4 H ^{+}+4 e^{-} \rightarrow 2 H _{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. $Fe ^{3+}$ oxidises $H _{2} O$ to $O _{2}$.
• D. $Mn ^{3+}$ oxidises $H _{2} O$ to $O _{2}$.

Answer 1

Answer: (D)

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