Answer: Option (a) is the correct answer.
Explanation:
The given overall reaction will be as follows.
[tex]2Co^{3+}(aq) + 2Cl^{-}(aq) \rightarrow 2Co^{2+}(aq) + Cl_{2}(g)[/tex]
It is given that standard potential of the cell is 0.46 V.
Hence, the oxidation and reduction half reactions of the cell are as follows.
Reduction-half reaction: [tex]Cl_{2}(g) + 2e^{-} \rightarrow 2Cl^{-}(aq)[/tex], [tex]E^{o} = 1.36 V[/tex]
Oxidation-half reaction: [tex]2Co^{3+}(aq) \rightarrow 2Co^{2+}(aq)[/tex], [tex]E^{o} = ?[/tex]
Now, expression of [tex]E^{o}_{cell}[/tex] of this reaction is as follows.
[tex]E^{o}_{cell} = E^{o}_{Co^{3+}/Co^{2+}} - E^{o}_{Cl_{2}/Cl^{-}}[/tex]
Putting the given values into the above formula as follows.
0.46 V = [tex]E^{o}_{Co^{3+}/Co^{2+}}[/tex] - 1.36 V
[tex]E^{o}_{Co^{3+}/Co^{2+}}[/tex] = 1.82 V
Thus, we can conclude that standard potential of the given half cell is 1.82 V.