Question

The E° in the given diagram is :-

• A. 0.5 V
• B. 0.63 V
• C. 0.75 V
• D. 0.85 V

Answer 1

Answer: (D)

0.85 V

Answer 2

The question asks for the value of E° marked in the Latimer diagram, which corresponds to the standard reduction potential for the conversion of $ClO^-$ to $Cl^-$. This overall reduction occurs in two steps:

1. $ClO^- \rightarrow \frac{1}{2} Cl_2$: This step involves a change in oxidation state from +1 to 0, meaning 1 electron ($n_1 = 1$) is transferred. The given $E_1^\circ = 0.40 V$.

2. $\frac{1}{2} Cl_2 \rightarrow Cl^-$: This step involves a change in oxidation state from 0 to -1, meaning 1 electron ($n_2 = 1$) is transferred. The given $E_2^\circ = 1.36 V$.

To find the overall standard potential (E°) for $ClO^- \rightarrow Cl^-$, we use the additivity of Gibbs free energy ($\Delta G^\circ = -nFE^\circ$).

The total number of electrons transferred for the overall reaction is $n_{total} = n_1 + n_2 = 1 + 1 = 2$.

The total Gibbs free energy change is the sum of the individual steps' Gibbs free energy changes:

$\Delta G_{total}^\circ = \Delta G_1^\circ + \Delta G_2^\circ = (-n_1 F E_1^\circ) + (-n_2 F E_2^\circ)$

$\Delta G_{total}^\circ = -(1 \times F \times 0.40) + -(1 \times F \times 1.36)$

$\Delta G_{total}^\circ = -0.40 F - 1.36 F = -1.76 F$

Now, relate the total Gibbs free energy change to the overall standard potential E°:

$\Delta G_{total}^\circ = -n_{total} F E^\circ$

$-1.76 F = -2 F E^\circ$

$E^\circ = \frac{1.76}{2} = 0.88 V$

The calculated value is 0.88 V. Among the given options, 0.85 V is the closest.