Identify Correct Statement for Spontaneous Electrochemical Cell Reaction

To determine the correct statement, we analyze the given standard reduction potentials for the two half-cell reactions:

1. AgBr/Ag half-cell: $\text{AgBr(s)} + \text{e}^- \rightarrow \text{Ag(s)} + \text{Br}^-\text{(aq)}, \quad E^\circ_1 = +0.07\text{ V}$

2. $\text{Fe(OH)}_2$/Fe half-cell: $\text{Fe(OH)}_2\text{(s)} + 2\text{e}^- \rightarrow \text{Fe(s)} + 2\text{OH}^-\text{(aq)}, \quad E^\circ_2 = -0.88\text{ V}$

For an electrochemical cell to operate spontaneously, the overall standard cell potential ($E^\circ_{\text{cell}}$) must be positive ($E^\circ_{\text{cell}} > 0$).

Since $E^\circ_1 (+0.07\text{ V}) > E^\circ_2 (-0.88\text{ V})$, reduction occurs at the half-cell with the higher standard reduction potential (Cathode), and oxidation occurs at the half-cell with the lower standard reduction potential (Anode).

• Reduction at Cathode: $2\text{AgBr(s)} + 2\text{e}^- \rightarrow 2\text{Ag(s)} + 2\text{Br}^-\text{(aq)}$

• Oxidation at Anode: $\text{Fe(s)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Fe(OH)}_2\text{(s)} + 2\text{e}^-$

Adding the two half-reactions gives the overall spontaneous cell reaction: $\text{Fe(s)} + 2\text{OH}^-\text{(aq)} + 2\text{AgBr(s)} \rightleftharpoons \text{Fe(OH)}_2\text{(s)} + 2\text{Ag(s)} + 2\text{Br}^-\text{(aq)}$

Now, let's evaluate the options:

1. Overall Reaction: As derived above, the reaction is $\text{Fe(s)} + 2\text{OH}^-\text{(aq)} + 2\text{AgBr(s)} \rightleftharpoons \text{Fe(OH)}_2\text{(s)} + 2\text{Ag(s)} + 2\text{Br}^-\text{(aq)}$. Thus, Option A is correct.
2. Cell EMF ($E^\circ_{\text{cell}}$): $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.07\text{ V} - (-0.88\text{ V}) = +0.95\text{ V}$ Since Option B states $E^\circ_{\text{cell}} = -0.95\text{ V}$, it is incorrect.
3. Oxidation State of Fe: $\text{Fe}$ is oxidized from an oxidation state of $0$ to $+2$ in $\text{Fe(OH)}_2$. Thus, Option C ("Fe is reduced") is incorrect.
4. Thermodynamic Property: $E^\circ_{\text{cell}}$ is an intensive property (it does not depend on the quantity of matter), whereas Gibbs free energy change ($\Delta G^\circ$) is an extensive property. Thus, Option D is incorrect.

Correct Answer: A