An electrochemical cell is constructed using half cells in the direction of spontaneous change
Fe(OH)$$_2$$(s) + 2e$$^-$$ → Fe(s) + 2OH$$^-$$(aq)       E$$^0$$ = −0.88 V
and AgBr(s) + e$$^-$$ → Ag(s) + Br$$^-$$(aq)                 E$$^0$$ = +0.07 V
Which of the following option is correct?

The two standard reduction half-reactions given are:
$$\text{(i)}\; Fe(OH)_2(s)+2e^- \rightarrow Fe(s)+2OH^- (aq),\; E^0 = -0.88\; \text{V}$$
$$\text{(ii)}\; AgBr(s)+e^- \rightarrow Ag(s)+Br^- (aq),\; E^0 = +0.07\; \text{V}$$

Step 1 : Identify the cathode and the anode.
The half-cell with the more positive $$E^0$$ is reduced; the other is forced to run in the opposite (oxidation) direction.
$$E^0_{AgBr/Ag} = +0.07\; \text{V} \gt E^0_{Fe(OH)_2/Fe} = -0.88\; \text{V}$$
Therefore:
• Cathode (reduction): $$AgBr(s)+e^- \rightarrow Ag(s)+Br^- (aq)$$
• Anode (oxidation): reverse of (i) ⇒ $$Fe(s)+2OH^- (aq) \rightarrow Fe(OH)_2(s)+2e^-$$

Step 2 : Balance the electrons.
The cathode process involves 1 e⁻, the anode 2 e⁻. Multiply the cathode equation by 2:

$$\begin{aligned} \text{Cathode: } & 2AgBr(s)+2e^- \rightarrow 2Ag(s)+2Br^- (aq) \\ \text{Anode: } & Fe(s)+2OH^- (aq) \rightarrow Fe(OH)_2(s)+2e^- \end{aligned}$$

Adding the two gives the overall cell reaction:

$$Fe(s)+2OH^- (aq)+2AgBr(s) \rightarrow Fe(OH)_2(s)+2Ag(s)+2Br^- (aq)$$

This is exactly the statement in Option A, so Option A is correct.

Step 3 : Calculate the standard cell potential $$E^0_{cell}$$.
Formula: $$E^0_{cell}=E^0_{\text{cathode (reduction)}}-E^0_{\text{anode (reduction form)}}$$
$$E^0_{cell}=0.07\; \text{V}-(-0.88\; \text{V})=+0.95\; \text{V}$$
Hence $$E^0_{cell}=+0.95\; \text{V}$$ (not −0.95 V), so Option B is wrong.

Step 4 : Examine the role of iron.
In the cell, $$Fe(s)$$ loses electrons to form $$Fe(OH)_2(s)$$, so iron is oxidised, not reduced. Option C is wrong.

Step 5 : Nature of $$E^0_{cell}$$.
A potential difference does not depend on the amount of substance; it is an intensive property, not extensive. Therefore Option D is wrong.

Final conclusion: Only Option A is correct: Fe(s)+2OH$$^-$$ (aq)+2AgBr(s) ⇌ Fe(OH)$$_2$$(s)+2Ag(s)+2Br$$^-$$ (aq).