Question

Consider following reaction :
$Fe(OH)_2 + 2e^- \rightarrow Fe(s) + 2OH^-$ , $E^{\circ} = -0.88 \text{ V}$
$AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)$ , $E^{\circ} = 0.07 \text{ V}$
Select the correct statement.
(1) $E^{\circ}_{cell} = -0.95 \text{V}$
(2) $E^{\circ}_{cell}$ is an extensive property
(3) $Fe$ is getting reduced
(4) Net cell reaction: $Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightarrow Fe(OH)_2 + 2Ag(s) + 2Br^-(aq)$

• A. (1)
• B. (2)
• C. (3)
• D. (4)

Answer 1

The Correct Option is D

Analyzing standard reduction potentials to determine cell spontaneity and the nature of the net reaction.

LOGIC:
1. Standard electrode potentials given are $E^{\circ}_{Fe^{2+}/Fe} = -0.88 \text{ V}$ and $E^{\circ}_{Ag^+/Ag} = 0.07 \text{ V}$ (in specific conditions).
2. In a spontaneous cell, the electrode with the more negative potential acts as the anode (undergoes oxidation). So, $Fe$ undergoes oxidation: $Fe \rightarrow Fe^{2+} + 2e^-$.
3. The electrode with more positive potential acts as the cathode (undergoes reduction). So, $AgBr$ is reduced: $AgBr + e^- \rightarrow Ag + Br^-$.
4. Spontaneous $E^{\circ}_{cell} = E^{\circ}_{red} - E^{\circ}_{ox} = 0.07 - (-0.88) = +0.95 \text{ V}$.
5. The overall reaction is the sum of these half-reactions: $Fe + 2OH^- + 2AgBr \rightarrow Fe(OH)_2 + 2Ag + 2Br^-$.
6. Comparing with options: Option (4) accurately describes this spontaneous net reaction.