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

Electrochemical cell reactions and potentials. $E^{\circ}_{cell}$ is calculated using the difference between cathode and anode potentials. A spontaneous reaction has a positive $E^{\circ}_{cell}$.
1. Identify the potential electrodes:
Both given reactions are written as reduction half-reactions.
$E^{\circ}_{red1} = -0.88 \text{ V}$ ($Fe$ system)
$E^{\circ}_{red2} = 0.07 \text{ V}$ ($Ag$ system)

2. Determine Cathode and Anode for spontaneity:
The system with the higher reduction potential will act as the cathode. Thus, the Silver electrode is the cathode ($E^{\circ} = 0.07 \text{ V}$) and the Iron electrode is the anode ($E^{\circ} = -0.88 \text{ V}$).

3. Calculate cell potential:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
$E^{\circ}_{cell} = 0.07 - (-0.88) = 0.95 \text{ V}$

4. Write the net reaction:
Anode (Oxidation): $Fe(s) + 2OH^-(aq) \rightarrow Fe(OH)_2 + 2e^-$
Cathode (Reduction): $2AgBr(s) + 2e^- \rightarrow 2Ag(s) + 2Br^-(aq)$
Net: $Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightarrow Fe(OH)_2 + 2Ag(s) + 2Br^-(aq)$

5. Evaluate statements:
(1) is incorrect ($E^{\circ}_{cell}$ is +0.95 V).
(2) is incorrect ($E^{\circ}$ is intensive).
(3) is incorrect ($Fe$ is oxidized).
(4) is correct.

Thus, option (4) is the correct statement.