Answer the following :
(i) Why is the Equilibrium Constant ($K_c$) related to $E^{\circ}_{cell}$ and not to $E_{cell}$?
(ii) Two metals ‘A’ and ‘B’ have standard electrode potential values of $-0.24$ V and $+0.80$ V respectively. Which of these will liberate hydrogen gas from dil. $H_2SO_4$?
(iii) Write the cell reaction which occurs in lead storage battery when it is in charging.

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

Answer 1

Step 1: Conceptual Overview:
Electrochemistry connects chemical energy to electrical potential. The standard potential ($ E^{\circ} $) represents the inherent driving force of a reaction, while $ E_{cell} $ reflects the current state of the system. Equilibrium is reached when the driving force ($ E_{cell} $) becomes zero, and no further electrical work can be performed.
Step 2: Detailed Explanation:
(i) Equilibrium Relation: At equilibrium, the rates of the forward and reverse reactions are equal, and no net chemical change occurs. As a result, the cell potential ($ E_{cell} $) drops to zero because the system can no longer perform electrical work.
This relationship is reflected in the Nernst equation: \[ 0 = E^{\circ}_{cell} - \frac{2.303 RT}{nF} \log K_c \] Rearranging the equation gives: \[ E^{\circ}_{cell} = \frac{2.303 RT}{nF} \log K_c \] This shows that the equilibrium constant $ K_c $ is determined by the standard cell potential ($ E^{\circ}_{cell} $), which remains constant for a given reaction. At equilibrium, $ E_{cell} $ is zero, which means it cannot distinguish between different values of $ K_c $.
(ii) Hydrogen Liberation: For a metal to liberate hydrogen gas ($ H_2 $) from an acid, it must be able to reduce $ H^+ $ ions to $ H_2 $. The standard electrode reaction for hydrogen is: \[ 2H^+ + 2e^- \rightarrow H_2 \quad (E^{\circ} = 0.00 \text{ V}) \] For a metal to liberate hydrogen gas, its standard reduction potential must be lower (more negative) than $ 0.00 \text{ V} $.
- Metal A has $ E^{\circ} = -0.24 \text{ V} $, which is less than $ 0.00 \text{ V} $, making it capable of liberating hydrogen. - Metal B has $ E^{\circ} = +0.80 \text{ V} $, which is greater than $ 0.00 \text{ V} $, so it cannot liberate hydrogen.
Therefore, only metal A is a strong enough reducing agent to liberate hydrogen gas.
(iii) Lead Storage Battery (Charging): During charging, an external power source reverses the chemical changes that took place during discharge. The lead sulfate ($ PbSO_4 $) formed at both electrodes during discharge is converted back to lead ($ Pb $) at the anode and lead dioxide ($ PbO_2 $) at the cathode.
The overall chemical equation for the charging process is: \[ 2PbSO_4(s) + 2H_2O(l) \xrightarrow{\text{Electrical energy}} Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \] Step 3: Final Conclusion:
(i) At equilibrium, $ E_{cell} $ is zero, and $ K_c $ is related to $ E^{\circ}_{cell} $.
(ii) Metal A ($ -0.24 $ V) can liberate $ H_2 $.
(iii) The overall reaction for the charging process is as given in Step 2.

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