Question

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.

Hint:

The standard cell potential is fixed at standard conditions and is directly linked to the equilibrium constant, while the actual cell potential depends on concentrations.

Answer 1

Step 1: Why K_c links to standard potential.
At equilibrium no net current flows, so \(E_{cell} = 0\). The relation \(\Delta G^\circ = -RT\ln K_c\) uses the standard free energy, and \(\Delta G^\circ = -nFE^\circ_{cell}\). Combining gives \(\ln K_c = \dfrac{nFE^\circ_{cell}}{RT}\). Since \(K_c\) is a fixed number at a given temperature, it must connect to the fixed quantity \(E^\circ_{cell}\), not to \(E_{cell}\), which keeps changing with concentration.

Step 2: Decide which metal releases hydrogen.
A metal can displace hydrogen from acid only if its standard electrode potential is negative (below hydrogen, which is \(0\) V). Metal A has \(E^\circ = -0.24\) V (negative) and metal B has \(E^\circ = +0.80\) V (positive).

Step 3: Pick the metal.
Only metal A, being more reactive than hydrogen, will liberate \(H_2\) gas from dilute \(H_2SO_4\). Metal B cannot.

Step 4: Lead storage battery on charging.
During charging the reactions of discharge are reversed, so lead sulphate on both plates is converted back to lead and lead dioxide.

Step 5: Write the charging reaction.
\[ 2PbSO_4(s) + 2H_2O(l) \rightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \]

Answer: \(K_c\) relates to \(E^\circ_{cell}\) because at equilibrium \(E_{cell}=0\) and \(\Delta G^\circ=-nFE^\circ_{cell}=-RT\ln K_c\). Metal A \((-0.24\ V)\) liberates hydrogen. On charging, \(2PbSO_4 + 2H_2O \rightarrow Pb + PbO_2 + 2H_2SO_4\).