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

Concept:
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. Whether a metal displaces hydrogen depends on its electrode potential being negative (more reducing than hydrogen). A lead storage battery reverses its discharge reaction during charging.

Step 1 (i):
At equilibrium the net cell reaction has stopped, so \(E_{cell} = 0\) and the cell does no useful work. The thermodynamic relation \(\Delta G^\circ = -nFE^\circ_{cell} = -RT\ln K_c\) uses standard conditions, giving \(\ln K_c = \frac{nFE^\circ_{cell}}{RT}\). Hence \(K_c\) is related to \(E^\circ_{cell}\) (a constant value), not to \(E_{cell}\) which varies with concentration and becomes zero at equilibrium.

Step 2 (ii):
A metal liberates \(H_2\) from dil. acid only if its standard electrode potential is less than that of hydrogen (0.00 V), i.e. it is negative. Metal A has \(E^\circ = -0.24\) V (negative) while metal B has \(+0.80\) V (positive). Therefore metal A will liberate hydrogen gas from dil. \(H_2SO_4\).

Step 3 (iii):
During charging the discharge reactions reverse. Overall charging reaction: \(2PbSO_4(s) + 2H_2O(l) \rightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq)\).

Answer: (i) Because at equilibrium \(E_{cell}=0\); \(K_c\) connects to the fixed standard potential through \(\Delta G^\circ = -nFE^\circ_{cell} = -RT\ln K_c\). (ii) Metal A (\(-0.24\) V) liberates hydrogen. (iii) \(2PbSO_4 + 2H_2O \rightarrow Pb + PbO_2 + 2H_2SO_4\).