(i) WHY $K_c$ IS RELATED TO $E^\circ_{cell}$ AND NOT TO $E_{cell}$

Step 1 - Write the two thermodynamic relations.
$$\Delta G^\circ = -nFE^\circ_{cell}$$
$$\Delta G^\circ = -2.303\,RT\log K_c$$

Step 2 - Equate them.
$$-nFE^\circ_{cell} = -2.303\,RT\log K_c$$
$$\log K_c = \frac{nFE^\circ_{cell}}{2.303\,RT}$$
At 298 K this becomes $\log K_c = \dfrac{nE^\circ_{cell}}{0.059}$

Step 3 - Reason it out.
$K_c$ is a CONSTANT at a given temperature. So it can only be related to a quantity
that is also constant at that temperature.
$E^\circ_{cell}$ is the STANDARD cell potential, measured when all species are in their
standard states (1 M solutions, 1 bar gases). It has a FIXED value at a given temperature.
$E_{cell}$, on the other hand, KEEPS CHANGING as the reaction proceeds, because the
concentrations of the reactants and products keep changing (Nernst equation). At
equilibrium $E_{cell}$ actually becomes ZERO.
A constant $K_c$ therefore cannot be linked to a continuously varying $E_{cell}$;
it can only be linked to the fixed quantity $E^\circ_{cell}$.

(ii) WHICH METAL LIBERATES HYDROGEN FROM DIL. $H_2SO_4$

Step 1 - State the criterion.
A metal can liberate $H_2$ from a dilute acid only if it is a STRONGER REDUCING AGENT
than hydrogen, i.e. only if its standard reduction potential is NEGATIVE (less than
$E^\circ_{H^+/H_2} = 0.00$ V). Such a metal lies ABOVE hydrogen in the electrochemical series.

Step 2 - Compare the two metals.
Metal A : $E^\circ = -0.24$ V (NEGATIVE, below hydrogen in reduction potential)
Metal B : $E^\circ = +0.80$ V (POSITIVE, above hydrogen in reduction potential)

Step 3 - Conclude.
METAL 'A' will liberate hydrogen gas from dil. $H_2SO_4$.
$$A(s) + 2H^+(aq) \rightarrow A^{2+}(aq) + H_2(g), \qquad E^\circ_{cell} = 0.00 - (-0.24) = +0.24\ V$$
$E^\circ_{cell}$ is positive, so the reaction is spontaneous.
For metal B, $E^\circ_{cell} = 0.00 - 0.80 = -0.80$ V (negative), so it cannot.

(iii) CELL REACTION DURING CHARGING OF A LEAD STORAGE BATTERY

During charging the discharge reactions are REVERSED - the battery acts as an
electrolytic cell.

At the electrode connected to the positive terminal (oxidation):
$$PbSO_4(s) + 2H_2O(l) \longrightarrow PbO_2(s) + SO_4^{2-}(aq) + 4H^+(aq) + 2e^-$$

At the electrode connected to the negative terminal (reduction):
$$PbSO_4(s) + 2e^- \longrightarrow Pb(s) + SO_4^{2-}(aq)$$

OVERALL CELL REACTION ON CHARGING:
$$2PbSO_4(s) + 2H_2O(l) \longrightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq)$$

The white $PbSO_4$ deposited during discharge is converted back into Pb and $PbO_2$,
and $H_2SO_4$ is regenerated (so the density of the acid rises again).