The correct answer is: A. Its voltage increases.
During the charging of a lead-acid cell, the following reactions occur:
- At the anode:
$$Pb(s) + SO_4^{2-}(aq) \to PbSO_4(s) + 2e^-$$ - At the cathode:
$$PbO_2(s) + 4H^+(aq) + 2e^- \to Pb(s) + 2H_2O(l)$$
The overall reaction is:
$$PbO_2(s) + 2H_2SO_4(aq) + Pb(s) \to 2PbSO_4(s) + 2H_2O(l)$$
As you can see from the overall reaction, the number of electrons on the left-hand side is equal to the number of electrons on the right-hand side. This means that the charge is conserved during the charging process.
The voltage of a lead-acid cell is determined by the Nernst equation:
$$E = E^0 – \frac{RT}{nF} \ln \frac{[Ox]^a}{[Red]^b}$$
where $E$ is the cell voltage, $E^0$ is the standard cell potential, $R$ is the ideal gas constant, $T$ is the temperature, $n$ is the number of electrons transferred in the reaction, $F$ is Faraday’s constant, $[Ox]$ is the concentration of the oxidized species, and $[Red]$ is the concentration of the reduced species.
As you can see from the Nernst equation, the voltage of a lead-acid cell is dependent on the concentration of the oxidized and reduced species. During the charging process, the concentration of the oxidized species (lead dioxide) decreases and the concentration of the reduced species (lead sulfate) increases. This causes the voltage of the cell to increase.
The other options are incorrect for the following reasons:
- Option B: During the charging process, the lead-acid cell takes in energy, not gives it out.
- Option C: The cathode of a lead-acid cell is made of lead dioxide, which is a dark chocolate brown solid. During the charging process, the lead dioxide is reduced to lead, which is a gray solid. Therefore, the color of the cathode does not change during the charging process.
- Option D: The specific gravity of sulfuric acid decreases during the discharging process, not the charging process.