On charging the lead storage battery, the oxidation state of lead changes from $$x_1$$ to $$y_1$$ at the anode and from $$x_2$$ to $$y_2$$ at the cathode. The values of $$x_1, y_1, x_2, y_2$$ are respectively:

The lead storage battery has two lead-based electrodes dipped in $$\mathrm{H_2SO_4}$$.
• Spongy lead plate  →  traditionally called the anode plate.
• Lead dioxide plate  →  traditionally called the cathode plate.

While discharging, the following galvanic half-reactions take place:

$$\mathrm{At\;anode\;(Pb):}\quad \mathrm{Pb\;(0)} + \mathrm{SO_4^{2-}} \;\rightarrow\; \mathrm{PbSO_4\;(Pb^{2+})} + 2e^-$$
$$\mathrm{At\;cathode\;(PbO_2):}\quad \mathrm{PbO_2\;(Pb^{4+})} + \mathrm{SO_4^{2-}} + 4\mathrm{H^+} + 2e^- \;\rightarrow\; \mathrm{PbSO_4\;(Pb^{2+})} + 2\mathrm{H_2O}$$

When the battery is CHARGED, an external emf forces the reactions to run in the reverse direction (electrolysis).
Therefore:

Reverse of the discharge reaction:

$$\mathrm{PbSO_4\;(Pb^{2+})} + 2e^- \;\rightarrow\; \mathrm{Pb\;(0)} + \mathrm{SO_4^{2-}}$$

Thus, at the anode plate the oxidation state of lead changes
from $$x_1 = +2$$ (in $$\mathrm{PbSO_4}$$) to $$y_1 = 0$$ (in metallic Pb).

Reverse of the discharge reaction:

$$\mathrm{PbSO_4\;(Pb^{2+})} + 2\mathrm{H_2O} \;\rightarrow\; \mathrm{PbO_2\;(Pb^{4+})} + \mathrm{SO_4^{2-}} + 4\mathrm{H^+} + 2e^-$$

Hence, at the cathode plate the oxidation state of lead changes
from $$x_2 = +2$$ (in $$\mathrm{PbSO_4}$$) to $$y_2 = +4$$ (in $$\mathrm{PbO_2}$$).

Collecting the four values:
$$x_1 = +2,\; y_1 = 0,\; x_2 = +2,\; y_2 = +4$$

These correspond to Option B.

Final answer: $$+2,\;0,\;+2,\;+4$$ (Option B).