The following reaction is responsible for producing electricity in your car battery (often called a lead storage battery): \(\mathrm{Pb}+\mathrm{PbO}_{2}+2 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{PbSO}_{4}+2 \mathrm{H}_{2} \mathrm{O}\) (a) Assign an oxidation state to each atom. (Hint: For \(\mathrm{PbSO}_{4}\), you can figure out the charge of the \(\mathrm{Pb}\) if you remember that the charge of the sulfate ion is \(-2\left(\mathrm{SO}_{4}^{2-}\right)\). Then use shortcut Rule 7 to get the oxidation state of the \(\mathrm{Pb}\) in \(\mathrm{PbSO}_{4}\).) (b) Identify the atom that gets oxidized and the atom that gets reduced. (c) Identify the reactant that is the oxidizing agent and the reactant that is the reducing agent.

To find the oxidation states of each atom, we can use the periodically increasing electronegativity and use 7 shortcut rules. Let's assign oxidation states for all atoms in the equation: \[Pb + PbO_{2} + 2H_{2}SO_{4} \rightarrow 2PbSO_{4} + 2H_{2}O\] For the atoms in their elemental form, the oxidation state is zero. Therefore, Pb (lead) has an oxidation state of 0. The oxidation states of hydrogen and oxygen in their compounds are generally +1 and -2, respectively. Therefore, in H₂SO₄, the oxidation state of H is +1, and the oxidation state of O is -2. Knowing that the charge of the sulfate ion (SO₄²⁻) is -2, we can find the oxidation state of Pb in PbSO₄ by using Rule 7, which says that in a polyatomic ion, the sum of oxidation states must equal the charge on the ion. In this case: \[Pb + (S + 4O) = +2\] We know that the oxidation state of S (sulfur) is generally +6, and the oxidation state of O (oxygen) is -2. That gives: \[Pb + (+6 - 8) = +2\] \[Pb - 2 = +2\] Therefore, the oxidation state of Pb in PbSO₄ is +4. In PbO₂, the oxidation state of O is -2, we can find the oxidation state of Pb by using the same method: \[Pb + (2O) = 0\] \[Pb - 4 = 0\] The oxidation state of Pb in PbO₂ is +4. For H₂O, the oxidation state of H is +1 and the oxidation state of O is -2. The assigned oxidation states for each atom are: \[0 + (+4)O_{2} + 2(+1)_{2} (+6)_{4}(-2)O \rightarrow 2(+4)(+6)(-2)O_{4} + 2(+1)_{2}(-2)O\]

Now we can identify the atoms that get oxidized and reduced by comparing their oxidation states on each side of the equation. The oxidation process involves an increase in oxidation state, while the reduction process involves a decrease in oxidation state. In this reaction, the oxidation state of Pb increases from 0 to +4. Thus, Pb gets oxidized. On the other hand, the oxidation state of Pb in PbO₂ decreases from +4 to +2 in PbSO₄. Therefore, Pb from PbO₂ gets reduced.

The oxidizing agent is the reactant that causes another substance to be oxidized (i.e., loses electrons) and is itself reduced. In this reaction, Pb from PbO₂ causes Pb to be oxidized and becomes reduced, so PbO₂ is the oxidizing agent. The reducing agent is the reactant that causes another substance to be reduced (i.e., gains electrons) and is itself oxidized. In this reaction, Pb causes Pb from PbO₂ to be reduced and becomes oxidized, so lead (Pb) is the reducing agent.