Balance each of the following oxidation-reduction reactions by using the oxidation states method. a. \(\mathrm{Cl}_{2}(g)+\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Pb}(s) \rightarrow \mathrm{Pb}(\mathrm{OH})_{2}(s)\) c. \(\mathrm{H}^{+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \rightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

First, identify the oxidation states of each species in the reaction: Cl₂, - Cl has an oxidation state of ₀ Al, - Al has an oxidation state of ₀ Al³⁺ - Al has an oxidation state of +3 Cl⁻ - Cl has an oxidation state of -1 The change in oxidation states: - Cl is gaining an electron (reduced): ₀ to -1 (1 electron gained) - Al is losing 3 electrons (oxidized): ₀ to +3 (3 electrons lost) Now we balance the number of atoms and the charges on both sides: ₂Cl₂ + 3Al → 2Al³⁺ + 6Cl⁻ Now the equation is balanced.

First, identify the oxidation states of each species in the reaction: O₂, - O has an oxidation state of ₀ H₂O, - O has an oxidation state of -2, H has an oxidation state of +1 Pb, - Pb has an oxidation state of ₀ Pb(OH)₂, - Pb has an oxidation state of +2, O has an oxidation state of -2, H has an oxidation state of +1 The change in oxidation states: - O is gaining 2 electrons (reduced): ₀ to -2 (2 electrons gained) - Pb is losing 2 electrons (oxidized): ₀ to +2 (2 electrons lost) Now we balance the number of atoms and the charges on both sides: 2H₂O + O₂ + 2Pb → 2Pb(OH)₂ Now the equation is balanced.

First, identify the oxidation states of each species in the reaction: H⁺, - H has an oxidation state of +1 MnO₄⁻, - Mn has an oxidation state of +7, O has an oxidation state of -2 Fe²⁺, - Fe has an oxidation state of +2 On the product side: Mn²⁺, - Mn has an oxidation state of +2 Fe³⁺, - Fe has an oxidation state of +3 H₂O, - O has an oxidation state of -2, H has an oxidation state of +1 The change in oxidation states: - Mn is gaining 5 electrons (reduced): +7 to +2 (5 electrons gained) - Fe is losing 1 electron (oxidized): +2 to +3 (1 electron lost) We need to multiply the Mn reduction by 1 and the Fe oxidation by 5 to balance the number of electrons lost and gained. Now we balance the number of atoms and the charges on both sides: 8H⁺ + MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O Now the equation is balanced.