1. Assign oxidation states to all atoms in the reaction.
BrO3-: Br = +5, O = -2
Pb: Pb = 0
Br-: Br = -1
Pb2+: Pb = +2
2. Identify the atoms that are oxidized and reduced.
BrO3- is reduced (loses oxygen) and Pb is oxidized (gains oxygen).
3. Write out separate half-reactions for oxidation and reduction.
Reduction half-reaction: BrO3- + 6H+ + 6e- → Br- + 3H2O
Oxidation half-reaction: Pb → Pb2+ + 2e-
4. Balance the number of electrons transferred in each half-reaction.
Multiply the oxidation half-reaction by 3 to balance the number of electrons transferred:
3Pb → 3Pb2+ + 6e-
5. Balance the overall charge in each half-reaction by adding H+ or OH- ions as necessary.
Add 6H+ to the reduction half-reaction:
BrO3- + 6H+ + 6e- → Br- + 3H2O
6. Combine the half-reactions and cancel out any common terms.
3Pb + BrO3- + 6H+ → 3Pb2+ + Br- + 3H2O
7. Verify that the atoms and charges are balanced on both sides of the equation.
The balanced equation is:
3Pb + BrO3- + 6H+ → 3Pb2+ + Br- + 3H2O
Therefore, the balanced redox reaction is 3Pb + BrO3- + 6H+ → 3Pb2+ + Br- + 3H2O.