Complete and balance the following redox reaction in basic solution. Be sure to include the proper phases for all species within the reaction. BrO3-(aq)+Zn(s)→Br-(aq)+Zn(OH)4 2-(aq)

Answer 1

1. List oxidation # of all species:

BrO₃^- (aq) + Zn(aq) → Br^- (aq) + Zn(OH)₄^2-

O₃: -2(3)= -6 Zn: 0 → Br: -1 (OH)₄: -1(4)= -4

Br: +5 Zn: +2

overall charge: -1 overall charge: 2-

1. Split half reaction

Oxidation half: Zn(s) → Zn(OH)₄^2- (aq)

Oxidation number of Zn increases from 0 to +2. This is oxidation

Reduction half: BrO₃^-(aq) →Br^- (aq)

Oxidation number Br decreases from +5 to -1. This is reduction

1. Balance other elements, leaving H and O last

BrO₃^- (aq) + Zn(aq) → Br^- (aq) + Zn(OH)₄^2-

Zn: 1 Zn: 1

Br: 1 Br:1

1. balance Oxygen with water molecules

Zn(s) + 4 H₂O → Zn(OH)₄^2- (aq)

BrO₃^-(aq) →Br^- (aq) + 3 H₂O

1. Balance Hydrogen with protons

Zn(s) + 4 H₂O → Zn(OH)₄^2- (aq) + 4H⁺

BrO₃^-(aq) + 6H⁺→Br^- (aq) + 3 H₂O

1. Balance charge with electrons

equation 1: Zn(s) + 4 H₂O(aq) → Zn(OH)₄^2- (aq) + 4H⁺ + 2e-

equation 2: BrO₃^-(aq) + 6H⁺+ 6e- →Br^- (aq) + 3 H₂O

1. multiply equation 1 by 3

equation 3: →3 Zn(s) + 12 H₂O(aq) → 3 Zn(OH)₄^2- (aq) + 12 H⁺ + 6e-

1. add equation 2 and 3

3Zn(s) + 12H₂O(aq)+ BrO₃^-(aq) + 6H⁺+ 6e- → 3 Zn(OH)₄^2- (aq) + 12 H⁺ + 6e- + Br^- + 3 H₂O (aq)

1. simplify

3Zn(s) + 9H₂O(aq)+ BrO₃^-(aq) → 3Zn(OH)₄^2- (aq) + 6H⁺ + Br^-

1. basic medium add 6 OH- on both sides

3Zn(s) + 9H₂O(aq)+ BrO₃^-(aq) + 6OH^-(aq) → 3Zn(OH)₄^2- (aq) + 6H⁺ + Br^- (aq)+ 6OH^-(aq)

1. simplify

3Zn(s) + 9H₂O(aq)+ BrO₃^-(aq) + 6OH^-(aq) → 3Zn(OH)₄^2- (aq) + Br^- (aq) + 6H₂O(aq)

1. cancel out waters

3Zn(s) + 3H₂O(aq)+ BrO₃^-(aq) + 6OH^-(aq) → 3Zn(OH)₄^2- (aq) + Br^-(aq)