Greatly need some assistance

Question

In a particular redox reaction, BrO- is oxidized to BrO3- and Cu2+ is reduced to Cu+. Complete and balance the equation for this reaction in acidic solution. Phases are optional.

Answer 1

Using the half-reaction method:

Oxidation half-reaction: BrO^- ==> BrO₃^- .. unbalanced

BrO^- + 2H₂O ==> BrO₃^- .. balanced for Br and O by adding 2H₂O

BrO^- + 2H₂O ==> BrO₃^- + 4H⁺.. balanced for Br, O and H by adding H⁺ (acid)

BrO^- + 2H₂O ==> BrO₃^- + 4H⁺ + 4e- .. balanced fro Br, O, H and charge = balanced half-reaction

Reduction half-reaction: Cu²⁺ ==> Cu⁺ .. unbalanced

Cu²⁺ + e- ==> Cu⁺ .. balanced for Cu and charge = balanced half-reaction

Multiply the reduction half-reaction by 4 in order to balanced the electrons being transferred:

4Cu²⁺ + 4e- ==> 4Cu⁺

Add the 2 half-reactions together:

BrO^- + 2H₂O ==> BrO₃^- + 4H⁺ + 4e-

4Cu²⁺ + 4e- ==> 4Cu⁺

----------------------------------------------------------

BrO^- + 2H₂O + 4Cu²⁺ + 4e- ==> BrO₃^- + 4H⁺ + 4e- + 4Cu⁺

BrO^- + 4Cu²⁺ + 2H₂O ==> BrO₃^- + 4Cu⁺ + 4H⁺ ... BALANCED REDOX REACTION