Chemistry Forum

Hi,
I know how to balance easier redox equations, but I'm struggling with this one:

Ag₂O₂ + Mn²⁺ + H⁺ = MnO₄^- + Ag⁺ + H₂O

I'm confused by the peroxide on the reactant side and I'm not sure what the half reaction of reduction should look like. Do I assume that all the oxygen in water has been reduced from the peroxide or does it just come from the acidic solution. Would it be easier to represent the acidic solution by H₃O+ ions?

I'd be grateful for any help and explanations.

Reduction O₂ 2- + 2 e- => 2 O 2-
Oxidation Mn₂+ + 4 H₂O => MnO₄ - + 8 H+ + 5 e-

SCM of electron = 10

5 O₂ 2- + 10 e- => 10 O 2-

2 Mn₂+ + 8 H₂O => 2 MnO₄ - + 16 H+ + 10 e-

Addition

2 Mn₂+ + 8 H₂O + 5 O₂ 2- => 2 MnO₄ - + 16 H+ + 10 O₂-

Elimination of water

2 Mn₂+ + 5 O₂ 2- => 2 MnO₄ - + 2 O₂-

Silver is spectator ion will not change.