Question

Hi, the following problem I'm working on is:

The hydroperoxide ion HO2- (aq) reacts with permanganate ion MnO4- (aq) to produce MnO2(s) and oxygen gas. Balance the equation for the oxidation of hydroperoxide ion to O2(g) by permanganate ion in a basic solution.

I know the beginning equation is: HO2^-(aq) + MNO4^-(aq) ---> MnO2(s) +O2(g) and that I have to write half rxns for each one. I also know the answer ends up being:

H2O(l) + 2MnO4- (aq) +3HO2-(aq) ---> 2MnO2(s) +3O2 (g) +5OH-(aq)

What I don't understand is why we use hydroxide ions (OH-) instead of hydrogen ions (H+).

How come the 1st half of the rxn is written 3e- + 4H2O(l) + MnO4-(aq) ---> MnO2(s) + 2H2O (l) +4OH-(aq) instead of:

4e- + 4H+(aq) + MnO4-(aq)---> MnO2(s) +2H2O(l) ?

I tried to write this as least confusing as possible, thank you for your help!

Answer 1

Balancing equation in basic (OH-) medium

half reactions,

MnO4- --> MnO2

balance O and H

MnO4- + 4H+ --> MnO2 + 2H2O

add OH- for basic medium

MnO4- + 2H2O --> MnO2 + 4OH-

add e-'s

MnO4- + 2H2O + 3e- --> MnO2 + 4OH-

multiply this by 2

2MnO4- + 4H2O + 6e- --> 2MnO2 + 8OH-

second half reaction

HO2- --> O2

balance H,

HO2- --> O2 + H+

add OH- for basic medium

HO2- + OH- --> O2 + H2O

add e-'s,

HO2- + OH- --> O2 + H2O + 2e-

multiply this by 3 and add both equations

2MnO4- + 4H2O + 6e- --> 2MnO2 + 8OH-

3HO2- + 3OH- --> 3O2 + 3H2O + 6e-

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2MnO4- + 3HO2- + H2O --> 2MnO2 + 3O2 + 5OH-

Is the net balanced equation as needed