The hydroperoxide ion, HO_2^- (aq), reacts with permanganate ion, MnO_4 ^- (aq)

Question

The hydroperoxide ion, HO_2^- (aq), reacts with permanganate ion, MnO_4 ^- (aq) to produce MnO_2 (s) and oxygen gas. Balance the equation for the oxidation of hydroperoxide ion to O_2 (g) by permanganate ion in a basic solution. 2MnO_4 ^- (aq) + 3HO_2 ^- (aq) + H_2 O (l) rightarrow 2MnO_2 (s) + 3O_2 (s) + 5OH^- (aq) Balancing an oxidation-reduction reaction entails five steps: 1. Separate the equations into half-reactions. 2a. Balance all masses except H and O. 2b. Balance O by adding water. 2c. Balance H by adding H^+ (aq). 3. Balance the charges. 4. Balance the number of electrons between the two half-reactions. 5. Add the two equations together.

Explanation / Answer

Oxidation half

HO2- (aq) ------------> O2(g)

Here oxidation no.of oxygen changes from -1 to zero ,that is one electron change

Reduction half

MnO4- (aq) ----------> MnO2(s)

Here oxidation no.of magnesium decreases from +7 to +4 ,that is three electron change.

add both equations and cross multiply by their electron changes.

Then add.

3HO2-(aq)   +MnO4- (aq) ----------> MnO2(s) + 3 O2(g)

Then by following above described steps

balanced equation is

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

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