The concentration of hydrogen peroxide (H2O2) can be determined by titrating it with an acidified MnO4− solution. The following is an unbalanced equation.

MnO4− (aq) + H2O2 (aq) → O2 (g) + Mn2+ (aq)

(a) Balance the above redox reaction.
(b) Determine the concentration of H2O2 solution in molarity if 10.00 mL of this solution requires 20.00 mL of 1.5 M MnO4− for a complete reaction.

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Answer:

A) [tex]{MnO_4}^- + 3 H^+ + \frac{5}{2} H_2O_2 \rightarrow Mn^{2+} + 4 H_2O + \frac{5}{2} O_2 [/tex]

B) 7.5 molar

Explanation:

A) Reduction

[tex]{MnO_4}^- + 8 H^+ + 5e^- \rightarrow Mn^{2+} + 4 H_2O[/tex]

Oxidation

[tex]H_2O_2 \rightarrow O_2 + 2 H^+ + 2e^-[/tex]

Multiplying the oxidation reaction by 5/2 and adding it to the reduction equation:

[tex]{MnO_4}^- + 8 H^+ + 5e^- \rightarrow Mn^{2+} + 4 H_2O[/tex]

+

[tex]\frac{5}{2} H_2O_2 \rightarrow \frac{5}{2} O_2 + 5 H^+ + 5e^-[/tex]

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[tex]{MnO_4}^- + 3 H^+ + \frac{5}{2} H_2O_2 \rightarrow Mn^{2+} + 4 H_2O + \frac{5}{2} O_2 [/tex]

B) 10 ml = 0.01 L

20 ml = 0.02 L

mol of MnO4− = molarity*volume = 1.5*0.02 = 0.03

1 mol of MnO4− reacts with 5/2 mol of H2O2, then:

mol of H2O2 = 0.03*5/2 = 0.075

molarity = mol/volume = 0.075/0.01 = 7.5 molar

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