Practice Problem 6

Write a balanced equation for the reaction between the permanganate ion and hydrogen peroxide in a basic solution to form manganese dioxide and oxygen.

MnO4-(aq) + H2O2(aq)

MnO2(s) + O2(g)

Solution

STEP 1: Write a skeleton equation for the reaction.

MnO₄^- + H₂O₂ MnO₂ + O₂

STEP 2: Assign oxidation numbers to atoms on both sides of the equation.

MnO4-	+	H2O2		MnO2	+	O2
+7 -2		+1 -1		+4 -2		0

STEP 3: Determine which atoms are oxidized and which are reduced.

 

STEP 4: Divide the reaction into oxidation and reduction half-reactions and balance these half-reactions. This reaction can be divided into the following half-reactions.

 

Reduction:			MnO4-		MnO2
			+7		+4

Oxidation:			H2O2		O2
			-1		0

Let's start by balancing the reduction half-reaction. It takes three electrons to reduce manganese from the +7 to the +4 oxidation state.

Reduction:

MnO4- + 3 e-

MnO2

 

We now try to balance either the number of atoms or the charge on both sides of the equation. Because the reaction is run in basic solution, we can add either OH^- ions or H₂O molecules to either side of the equation, as needed. The key to deciding which side of the equation gets each of these reagents is simple: The only way to balance the net charge of -4 on the left side of the equation is to add four OH^- ions to the products.

 

Reduction:

MnO4- + 3 e-

MnO2 + 4 OH-

 

We can then balance the number of hydrogen and oxygen atoms by adding two H₂O molecules to the reactants.

 

Reduction:

MnO4- + 3 e- + 2 H2O

MnO2 + 4 OH-

 

We now turn to the oxidation half-reaction. Two electrons are lost when hydrogen peroxide is oxidized to form O₂ molecules.

 

Oxidation:

H2O2

O2 + 2 e-

 

We can balance the charge on this half-reaction by adding a pair of OH^- ions to the reactants.

 

Oxidation:

H2O2 + 2 OH-

O2 + 2 e-

 

The only way to balance the number of hydrogen and oxygen atoms is to add two H₂O molecules to the products,

Oxidation:

H2O2 + 2 OH-

O2 + 2 H2O + 2 e-

 

as shown in the figure below.

 

STEP 5: Combine the two half-reactions so that electrons are neither created nor destroyed. Two electrons are given off during the oxidation half-reaction and three electrons are consumed in the reduction half-reaction. We can therefore combine these half-reactions by using the lowest common multiple of 2 and 3.

2(MnO4- + 3 e- + 2 H2O MnO2 + 4 OH-)

+ 3(H2O2 + 2 OH- O2 + 2 H2O + 2 e-)

________________________________________________________

2 MnO4- + 3 H2O2 + 6 OH- + 4 H2O 2 MnO2 + 3 O2 + 8 OH- + 6 H2O

 

STEP 6: Balance the remainder of the equation by inspection, if necessary. The simplest balanced equation is obtained by subtracting four H₂O molecules and six OH^- ions from each side of the equation derived in the previous step.

 

2 MnO₄^-(aq) + 3 H₂O₂(aq) 2 MnO₂(s) + 3 O₂(g) + 2 OH^-(aq) + 2 H₂O(l)

Note that the ratio of moles of MnO₄^- to moles of H₂O₂ consumed is different in acidic and basic solutions. This difference results from the fact that MnO₄^- is reduced all the way to Mn²⁺ in acid, but the reaction stops at MnO₂ in base.