Practice Problem 4

We can determine the concentration of an acidic permanganate ion solution by titrating this solution with a known amount of oxalic acid until the charactistic purple color of the MnO₄^- ion disappears.

H₂C₂O₄(aq) + MnO₄^-(aq) CO₂(g) + Mn²⁺(aq)

Use the half-reaction method to write a balanced equation for this reaction.

Solution

STEP 1: Write a skeleton equation for the reaction.

H₂C₂O₄ + MnO₄^- CO₂ + Mn²⁺

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

H2C2O4	+	MnO4-		CO2	+	Mn2+
+1 +3 -2		+7 -2		+4 -2		+2

STEP 3: Determine which atoms are oxidized and which are reduced. The only elements that undergo a change in oxidation state are the carbon atoms in oxalic acid and the manganese atom in the MnO₄^- ion.  

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.

Oxidation:			H2C2O4		CO2
			+3		+4

Reduction:			MnO4-		Mn2+
			+7		+2

We'll balance the reduction half-reaction first. It takes five electrons to reduce manganese from +7 to +2.  

Reduction:

MnO4- + 5 e-

Mn2+

Because the reaction is run in acid, we can add H⁺ ions or H₂O molecules to either side of the equation, as needed. There are two hints that tell us which side of the equation gets H⁺ ions and which side gets H₂O molecules. The only way to balance the charge on both sides of this equation is to add eight H⁺ ions to the left side of the equation.  

Reduction:

MnO4- + 8 H+ + 5 e-

Mn2+

The only way to balance the number of oxygen atoms is to add four H₂O molecules to the right side of the equation.  

Reduction:

MnO4- + 8 H+ + 5 e-

Mn2+ + 4 H2O

We can now turn to the oxidation half-reaction. We might start by assuming that both of the carbon atoms in oxalic acid end up as carbon dioxide. This is therefore a two-electron oxidation half-reaction.  

Oxidation:			H2C2O4		2 CO2	+	2 e-
			+3		+4

We can balance both charge and mass by noting that two H⁺ ions are given off when oxalic acid is oxidized to carbon dioxide,

Oxidation:

H2C2O4

2 CO2 + 2 e- + 2 H+

as shown in the figure below.

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

2(MnO4- + 8 H+ + 5 e- Mn2+ + 4 H2O)

+ 5(H2C2O4 2 CO2 + 2 e- + 2 H+)

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2 MnO4- + 16 H+ + 5 H2C2O4 10 CO2 + 2 Mn2+ + 8 H2O + 10 H+

  STEP 6: Balance the remainder of the equation by inspection, if necessary. We can generate the simplest balanced equation for this reaction by subtracting 10 H⁺ ions from both sides of the equation derived in the previous step.  

2 MnO₄^-(aq) + 5 H₂C₂O₄(aq) + 6 H⁺(aq) 10 CO₂(g) + 2 Mn²⁺(aq) + 8 H₂O(l)