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.
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.
|+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.
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-|
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+)|
|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)