Redox Reactions - Balanced Equation Example Problem

Worked Chemistry Problems

Redox reactions involve charge as well as mass.
Redox reactions involve charge as well as mass. Rafe Swan, Getty Images

This is a worked example redox reaction problem showing how to calculate volume and concentration of reactants and products using a balanced redox equation.

Quick Redox Review

A redox reaction is a type of chemical reaction in which reduction and oxidation occur. Because electrons are transferred between chemical species, ions form. So, to balance a redox reaction requires not only balancing mass (number and type of atoms on each side of the equation), but also charge.

In other words, the number of positive and negative electrical charges on both sides of the reaction arrow are the same in a balanced equation.

Once the equation is balanced, the mole ratio may be used to determine the volume or concentration of any reactant or product as long as the volume and concentration of any species is known.

Redox Reaction Problem

Given the following balanced redox equation for the reaction between MnO4- and Fe2+ in an acidic solution:

MnO4-(aq) + 5 Fe2+(aq) + 8 H+(aq) → Mn2+(aq) + 5 Fe3+(aq) + 4 H2O

Calculate the volume of 0.100 M KMnO4 needed to react with 25.0 cm3 0.100 M Fe2+ and the concentration of Fe2+ in a solution if you know that 20.0 cm3 of solution reacts with 18.0 cm3 of 0.100 KMnO4.

How to Solve

Since the redox equation is balanced, 1 mol of MnO4- reacts with 5 mol of Fe2+. Using this, we can obtain the number of moles of Fe2+:

moles Fe2+ = 0.100 mol/L x 0.0250 L

moles Fe2+ = 2.50 x 10-3 mol

Using this value:

moles MnO4- = 2.50 x 10-3 mol Fe2+ x (1 mol MnO4-/ 5 mol Fe2+)

moles MnO4- = 5.00 x 10-4 mol MnO4-

volume of 0.100 M KMnO4 = (5.00 x 10-4 mol) / (1.00 x 10-1 mol/L)

volume of 0.100 M KMnO4 = 5.00 x 10-3 L = 5.00 cm3

To obtain the concentration of Fe2+ asked in the second part of this question, the problem is worked the same way except solving for the unknown iron ion concentration:

moles MnO4- = 0.100 mol/L x 0.180 L

moles MnO4- = 1.80 x 10-3 mol

moles Fe2+ = (1.80 x 10-3 mol MnO4-) x (5 mol Fe2+ / 1 mol MnO4)

moles Fe2+ = 9.00 x 10-3 mol Fe2+

concentration Fe2+ = (9.00 x 10-3 mol Fe2+) / (2.00 x 10-2 L)

concentration Fe2+ = 0.450 M

Tips for Success

When solving this type of problem, it's important to check your work:

  • Check to make certain the ionic equation is balanced. Make certain the number and type of atoms is the same on both sides of the equation. Make certain the net electrical charge is the same on both sides of the reaction.
  • Be careful to work with the mole ratio between reactants and products and not the gram quantities. You may be asked to provide a final answer in grams. If so, work the problem using moles and then use the molecular mass of the species to convert between units. The molecular mass is the sum of the atomic weights of the elements in a compound. Multiply the atomic weights of atoms by any subscripts following their symbol. Don't multiply by the coefficient in front of the compound in the equation because you've already taken that into account by this point!
  • Be careful to report moles, grams, concentration, etc., using the correct number of significant figures.