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

### Key Takeaways: Redox Reaction Chemistry Problem

- A redox reaction is a chemical reaction in which reduction and oxidation occur.
- The first step in solving any redox reaction is to balance the redox equation. This is a chemical equation that must be balanced for charge as well as mass.
- Once the redox equation is balanced, use the mole ratio to find the concentration or volume of any reactant or product, provided the volume and concentration of any other reactant or product is known.

## Quick Redox Review

A redox reaction is a type of chemical reaction in which **red**uction and **ox**idation 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 is 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 are known.

## Redox Reaction Problem

Given the following balanced redox equation for the reaction between MnO_{4}^{-} and Fe^{2+} in an acidic solution:

- MnO
_{4}^{-}(aq) + 5 Fe^{2+}(aq) + 8 H^{+}(aq) → Mn^{2+}(aq) + 5 Fe^{3+}(aq) + 4 H_{2}O

Calculate the volume of 0.100 M KMnO_{4} needed to react with 25.0 cm^{3} 0.100 M Fe^{2+} and the concentration of Fe^{2+} in a solution if you know that 20.0 cm^{3} of solution reacts with 18.0 cm^{3} of 0.100 KMnO_{4}.

## How to Solve

Since the redox equation is balanced, 1 mol of MnO_{4}^{-} reacts with 5 mol of Fe^{2+}. Using this, we can obtain the number of moles of Fe^{2+}:

- moles Fe
^{2+}= 0.100 mol/L x 0.0250 L - moles Fe
^{2+}= 2.50 x 10^{-3}mol - Using this value:
- moles MnO
_{4}^{-}= 2.50 x 10^{-3}mol Fe^{2+}x (1 mol MnO_{4}^{-}/ 5 mol Fe^{2+}) - moles MnO
_{4}^{-}= 5.00 x 10^{-4}mol MnO_{4}^{-} - volume of 0.100 M KMnO
_{4}= (5.00 x 10^{-4}mol) / (1.00 x 10^{-1}mol/L) - volume of 0.100 M KMnO
_{4}= 5.00 x 10^{-3}L =**5.00 cm**^{3}

To obtain the concentration of Fe^{2+} asked in the second part of this question, the problem is worked the same way except solving for the unknown iron ion concentration:

- moles MnO
_{4}^{-}= 0.100 mol/L x 0.180 L - moles MnO
_{4}^{-}= 1.80 x 10^{-3}mol - moles Fe
^{2+}= (1.80 x 10^{-3}mol MnO_{4}^{-}) x (5 mol Fe^{2+}/ 1 mol MnO_{4}) - moles Fe
^{2+}= 9.00 x 10^{-3}mol Fe^{2+} - concentration Fe
^{2+}= (9.00 x 10^{-3}mol Fe^{2+}) / (2.00 x 10^{-2}L) - concentration Fe
^{2+}=**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.

## Sources

- Schüring, J., Schulz, H. D., Fischer, W. R., Böttcher, J., Duijnisveld, W. H., eds (1999).
*Redox: Fundamentals, Processes and Applications*. Springer-Verlag, Heidelberg ISBN 978-3-540-66528-1. - Tratnyek, Paul G.; Grundl, Timothy J.; Haderlein, Stefan B., eds. (2011).
*Aquatic Redox Chemistry*. ACS Symposium Series. 1071. ISBN 9780841226524.