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

### 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**