Module 5

Lesson 3.3  Practical Applications of Redox Stoichiometry



Key Concepts


Redox stoichiometry has many practical applications. For example, iron can be a problem in well water in Alberta and elsewhere. Iron(II) ions give well water a pale yellow-brown colour and iron(III) ions can produce insoluble precipitates. Iron can give water a metallic taste and can stain dishes and laundry.


Many people using well water choose to install a water treatment system to remove iron. A popular method to remove iron from water involves treating iron with potassium permanganate, KMnO4(s). How might this system be able to remove iron(II) ions from well water?


You may recall that the acidified permanganate ion is a strong oxidizing agent. Potassium permanganate brings about the oxidation of iron(II) ions to iron(III) ions. Iron (III) ions are poorly soluble in water and form a precipitate. Hence, the precipitate can be easily removed from the water by using a filter.


Check Your Understanding


Calculate the volume of water that can be treated if 50 g of potassium permanganate is added to the system. Assume that the concentration of iron (II) ions in well water is 5.0 ppm.

Hint: 5.0 ppm = 5.0 mg/kg = 5.0 mg/L since water has a density of 1 kg/L.

Determine the balanced redox reaction:

\( \mathrm { MnO_4^-(aq) + 8 H^+(aq) + 5 e^- \rightarrow Mn^{2+}(aq) + 4 H_2O(l) } \)

\( \mathrm { 5(Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + 1e^-) } \)

\( \mathrm { MnO_4^-(aq) + 8 H^+(aq) + 5Fe^{2+}(aq) \rightarrow 5 Fe^{3+}(aq) + 4 H_2O(l) + Mn^{2+}(aq) } \)


Determine # of moles of KMnO4:

\( \mathrm { n = \dfrac{m}{M} = \dfrac{50 g }{ 158.04 \frac{g}{mol}} = 0.3164~ mol } \)

KMnO4dissociates completely into K+ and MnO4-, so n of MnO4- is also 0.3164 mol.


Determine # of moles of Fe2+:

\( \mathrm { n = 0.3164~mol \times (5/1) = 1.582~ mol } \)


Determine concentration of Fe2+ in mol/L:

5.0 ppm = 5.0 mg/kg = 5.0 mg/L since water has a density of 1 kg/L.

\( \mathrm { c = \dfrac{n}{V} = \dfrac{\dfrac{0.0050 g }{ 55.85 \frac{g}{mol}} }{ 1~ L} = 8.95 \times 10^{-5}~\frac{ mol}{L} } \)


Determine volume of waste:

\( \mathrm { V = \dfrac{n}{c} = \dfrac{1.582 ~mol }{ 8.95 \times 10^{-5}~\frac{mol}{L}} = 17 676~ L = 1.8 \times 10^4~ L } \)