| Themes > Science > Chemistry > Inorganic Chemistry > Chemical Reaction , Chemical Formulas , Chemical Equation > Chemical Equation > Redox Reaction Stoichiometry |
Stoichiometric calculations involving redox reactions always begin with the balanced redox reaction. Once the balanced redox reaction has been obtained, the mole ratios will give the desired stoichiometric information. Example. We can calculate the mass of KMnO4 needed to oxidize 15.0 g of As4O6 in acidic solution, given that the products of the reaction are Mn2+ and arsenic acid, H3AsO4. It is first necessary to obtain the balanced redox reaction. The balanced half-reactions or electrode reactions are: 5e- + 8H+ + KMnO4 --> Mn2+ + K+ + 4H2O 10H2O + As4O6 --> 4H3AsO4 + 8H+ + 8e- Multiplying by eight and by five, adding the reactions, and cancelling electrons, protons, and water to the extent they appear on both sides of the equation gives: 24H+ + 8KMnO4 + 18H2O + 5As4O6 --> 8Mn2+ + 8K+ + 20H3AsO4 The molar mass of KMnO4 is 158.04 g/mol; in the balanced reaction there appears 8 x 158.04 g. The molar mass of As4O6 is 395.6 g/mol; in the balanced reaction there appears 5 x 395.6 g. The molar ratio in the reaction between them is 8 KMnO4 to 5 As4O6. Written out, this ratio is 8 x 158.04 : 5 x 395.6 = x : 15.0. As a consequence x = 15.0 g As4O6 (8 x 158.04) g KMnO4/(5 x 395.6) g As4O6 = 9.58 g KMnO4 |
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