Balance the following, in basic solution, by the method of half reactions. C2H5OH + MnO4- --> C2H3O2- + MnO2. What is the coefficient on water (H2O) in the product of the balanced equation?

Answer

To balance the redox reaction in a basic solution, we split it into two half-reactions. 1) Oxidation: C2H5OH -> C2H3O2-. First, balance oxygen by adding H2O: C2H5OH + H2O -> C2H3O2-. Then balance hydrogen by adding H+: C2H5OH + H2O -> C2H3O2- + 5H+. Balance charge by adding 4e-: C2H5OH + H2O -> C2H3O2- + 5H+ + 4e-. 2) Reduction: MnO4- -> MnO2. Balance oxygen with 2H2O: MnO4- -> MnO2 + 2H2O. Balance hydrogen with 4H+: MnO4- + 4H+ -> MnO2 + 2H2O. Balance charge with 3e-: MnO4- + 4H+ + 3e- -> MnO2 + 2H2O. Multiply the oxidation half-reaction by 3 and the reduction by 4 to equalize electrons (12e-). 3(C2H5OH + H2O -> C2H3O2- + 5H+ + 4e-) gives 3C2H5OH + 3H2O -> 3C2H3O2- + 15H+ + 12e-. 4(MnO4- + 4H+ + 3e- -> MnO2 + 2H2O) gives 4MnO4- + 16H+ + 12e- -> 4MnO2 + 8H2O. Combine the half-reactions: 3C2H5OH + 4MnO4- + H+ -> 3C2H3O2- + 4MnO2 + 5H2O. To convert to basic solution, add 1 OH- to both sides to neutralize H+: 3C2H5OH + 4MnO4- + H2O -> 3C2H3O2- + 4MnO2 + 5H2O + OH-. Simplify water by subtracting 1 H2O from both sides: 3C2H5OH + 4MnO4- -> 3C2H3O2- + 4MnO2 + 4H2O + OH-. The coefficient for water (H2O) in the products is 4.