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

Step 1: Divide the reaction into two half-reactions: Oxidation: C2H5OH -> C2H3O2-, Reduction: MnO4- -> MnO2. Step 2: Balance the oxidation half-reaction. Balance elements other than O and H: 2C on both sides. Balance O by adding H2O: C2H5OH + H2O -> C2H3O2-. Balance H by adding H+: C2H5OH + H2O -> C2H3O2- + 5H+. Balance charge by adding e-: C2H5OH + H2O -> C2H3O2- + 5H+ + 4e-. Step 3: Balance the reduction half-reaction. Balance O by adding H2O: MnO4- -> MnO2 + 2H2O. Balance H by adding H+: MnO4- + 4H+ -> MnO2 + 2H2O. Balance charge by adding e-: MnO4- + 4H+ + 3e- -> MnO2 + 2H2O. Step 4: Equalize the number of electrons. Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 4: 3(C2H5OH + H2O -> C2H3O2- + 5H+ + 4e-) => 3C2H5OH + 3H2O -> 3C2H3O2- + 15H+ + 12e-; 4(MnO4- + 4H+ + 3e- -> MnO2 + 2H2O) => 4MnO4- + 16H+ + 12e- -> 4MnO2 + 8H2O. Step 5: Add the half-reactions and simplify: 3C2H5OH + 3H2O + 4MnO4- + 16H+ -> 3C2H3O2- + 15H+ + 4MnO2 + 8H2O. Simplifying: 3C2H5OH + 4MnO4- + H+ -> 3C2H3O2- + 4MnO2 + 5H2O. Step 6: Convert to basic solution by adding OH- to both sides to neutralize H+: 3C2H5OH + 4MnO4- + H+ + OH- -> 3C2H3O2- + 4MnO2 + 5H2O + OH- => 3C2H5OH + 4MnO4- + H2O -> 3C2H3O2- + 4MnO2 + 5H2O + OH-. Step 7: Final simplification: 3C2H5OH + 4MnO4- -> 3C2H3O2- + 4MnO2 + 4H2O + OH-. The final balanced equation in basic solution demonstrates that the coefficient for water (H2O) in the product is 4.