How do you balance acidic and basic solutions?

First, always pretend the solutions are acidic, because those are easier to balance since you only have hydrogen to worry about. Then you can convert to basic, which is a simple enough step if the acidic balancing is done. Figure out which element in each reaction is oxidised, and which is reduced. Write half reactions for those elements, balance the electrons and other elements, then combine them, eliminating the electrons. To balance O, add H2O to the other side of the arrow displaying the O atoms, and to balance H, add H^+. Cu + NO3^- → Cu^2+ + NO Cu is oxidised and changes in oxidation number from 0 to +2, losing 2 electrons. N is reduced, lowering from +5 to +2, gaining 3 electrons. Cu → Cu^2+ + 2e^- (balanced) NO3^- + 3e^- → NO (not balanced, add H2O and H^+) NO3^- + 4H^+ + 3e^- → NO + 2H2O (balanced) Multiply the first equation by 3 and the second by 2 to balance the electrons so they cancel. You end up with: 3Cu + 2NO3^- + 8H^+ → 3Cu^2+ + 2NO + 4H2O For the second equation, do the same in acid first. C goes from -2 to +4, and Mn goes from +7 to +4. CH3OH + H2O → CO2 + 6H^+ + 6e^- MnO4^- + 4H^+ + 3e^- → MnO2 + 2H2O Add, and cancel identical formulae on either side (e^-, H^+ and H2O in this case). CH3OH + H2O + 2MnO4^- + 8H^+ + 6e^- → CO2 + 6H^+ + 6e^- + 2MnO2 + 4H2O CH3OH + 2MnO4^- + 2H^+ → CO2 + 2MnO2 + 3H2O To convert acidic to basic, add just as many OH^- ions to both sides of the equation as there are H^+ ions. This will form H2O and allow you to remove the acidic conditions. There are 2 H^+ ions here, so add 2 OH^- ions. Then simplify again. CH3OH + 2MnO4^- + 2H^+ + 2OH^- → CO2 + 2MnO2 + 3H2O + 2OH^- CH3OH + 2MnO4^- + 2H2O → CO2 + 2MnO2 + 3H2O + 2OH^- CH3OH + 2MnO4^- → CO2 + 2MnO2 + H2O + 2OH^- This is now balanced.