How do you balance redox reactions in basic solution?

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    WARNING this is a long answer.
    In basic solution you balance redox equations as if they were in acid. At the end you use OH to convert to base.
    EXAMPLE:
    Balance the following equation in basic solution:
    MnO + CN MnO + CNO
    Solution:
    Step 1: Separate the equation into two half-reactions.
    MnO MnO
    CN CNO
    Step 2: Balance all atoms other than H and O.
    Done
    Step 3: Balance O by adding HO to the deficient side.
    MnO MnO+ 2HO
    CN+ HO CNO
    Step 4: Balance H by adding H to the deficient side.
    MnO+ 4H MnO+ 2HO
    CN+ HO CNO + 2H
    Step 5: Balance charge by adding electrons to the more positive side.
    MnO+ 4H + 3e MnO+ 2HO
    CN+ HO CNO + 2H + 2e
    Step 6: Multiply each half-reaction by a factor that gives the lowest common multiple of the electrons transferred.
    In this case the lowest common multiple of 3 and 2 is 6.
    We multiply the first half-reaction by 2 and the second half-reaction by 3.
    2 [MnO+ 4H + 3e MnO+ 2HO]
    3 [CN+ HO CNO + 2H + 2e]
    Step 7: Add the two half-reactions cancelling any like terms.
    2MnO+ 3CN+ 2H 2MnO + 3CNO + HO
    This is the balanced equation in acid solution. We must now convert to base solution.
    Step 8: Add enough multiples of the equations H + OH HO or
    HO H + OH to cancel the H in the redox equation cancelling like terms.
    2MnO+ 3CN+ 2H 2MnO + 3CNO + HO
    2HO 2H + 2OH
    2MnO+ 3CN+ HO 2MnO + 3CNO + 2OH
    Step 9: Check that atoms balance.
    On the left: 2 Mn; 9 O; 3 C; 3 N; 2 H
    On the right: 2 Mn; 9 O; 3 C; 3 N; 2 H
    Step 10: Check that charges balance.
    On the left: 2- + 3- = 5-
    On the right: 3- + 2- = 5-
    The balanced equation is
    2MnO+ 3CN+ HO 2MnO + 3CNO + 2OH

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