I'm having trouble balancing this equation in a basic solution.

Cr(OH)3 + Br2 -----> CrO4{2-} + Br{-}

Do I do start this with half reactions


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  1. I always revert to half reactions when I have trouble. Most students forget to compare apples with applies; i.e., you should make that 2Br^ on the right to sstart. But let's do try this way.
    Cr is +3 on the left and +6 on the right so a gain of 3e.
    Br is zero on the left and 2Br^- is -2 on the right or a loss of 2. Cr x 2 and Br x 3 gives
    2Cr(OH)3 + 3Br2 ==> 2CrO4^2- + 6Br^-

    The charge on the right is -10 and zero on the left so add 10 OH to the left.
    10 OH^- + 2Cr(OH)3 + 3Br2 ==> 2CrO4^2- + 6Br^-.

    On the left there are 16H so on the right place 8 H2O
    10 OH^- + 2Cr(OH)3 + 3Br2 ==> 2CrO4^2- + 6Br^- + 8H2O

    See if that checks. It looks ok to me.

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