An equation like this is confusing because I^-(on the left) is going to I2 (on the right) but it then dissolves in KI to make KI3 so let's simplify it by taking the KI out of it, balance it, then put the KI back in.
2I^- --> I2
I is -2 on left and 0 on right. Balance with e change.
2I^- ==> I2 + 2e
Charge on left is 2- and 2- on right. No OH^- needed.
IO4^- --> IO3^-
I on left is +7 and +5 on right. Add e
IO4^- + 2e ==> IO3^-
Charge on left is -3 and on right -1. Add OH to balance
IO4^- + 2e ==> IO3^- + 2OH^-
Now add H2O to balance.
IO4^- + 2e + H2O ==> IO3^- + 2OH^-
Both half rxns have 23 change; therefore, we can add the two to obtain
2I^- + IO4^- + H2O ==> I2 + IO3^- + 2OH^-
Now we add the I^- back in to make that I2 on the right I3^- and the final becomes
3I^- + IO4^- + H2O ==> I3^- + IO3^- + 2OH^-
Balance the following equation in basic solution. Phases are optional.
I{-}+IO4{-} --> I3{-}+IO3{-}
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