In the balanced chemical equation MnO_4^- + B | vedclass

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(C) The oxidation half-reaction is: $Br^- + 6OH^- 	o BrO_3^- + 3H_2O + 6e^-$The reduction half-reaction is: $MnO_4^- + 2H_2O + 3e^- 	o MnO_2 + 4OH^-

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$2, 1, 2$

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(C) The oxidation half-reaction is: $Br^- + 6OH^- 	o BrO_3^- + 3H_2O + 6e^-$The reduction half-reaction is: $MnO_4^- + 2H_2O + 3e^- 	o MnO_2 + 4OH^-$To balance the electrons,multiply the reduction half-reaction by $2$:$2MnO_4^- + 4H_2O + 6e^- 	o 2MnO_2 + 8OH^-$Adding the two half-reactions:$2MnO_4^- + Br^- + 4H_2O 	o 2MnO_2 + BrO_3^- + 3H_2O + 2OH^-$Simplifying the water molecules:$2MnO_4^- + Br^- + H_2O 	o 2MnO_2 + BrO_3^- + 2OH^-$The coefficients of $MnO_4^-$,$BrO_3^-$,and $OH^-$ are $2, 1, 2$ respectively.

In the balanced chemical equation $MnO_4^- + Br^- + H_2O \to MnO_2 + BrO_3^- + OH^-$,the coefficients of $MnO_4^-$,$BrO_3^-$,and $OH^-$ are respectively:

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