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(D) The given skeletal equation is $2MnO_4^- (aq) + Br^- (aq) 	o 2MnO_2 (s) + BrO_3^- (aq)$.Step $1$: Balance the charge. The total charge on the lef

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Both $(a)$ and $(b)$

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(D) The given skeletal equation is $2MnO_4^- (aq) + Br^- (aq) 	o 2MnO_2 (s) + BrO_3^- (aq)$.Step $1$: Balance the charge. The total charge on the left is $-3$ and on the right is $-1$. To balance the charge in a basic medium,we add $2 \ OH^-$ ions to the right side.$2MnO_4^- (aq) + Br^- (aq) 	o 2MnO_2 (s) + BrO_3^- (aq) + 2 \ OH^- (aq)$Step $2$: Balance the hydrogen atoms. There are $2 \ H$ atoms on the right,so we add one $H_2O$ molecule to the left side.$2MnO_4^- (aq) + Br^- (aq) + H_2O (l) 	o 2MnO_2 (s) + BrO_3^- (aq) + 2 \ OH^- (aq)$Thus,both $(a)$ and $(b)$ are correct steps to balance the reaction.

Consider the following reaction occurring in basic medium $2MnO_4^- (aq) + Br^- (aq) \to 2MnO_2 (s) + BrO_3^- (aq)$. How can the above reaction be balanced further?

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