– (aq)+Br – (aq) → 2MnO (s)+BrO – (aq) Step : As the reaction occurs in the basic medium, and the ionic charges are not equal on both sides, add OH – ions on the right to make ionic charges equal. 2MnO – (aq) + Br – (aq) → 2MnO (s) + BrO – (aq) + 2OH – (aq) Step : Finally, count the hydrogen atoms and add appropriate number of water molecules (i.e. one H O molecule) on the left side to achieve balanced redox change. 2MnO – (aq) + Br – (aq) + H O(l) → 2MnO (s) + BrO – (aq) + 2OH – (aq) (b) Half Reaction Method: In this method, the two half equations are balanced separately and then added together to give balanced equation.
Suppose we are to balance the equation showing the oxidation of Fe + ions to Fe + ions by dichromate ions (Cr O ) – in acidic medium, wherein, Cr O – ions are reduced to Cr + ions. The following steps are involved in this task. Step : Produce unbalanced equation for the reaction in ionic form : Fe + (aq) + Cr O – (aq) → Fe + (aq) + Cr + (aq) ( . ) Step : Separate the equation into half- reactions: + + Oxidation half : Fe + (aq) → Fe + (aq) ( .
) + – + Reduction half : Cr O – (aq) → Cr + (aq) ( . ) Step : Balance the atoms other than O and H in each half reaction individually. Here the oxidation half reaction is already balanced with respect to Fe atoms. For the reduction half reaction, we multiply the Cr + by to balance Cr atoms.
– (aq) → Cr + (aq) ( . ) Step : For reactions occurring in acidic medium, add H O to balance O atoms and H + to balance H atoms. Thus, we get : – (aq) + 14H + (aq) →