Question 9 of 31intermediate🔧 ApplyShort Answer3 marks

Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.

Correct Answer

2MnO4−(aq)+Br−(aq)+H2O(l)→2MnO2(s)+BrO3−(aq)+2OH−(aq)2\text{MnO}_4^{-}(\text{aq}) + \text{Br}^{-}(\text{aq}) + \text{H}_2\text{O} (\text{l}) \rightarrow 2\text{MnO}_2(\text{s}) + \text{BrO}_3^{-}(\text{aq}) + 2\text{OH}^-(\text{aq})

Exercise: Problem 7.9 | Q: Problem 7.9 | (Page Page 13)
For More Understanding

Explanation

To balance the redox reaction in a basic medium, we first write the skeletal equation and balance the atoms and charges using the oxidation number method. The oxidation number of Mn decreases from +7 to +4, while Br increases from -1 to +5. To equalize the electron transfer, Mn species are multiplied by 2. Finally, hydroxide ions and water molecules are added to balance the charges and hydrogen atoms for the basic medium.

Solution Steps

  1. Step 1: The skeletal ionic equation is MnO4−(aq)+Br−(aq)→MnO2(s)+BrO3−(aq)\text{MnO}_4^{-}(\text{aq}) + \text{Br}^{-}(\text{aq}) \rightarrow \text{MnO}_2(\text{s}) + \text{BrO}_3^{-}(\text{aq})

  2. Step 2: Assign oxidation numbers for Mn and Br. Mn decreases from +7 to +4, and Br increases from -1 to +5.

  3. Step 3: Calculate the increase and decrease of oxidation number, and make the increase equal to the decrease. Multiply MnO4−\text{MnO}_4^{-} and MnO2\text{MnO}_2 by 2: 2MnO4−(aq)+Br−(aq)→2MnO2(s)+BrO3−(aq)2\text{MnO}_4^{-}(\text{aq}) + \text{Br}^{-}(\text{aq}) \rightarrow 2\text{MnO}_2(\text{s}) + \text{BrO}_3^{-}(\text{aq})

  4. Step 4: As the reaction occurs in the basic medium, and the ionic charges are not equal on both sides, add 2 OH−\text{OH}^- ions on the right to make ionic charges equal: 2MnO4−(aq)+Br−(aq)→2MnO2(s)+BrO3−(aq)+2OH−(aq)2\text{MnO}_4^{-}(\text{aq}) + \text{Br}^{-}(\text{aq}) \rightarrow 2\text{MnO}_2(\text{s}) + \text{BrO}_3^{-}(\text{aq}) + 2\text{OH}^-(\text{aq})

  5. Step 5: Finally, count the hydrogen atoms and add appropriate number of water molecules (i.e., one H2O\text{H}_2\text{O} molecule) on the left side to achieve balanced redox change: 2MnO4−(aq)+Br−(aq)+H2O(l)→2MnO2(s)+BrO3−(aq)+2OH−(aq)2\text{MnO}_4^{-}(\text{aq}) + \text{Br}^{-}(\text{aq}) + \text{H}_2\text{O} (\text{l}) \rightarrow 2\text{MnO}_2(\text{s}) + \text{BrO}_3^{-}(\text{aq}) + 2\text{OH}^-(\text{aq})

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