Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.

Use the half-reaction method to balance the following equations :
MnO4- + Br- = MnO2 + BrO3- (Acidic Solution )


Sagot :

Br- + 2MnO4- + 2H+ → BrO3- + 2MnO2 + H2O

Further explanation

Given

MnO4- + Br- = MnO2 + BrO3-

Required

The half-reaction

Solution

In acidic conditions :

1. Add the coefficient

2. Equalization O atoms (add H₂O) on the O-deficient side.  

3. Equalization H atoms (add H⁺ ) on the H -deficient side.  .  

4. Equalization of charge (add electrons (e) )

5. Equalizing the number of electrons and then adding the two half -reactions together

Oxidation : Br- → BrO3-  

Reduction : MnO4- → MnO2

  • Equalization O atoms

Br- + 3H2O → BrO3-  

MnO4- → MnO2 + 2H2O

  • Equalization H atoms

Br- + 3H2O → BrO3- + 6H+  

MnO4- + 4H+ → MnO2 + 2H2O

  • Equalization of charge

Br- + 3H2O → BrO3- + 6H+ + 6e-         x1

MnO4- + 4H+ + 3e- → MnO2 + 2H2O x2

  • Equalizing the number of electrons and then adding the two half -reactions together

Br- + 3H2O → BrO3- + 6H+ + 6e-  

2MnO4- + 8H+ + 6e- → 2MnO2 + 4H2O

Br- + 2MnO4- + 3H2O + 8H+ + 6e- → BrO3- + 2MnO2 + 6H+ + 4H2O + 6e-

Br- + 2MnO4- + 2H+ → BrO3- + 2MnO2 + H2O