Balance the following redox equations by the ion-electron method. (Use the lowest possible coefficients.)?

Question

Br2 + OH - BrO3- + Br - + H2O (in basic solution)

Answer 1

This is called a disproportionation reaction, since there is one reactant (elemental Br, valence 0) which is both oxidized (Br+5 in bromate) and reduced (Br-1 in bromide). This is done in BASIC solution, so if there is a shortfall in O and H in the reaction, OH- will balance it out. That's the game plan for the method.
Oxidize: Br -> Br+5 + 5e
Reduce: Br + e -> Br-
Balance e: 5Br + 5e-> 5 Br-
OH 6 (OH)- -> 3O-2 + 3 H2O
Put it together:

3 Br2 + 6 OH- -> BrO3- +5 Br- + 3H2O

Answer 2

an prolonged technique of fixing those by ability of ion electron technique : (d)STEP1: CL- = CL2 CLO3- + CLO2 STEP2: 2CL- = CL2 CLO3- + 2H+ = CLO2 + H2O STEP3: 2CL- = CL2 + 2e- CLO3- + 2H+ +4e- = CLO2 + H2O STEP4: multiplying first equation by ability of two for canceling the electron 4CL- = 2CL2 + 4e- CLO3- + 2H+ +4e- = CLO2 + H2O STEP5: 4CL- CLO3- + 2H+ = 2CL2 + CLO2 + H2O