Table of contents

This is an oxidation-reduction (redox) reaction:

6 Br0 + 6 e- → 6 Br-I (reduction)

2 B0 - 6 e- → 2 BIII (oxidation)

Br2 is an oxidizing agent, B is a reducing agent.

Word equation

boron + bromine → boron tribromide

Input interpretation

2B + 3Br2 → BBr3

Boron bromine tribromide

Balanced equation

Balance the chemical equation algebraically:

2B + 3Br2 → BBr3

Add stoichiometric coefficients, c, to the reactants and products:

2B + 3Br2 → BBr3

Set the number of atoms in the reactants equal to the number of atoms in the products for B and Br:

B: c1 = c3

Br: 2c2 = 3c3

Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c1 = 1 and solve the system of equations for the remaining coefficients:

c1 = 1

c2 = 3/2

c3 =1

Multiply by the least common denominator, 2, to eliminate fractional coefficients:

c1 = 2

c2 = 3 

c3 = 2

Substitute the coefficients into the chemical reaction to obtain the balanced



2B + 3Br2 → BBr3