Question

When the following equation is balanced properly under acidic conditions, what are the coefficients of the species shown? Xe + BrO_3^+ rightarrow HXeO_4^- + Br_2 Water appears in the balanced equation as a (reactant, product, neither) with a coefficient of How many electrons are transferred in this reaction?

Answer 1

Split

Xe --> HXeO4-

BrO3- --> Br2

Balance Br

Xe --> HXeO4-

2BrO3- --> Br2

Balance O

4H2O + Xe --> HXeO4-

2BrO3- --> Br2 + 6H2O

Balanc eH

4H2O + Xe --> HXeO4- + 7H+

12H+ + 2BrO3- --> Br2 + 6H2O

Balance charges

4H2O + Xe --> HXeO4- + 7H+ + 6e-

10e- + 12H+ + 2BrO3- --> Br2 + 6H2O

Balance e-

20H2O + 5Xe --> 5HXeO4- + 35H+ + 30e-

30e- + 36H+ + 6BrO3- --> 3Br2 + 18H2O

Add both equations

20H2O + 5Xe + 30e- + 36H+ + 6BrO3 --> 5HXeO4- + 35H+ + 30e- + 3Br2 + 18H2O

cancel common terms

2H2O + 5Xe + H+ + 6BrO3 --> 5HXeO4- + 3Br2

the water apears with a coefficient of +2 in the reactants

the electrons transferred were 30 e-