Question

Balance the following oxidation-reduction reaction equations using 1/2 reaction method:

please show work

Sb+NO3= Sb2O5+NO

NO2 = NO3 + NO

Br2+NH3=NH4BR+N2

Answer 1

The first redox reaction:

Sb + NO3= Sb2O5 + NO

Oxidation half reaction:

Sb = Sb2O5

2Sb = Sb2O5

2Sb + 5H2O = Sb2O5

2Sb + 5H2O = Sb2O5 + 10H⁺

(2Sb + 5H2O = Sb2O5 + 10H⁺ + 10e^-)*2

Redox half reaction:

NO3 = NO

NO3 = NO + 2H2O

NO3 + 4H⁺ = NO + 2H2O

(NO3 + 4H⁺ + 4e^- = NO + 2H2O)*5

The complete balanced redox reaction:

4Sb + 5NO3= 2Sb2O5 + 5NO

The second redox reaction:

NO2 = NO3^- + NO

The oxidation half reaction:

NO2 = NO3^-

NO2 + H2O = NO3^-

NO2 + H2O = NO3^- + 2H⁺

(NO2 + H2O = NO3^- + 2H⁺ + e^-)*2

The Redox half reaction:​

NO2 = NO

NO2 = NO + H2O

NO2 + 2H⁺ = NO + H2O

(NO2 + 2H⁺ + 2e^- = NO + H2O)*1

The complete balanced redox reaction:

3NO2 + H2O = 2NO3^- + NO + 2H⁺

The third redox reaction:

Br2 + NH3 = NH4Br + N2

Oxidation half reaction:

NH3 = N2

2NH3 = N2

2NH3 = N2 + 6H⁺

(2NH3 = N2 + 6H⁺ + 6e^-)*1

Redox half reaction:

Br2 = Br^-

Br2 = 2Br^-

(Br2 + 2e^- = 2Br^-)*3

The complete balanced redox reaction:

3Br2 + 2NH3 = 6Br^- + N2 + 6H⁺

(Or)

3Br2 + 8NH3 = 6NH4Br + N2