Question

Balance the following Redox reactions in basic solution

a) NiO2 (s) + Fe (s) → Ni(OH)2 (s)+ Fe(OH)2 (s)

b) OH- (aq) + NO2 (g) → NO3- (aq) + NO2- (aq) + H20 (l)

Answer 1

a) NiO2 (s) + Fe (s) → Ni(OH)2 (s)+ Fe(OH)2 (s)

split

NiO2 = Ni(OH)2

Fe = Fe(OH)2

balance O

NiO2 = Ni(OH)2

2H2O + Fe = Fe(OH)2

balace H+

2H+ + NiO2 = Ni(OH)2

2H2O + Fe = Fe(OH)2 + 2H+

balance charges

2e- 2H+ + NiO2 = Ni(OH)2

2H2O + Fe = Fe(OH)2 + 2H+ + 2e-

add all

2e- 2H+ + NiO2+2H2O + Fe = Ni(OH)2+ Fe(OH)2 + 2H+ + 2e-

cancel common terms

NiO2+2H2O + Fe  = Ni(OH)2+ Fe(OH)2

therefore, we only require 2 H2O molecules

b)

b) OH- (aq) + NO2 (g) → NO3- (aq) + NO2- (aq) + H20 (l)

NO2 = NO3-

NO2 = NO2-

balance O

H2O + NO2 = NO3-

NO2 = NO2-

baalnce H

H2O + NO2 = NO3- + 2H+

NO2 = NO2-

balance e-

H2O + NO2 = NO3- + 2H+ +e-

e-+ NO2 = NO2-

add all

e-+ NO2 + H2O + NO2 = NO3- + 2H+ +e- +  NO2-

cancel common terms

H2O + 2NO2 = NO3- + 2H+NO2-