Balancing Redox Solutions

1.       a. VO+2 + 2H+1 + 1e àV+3 + H2O

          Why?

          (1)Participating atom (V) is already balanced. (one V on each side)

          (2)Oxidation number has to be balanced with number of electrons.

In VO+2 the oxidation number of V is +4 because v +2(-2) = +2. To move to +3 it has to gain one electron.

(3)Charge must be balanced. By adding one electron to the left side, we have created an overall charge of +1 (+2 from VO+2 and –1 from 1e add up to +1) on the left hand side of the equation, but there is a +3 charge on the right hand side. So we add 2H+1 on the left hand side to create a charge of 3 on each side.

(4) Hydrogens have to be balanced. So we add 1 H2O on the right hand side.

 

b.  H3PO3 + H2O à H3PO4 + 2H+1 + 2e

Why are two electrons being lost? In H3PO3, P is at (+3); but in H3PO4 it's at ( + 5), so it's losing two electrons.

c.     MnO2 + 4H+1 + 2eàMn+2 + 2H2O

 

 2.    a.          C2O4-2 + MnO4-1 à Mn+2 + CO2

oxidation:       C2O4-2 à 2CO2 + 2 e

reduction:       MnO4-1 + 5 e + 8H+1à Mn+2 + 4 H2O

           

5 C2O4-2 + 2 MnO4-1 + 16 H+1 à 10 CO2 + 2 Mn+2 + 8 H2O

b.         MnO2 + H+1 + NO2-1 à NO3-1 + Mn+2 + H2O

oxidation:       NO2-1 + H2O à NO3-1 + 2 e + 2 H+1

reduction:       MnO2 + 2 e + 4 H+1 à Mn+2+ 2 H2O

 

NO2-1 + MnO2 + 2 H+1 à NO3-1 + Mn+2 + H2O

c.         Sn+2 + Cr2O7-2 à Sn+4 + Cr+3

oxidation:       Sn+2 à Sn+4 + 2 e

reduction:       Cr2O7-2 + 6 e + 14H+1 à 2Cr+3+ 7H2O

3Sn+2 + Cr2O7-2 + 14 H+1 à 3Sn+4 + 2 Cr+3 + 7 H2O

 

d.         I2 + NO3-1 à IO3-1 + NO

                        oxidation:       6 H2O + I2 à 2 IO3-1 + 10 e + 12H+1

                        reduction:       NO3-1 + 3 e- + 4H+1à NO + 2 H2O

                        3 I2 + 10 NO3-1 + 4 H+1 à 6 IO3-1 + 10 NO + 2 H2O

e.         MnO4-1 + NO2-1 à MnO2 + NO3-1

            oxidation:       H2O  + NO2-1 à NO3-1 + 2e + 2 H+

reduction:       4 H+ + 3 e + MnO4-1à MnO2 + 2 H2O

                        2 H+ + 2 MnO4-1 + 3NO2-1 à 3NO3-1 + 2 MnO2 + H2O

f.          NiO2 + S2O3-2 à Ni+2 + SO3-2

                        oxidation:       3 H2O + S2O3-2 à 2 SO3-2 + 4e + 6 H+

                                        reduction:       4 H+ + 2e + NiO2à Ni+2 + 2 H2O

                        2 H+ + S2O3-2 + 2 NiO2à 2 Ni+2 + 2 SO3-2 + H2O

g.         NO2-1 + Al à NH3 + AlO2-1

oxidation:       2 H2O +Al à AlO2-1 + 3e + 4 H+

reduction:       7H+ + NO2-1 + 6e à NH3 + 2 H2O

2 H2O + 2 Al + NO2-1 à NH3 +  2 AlO2-1 + H+

 3. In basic solution:

a.         MnO2 + NO2-1 à Mn+2 + NO3-1

                        oxidation:       2OH-1 + NO2-1 à NO3-1 + 2 e + H2O

                        reduction:       MnO2+ 2 e + 2H2O à Mn+2 + 4OH-1

NO2-1 + MnO2+ H2O à NO3-1 + Mn+2+ 2 OH-1

b.         Sn+2 + Cr2O7-2  Sn+4 + Cr+3

oxidation:       Sn+2 à Sn+4 + 2 e

reduction:       Cr2O7-2 + 6 e + 7H2O à 2Cr+3+ 14OH-1

3Sn+2 + Cr2O7-2  + 7 H2O à  3 Sn+4 + 2 Cr+3 + 14OH-1

c.      I2 + NO3-1 à IO3-1 + NO 

               oxidation:       I2 + 12 OH- à 2 IO3-1 + 10 e + 6 H2O

                        reduction:       NO3-1 + 3e + 2 H2Oà NO + 4 OH-

               3 I2 + 10 NO3-1 + 2 H2Oà 6 IO3-1 + 10 NO + 4 OH-

d.         NiO2 + S2O3-2 à Ni(OH)2 + SO3-2

                        oxidation:       S2O3-2 + 6 OH-1 à 2SO3-2 + 4 e + 3 H2O

                        reduction:       NiO2 + 2 e + 2 H2O à Ni(OH)2 + 2 OH-1

                        S2O3-2 + 2 NiO2 + H2O + 2 OH-1à 2 SO3-2 + 2 Ni(OH)2

e.         Cr + CrO4-2 à Cr(OH)3

                        oxidation:       Cr + 3 OH-1à Cr(OH)3 + 3 e

                        reduction:       CrO4-2 + 3 e + 4H2O à Cr(OH)3 +5 OH-1

                        Cr + CrO4-2 + 4 H2O à 2 Cr(OH)3 + 2OH-1