Balancing Redox Solutions

1. a. VO+2 + 2H+1 + 1e V+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 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 + 2eMn+2 + 2H2O

 

 2. a. C2O4-2 + MnO4-1 Mn+2 + CO2

oxidation: C2O4-2 2CO2 + 2 e

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

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

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

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

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

 

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

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

oxidation: Sn+2 Sn+4 + 2 e

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

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

 

d. I2 + NO3-1 IO3-1 + NO

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

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

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

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

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

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

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

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

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

reduction: 4 H+ + 2e + NiO2 Ni+2 + 2 H2O

2 H+ + S2O3-2 + 2 NiO2 2 Ni+2 + 2 SO3-2 + H2O

g. NO2-1 + Al NH3 + AlO2-1

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

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

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

 3. In basic solution:

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

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

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

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

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

oxidation: Sn+2 Sn+4 + 2 e

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

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

c. I2 + NO3-1 IO3-1 + NO 

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

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

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

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

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

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

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

e. Cr + CrO4-2 Cr(OH)3

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

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

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