Question

4. Balance the following redox equations using the half reaction method under basic conditions:

a. H+ + Cu(s) + NO3 - → Cu2+ + NO2 + H2O

b. MnO4 - + Cl- + H+ → Mn2+ + Cl2 + H2O

c. Zn(s) + ClO3 - + H+ → Zn2+ + Cl- + H2O

d. Cl- + H+ + Cr2O7 2- → Cr3+ + Cl2 + H2O

e. Mn2+ + NaBiO3 + H+ → MnO4 - + Bi3+ + Na+ + H2O

Answer 1

a)

Cu(s) + H⁺(aq) + NO₃⁻(aq) −→ Cu²⁺(aq) + NO(g)

Cu(s) + H (aq) + NO₃ (aq) −→ Cu     (aq) + NO(g)

ox:       Cu −→ Cu²⁺red: NO₃⁻−→ NO

ox:       Cu −→ Cu²⁺

red:     NO₃⁻−→ NO + 2H₂O

red:     NO₃⁻ + 4H⁺ −→ NO + 2H₂O

ox: Cu −→ Cu²⁺ + 2e⁻ red:            NO₃⁻ + 4H⁺ + 3e⁻−→ NO + 2H₂O

3 × ox: 3Cu −→ 3Cu²⁺ + 6e⁻

2 × red:           2NO₃⁻ + 8H⁺ + 6e⁻−→ 2NO + 4H₂O

3Cu + 2NO₃⁻ + 8H⁺ + 6e⁻−→ 3Cu²⁺ + 6e⁻ + 2NO + 4H₂O

3Cu + 2NO₃⁻ + 8H⁺ −→ 3Cu²⁺ + 2NO + 4H₂O

3Cu(s) + 2NO₃⁻(aq) + 8H⁺(aq) −→ 3Cu²⁺(aq) + 2NO(g) + 4H₂O(l)

b)

MnO4- ----> Mn2+ (reduction)
Cl- --------> Cl2 (oxidation) .

MnO4- ----> Mn2+
MnO4- -----> Mn2+ + 4H2O (add water to the side that needs O)
8H+ + MnO4- -----> Mn2+ + 4H2O (add H+ to the side that needs H)
5e- + 8H+ + MnO4- -----> Mn2+ + 4H2O (add e- to balance charge)

Cl- ----> Cl2
2Cl- ----> Cl2 + 2e-

2(5e- + 8H+ + MnO4- -----> Mn2+ + 4H2O )
5(2Cl- ----> Cl2 + 2e-)

16H+ + 2MnO4- + 10Cl- ---> 2Mn2+ + 5Cl2 + 8H2O

d)

Cr2O7(2-)(aq) -> 2Cr3+
Cr = 6+ -> since each each O is -2. 2X -2(7) =-2 -> 2X = 12 -> X = 6
Cr = 3+ ->the only molecule in it.
Cl-(aq) -> Cl2(g)
Cl-(aq) = -1
Cl2(g) = 0

6e- + Cr2O7(2-)(aq) -> Cr3+

2Cl-(aq) -> Cl2(g) + 2e-


6e- + Cr2O7(2-)(aq) -> 2Cr3+


6e- + Cr2O7(2-)(aq) + 14 H+ -> 2Cr3+ + 7 H2O.


6Cl-(aq) -> 3Cl2(g) + 6e-

6e- + Cr2O7(2-)(aq) + 14 H+ -> 2Cr3+ + 7 H2O.

Cr2O7(2-)(aq) + 14 H+ + 6Cl-(aq) > 2Cr3+ + 7 H2O + 3Cl2(g)

e)

Start with the Mn2+

Mn2+ --> MnO4-

Mn2+ + H2O --> MnO4- + H+

Mn2+ + 4H20 --> MnO4- + 8H+

Mn2+ + 4H2O --> MnO4- + 8H+ + 5e-

NaBiO3 --> Bi3+ + Na+ + 3H2O

NaBiO3 + 6H+ --> Bi3+ + Na+ + 3H2O

NaBiO3 + 6H+ + 2e- --> Bi3+ + Na+ + 3H2O

2Mn2+ + 8H2O + 5NaBiO3 + 30H+ --> 2MnO4- + 16H+ 5Bi3+ + 5Na+ + 15H2O

2Mn2+ + 5NaBiO3 + 14H+ --> 2MnO4- + 5Bi3+ + 5Na+ + 7H2O