Balancing redox reactions in basic solution
Problems 11 - 25

Problem #11: ClO₃¯ + N₂H₄ ---> NO + Cl¯

Solution:

1) Half-reactions:

ClO₃¯ ---> Cl¯
N₂H₄ ---> NO

2) Balance in acidic solution first:

6e¯ + 6H⁺ + ClO₃¯ ---> Cl¯ + 3H₂O
2H₂O + N₂H₄ ---> 2NO + 8H⁺ + 8e¯

3) Equalize electrons:

24e¯ + 24H⁺ + 4ClO₃¯ ---> 4Cl¯ + 12H₂O
6H₂O + 3N₂H₄ ---> 6NO + 24H⁺ + 24e¯

4) Add:

4ClO₃¯ + 3N₂H₄ ---> 4Cl¯ + 6NO + 6H₂O

Note that there is no hydroxide in the final answer. Does that mean this reaction does not require a basic solution? The answer is no. The hydroxide is functioning as a catalyst to the reaction. Some hydroxide is used up to start the rection, but it is regenerated by the end of the reaction. So, while hydroxide is necessary for the reaction, it does not appear in the overall equation.

Problem #12: H₂O₂ + Cl₂O₇ ---> ClO₂¯ + O₂

Solution:

1) Half-reactions:

H₂O₂ ---> O₂
Cl₂O₇ ---> ClO₂¯

2) Balance in acidic solution:

H₂O₂ ---> 2H⁺ + O₂ + 2e¯
8e¯ + 6H⁺ + Cl₂O₇ ---> 2ClO₂¯ + 3H₂O

3) Equalize electrons:

4H₂O₂ ---> 8H⁺ + 4O₂ + 8e¯
8e¯ + 6H⁺ + Cl₂O₇ ---> 2ClO₂¯ + 3H₂O

4) Add:

Cl₂O₇ + 4H₂O₂ ---> 2ClO₂¯ + 4O₂ + 3H₂O + 2H⁺

5) Add two hydroxides to each side:

2OH¯ + Cl₂O₇ + 4H₂O₂ ---> 2ClO₂¯ + 4O₂ + 5H₂O

Problem #13: HXeO₄¯ + Pb ---> Xe + HPbO₂¯

Solution:

1) Half-reactions:

HXeO₄¯ ---> Xe
Pb ---> HPbO₂¯

2) Balance in acid:

6e¯ + 7H⁺ + HXeO₄¯ ---> Xe + 4H₂O
2H₂O + Pb ---> HPbO₂¯ + 3H⁺ + 2e¯

3) Equalize electrons:

6e¯ + 7H⁺ + HXeO₄¯ ---> Xe + 4H₂O
6H₂O + 3Pb ---> 3HPbO₂¯ + 9H⁺ + 6e¯

4) Add:

2H₂O + HXeO₄¯ + 3Pb ---> Xe + 3HPbO₂¯ + 2H⁺

5) Convert to basic with two hydroxides on each side:

2OH¯ + HXeO₄¯ + 3Pb ---> Xe + 3HPbO₂¯

Problem #14: HPbO₂¯ + Cr(OH)₃ ---> Pb + CrO₄²¯

Solution:

1) Half-reactions:

HPbO₂¯ ---> Pb
Cr(OH)₃ ---> CrO₄²¯

2) Balance in acidic solution:

2e¯ + 3H⁺ + HPbO₂¯ ---> Pb + 2H₂O
H₂O + Cr(OH)₃ ---> CrO₄²¯ + 5H⁺ + 3e¯

3) Equalize electrons:

6e¯ + 9H⁺ + 3HPbO₂¯ ---> 3Pb + 6H₂O
2H₂O + 2Cr(OH)₃ ---> 2CrO₄²¯ + 10H⁺ + 6e¯

4) Add:

3HPbO₂¯ + 2Cr(OH)₃ ---> 3Pb + 2CrO₄²¯ + H⁺ + 4H₂O

5) Add one hydroxide to each side:

3HPbO₂¯ + 2Cr(OH)₃ + OH¯ ---> 3Pb + 2CrO₄²¯ + 5H₂O

Problem #15: Basic solutions of thallium(III) oxide and hydroxylamine are mixed together to produce thallium(I) hydroxide and nitrogen gas

Solution:

1) The full equation followed by the half-reactions:

Tl₂O₃ + NH₂OH ---> TlOH + N₂

Tl₂O₃ ---> TlOH
NH₂OH ---> N₂

2) Balance in acidic solution:

4e¯ + 4H⁺ + Tl₂O₃ ---> 2TlOH + H₂O
2NH₂OH ---> N₂ + 2H₂O + 2H⁺ + 2e¯

3) Equalize electrons:

4e¯ + 4H⁺ + Tl₂O₃ ---> 2TlOH + H₂O
4NH₂OH ---> 2N₂ + 4H₂O + 4H⁺ + 4e¯

4) Add:

Tl₂O₃ + 4NH₂OH ---> 2TlOH + 2N₂ + 5H₂O

Problem #16: Mg + NiO₂ ---> Mg(OH)₂ + Ni(OH)₂

Solution:

1) Half-reactions:

Mg ---> Mg(OH)₂
NiO₂ ---> Ni(OH)₂

2) Balance first in basic and second in acidic:

2OH¯ + Mg ---> Mg(OH)₂ + 2e-
2e¯ + 2H⁺ + NiO₂ ---> Ni(OH)₂

3) Note that the electrons are already equalized. Note also that hydrogen ion and hydroxide ion (both on the left-hand side of the two half-reactions) react immediately to produce water. That leads to the final answer:

Mg + NiO₂ + 2H₂O ---> Mg(OH)₂ + Ni(OH)₂

Problem #17: Cr(OH)₄¯ + H₂O₂ ---> CrO₄²¯ + H₂O

Solution:

1) Half-reactions:

Cr(OH)₄¯ ---> CrO₄²¯
H₂O₂ ---> H₂O

2) Balance as if in acidic solution:

Cr(OH)₄¯ ---> CrO₄²¯ + 4H⁺ + 3e¯
2e¯ + 2H⁺ + H₂O₂ ---> 2H₂O

3) Equalize electrons:

2Cr(OH)₄¯ ---> 2CrO₄²¯ + 8H⁺ + 6e¯
6e¯ + 6H⁺ + 3H₂O₂ ---> 6H₂O

4) Add:

2Cr(OH)₄¯ + 3H₂O₂ ---> 2CrO₄²¯ + 2H⁺ + 6H₂O

5) Change to basic by adding two hydroxides to each side:

2OH¯ + 2Cr(OH)₄¯ + 3H₂O₂ ---> 2CrO₄²¯ + 8H₂O

The two H⁺ and the two OH¯ on the right combine to form two waters to make a total of eight waters.

Problem #18: CrO₄²¯ + HSnO₂¯ + H₂O ---> CrO₂(s) + H₂SnO₃(aq) + OH¯(aq)

1) Balanced half-reactions:

4H⁺ + CrO₄²¯ + 2e¯ ---> CrO₂ + 2H₂O
HSnO₂¯ + H₂O ---> H₂SnO₃ + H⁺ + 2e¯

2) Add, then simplify:

4H⁺ + CrO₄²¯ + HSnO₂¯ + H₂O ---> CrO₂ + H₂SnO₃ + H⁺ + 2H₂O

3H⁺ + CrO₄²¯ + HSnO₂¯ ---> CrO₂ + H₂SnO₃ + H₂O

3) Add 3OH¯ to each side to neutralize H⁺ since the reaction is in basic solution, then simplify:

3H₂O + CrO₄²¯ + HSnO₂¯ ---> CrO₂ + H₂SnO₃ + H₂O + 3OH¯

2H₂O + CrO₄²¯ + HSnO₂¯ ---> CrO₂ + H₂SnO₃ + 3OH¯ <--- that's the balanced equation

Problem #19: HXeO₄¯(aq) ---> XeO₆⁴¯(aq) + Xe(g)

Solution:

1) Half-reactions:

HXeO₄¯(aq) ---> XeO₆⁴¯(aq)
HXeO₄¯(aq) ---> Xe(g)

2) Balance in acidic:

2H₂O + HXeO₄¯(aq) ---> XeO₆⁴¯(aq) + 5H⁺ + 2e¯
6e¯ + 7H⁺ + HXeO₄¯(aq) ---> Xe(g) + 4H₂O

3) Equalize electrons:

6H₂O + 3HXeO₄¯(aq) ---> 3XeO₆⁴¯(aq) + 15H⁺ + 6e¯
6e¯ + 7H⁺ + HXeO₄¯(aq) ---> Xe(g) + 4H₂O

4) Add:

2H₂O + 4HXeO₄¯(aq) ---> 3XeO₆⁴¯(aq) + Xe(g) + 8H⁺

5) Add eight hydroxides to each side:

8OH¯ + 2H₂O + 4HXeO₄¯(aq) ---> 3XeO₆⁴¯(aq) + Xe(g) + 8H₂O

6) Remove water for the final answer (with state symbols for everything):

8OH¯(aq) + 4HXeO₄¯(aq) ---> 3XeO₆⁴¯(aq) + Xe(g) + 6H₂O(ℓ)

Problem #20: S₂O₄²¯ + O₂(g) ---> SO₄²¯

Solution:

1) Half-reactions:

S₂O₄²¯ ---> SO₄²¯
O₂(g) ---> OH¯

No hydroxide is indicated in the problem but, since we know the reaction to be in basic solution, we can add it in.

2) Balance:

4H₂O + S₂O₄²¯ ---> 2SO₄²¯ + 8H⁺ + 6e¯
4e¯ + 2H⁺ + O₂(g) ---> 2OH¯ <--- note the hydrogen ion on one side and the hydroxide on the other. Unusual.

3) Equalize electrons:

8H₂O + 2S₂O₄²¯ ---> 4SO₄²¯ + 16H⁺ + 12e¯
12e¯ + 6H⁺ + 3O₂(g) ---> 6OH¯

4) Add:

8H₂O + 2S₂O₄²¯ + 3O₂(g) ---> 4SO₄²¯ + 10H⁺ + 6OH¯

5) Add four hydroxides:

4OH¯ + 8H₂O + 2S₂O₄²¯ + 3O₂(g) ---> 4SO₄²¯ + 10H₂O

Note that I combined the hydrogen ion and hydroxide ion on the right into water.

6) Remove water for the final answer:

4OH¯ + 2S₂O₄²¯ + 3O₂(g) ---> 4SO₄²¯ + 2H₂O

Here's an incorrect answer to this problem. The answerer failed to reduce the O₂ properly and wound up with an answer that is not balanced for charge or for oxygen

Problem #21: SO₃^2- + Cr₂O₇^2- ---> Cr³⁺ + SO₄^2-

Solution:

1) Half-reactions:

SO₃^2- ---> SO₄^2-
Cr₂O₇^2- ---> Cr³⁺

2) Balance in acidic solution first:

H₂O + SO₃^2- ---> SO₄^2- + 2H⁺ + 2e¯
6e¯ + 14H⁺ + Cr₂O₇^2- ---> 2Cr³⁺ + 7H₂O

3) Equalize electrons:

3H₂O + 3SO₃^2- ---> 3SO₄^2- + 6H⁺ + 6e¯
6e¯ + 14H⁺ + Cr₂O₇^2- ---> 2Cr³⁺ + 7H₂O

4) Add:

8H⁺ + 3SO₃^2- + Cr₂O₇^2- ---> 2Cr³⁺ + 3SO₄^2- + 4H₂O

5) Convert to basic:

4H₂O + 3SO₃^2- + Cr₂O₇^2- ---> 2Cr³⁺ + 3SO₄^2- + 8OH¯

6) Since chromium(III) hydroxide precipitates, this may the desired answer:

4H₂O + 3SO₃^2- + Cr₂O₇^2- ---> 2Cr(OH)₃ + 3SO₄^2- + 2OH¯

Problem #22: Sb₂O₃(s) + NO₃¯(aq) ---> H₃SbO₄(aq) + NO(g)

Solution:

1) Half-reactions:

Sb₂O₃(s) ---> H₃SbO₄(aq)
NO₃¯(aq) ---> NO(g)

2) Balance in acidic:

5H₂O + Sb₂O₃(s) ---> 2H₃SbO₄(aq) + 4H⁺ + 4e¯
3e¯ + 4H⁺ + NO₃¯(aq) ---> NO(g) + 2H₂O

3) Equalize electrons:

15H₂O + 3Sb₂O₃(s) ---> 6H₃SbO₄(aq) + 12H⁺ + 12e¯
12e¯ + 16H⁺ + 4NO₃¯(aq) ---> 4NO(g) + 8H₂O

4) Add:

4H⁺ + 7H₂O + 3Sb₂O₃(s) + 4NO₃¯(aq) ---> 6H₃SbO₄(aq) + 4NO(g)

5) Convert to basic:

11H₂O + 3Sb₂O₃(s) + 4NO₃¯(aq) ---> 6H₃SbO₄(aq) + 4NO(g) + 4OH¯

Problem #23: Al(s) + NO₃¯ ---> Al(OH)₄¯ + NH₃(g)

Solution:

I propose to balance one half-reaction in basic, the other in acidic. I'll then add and change over to basic at the end.

1) Half-reactions:

Al(s) ---> Al(OH)₄¯
NO₃¯ ---> NH₃(g)

2) Balance:

Al(s) + 4OH¯ ---> Al(OH)₄¯ + 3e¯
8e¯ + 9H⁺ + NO₃¯ ---> NH₃(g) + 3H₂O

3) Equalize electrons:

8Al(s) + 32OH¯ ---> 8Al(OH)₄¯ + 24e¯
24e¯ + 27H⁺ + 3NO₃¯ ---> 3NH₃(g) + 9H₂O

4) Add:

8Al(s) + 32OH¯ + 27H⁺ + 3NO₃¯ ---> 8Al(OH)₄¯ + 3NH₃(g) + 9H₂O

5) Make twenty-seven waters from 27 hydrogen ions and 27 hydroxide ions:

8Al(s) + 5OH¯ + 27H₂O + 3NO₃¯ ---> 8Al(OH)₄¯ + 3NH₃(g) + 9H₂O

6) Eliminate nine waters for the final answer:

8Al(s) + 3NO₃¯ + 5OH¯ + 18H₂O ---> 8Al(OH)₄¯ + 3NH₃(g)

I rearranged the left-hand side a bit, simply because it looks better to me. Also, note how I never had to change over to basic, it just happened naturally.

Problem #24: MnO₄¯ + I¯ ----> MnO₄²¯ + IO₃¯

Solution:

1) Half-reactions:

MnO₄¯ ---> MnO₄²¯
I¯ ---> IO₃¯

2) Balance in acidic solution:

e¯ + MnO₄¯ ---> MnO₄²¯
3H₂O + I¯ ---> IO₃¯ + 6H⁺ + 6e¯

3) Equalize electrons:

6e¯ + 6MnO₄¯ ---> 6MnO₄²¯
3H₂O + I¯ ---> IO₃¯ + 6H⁺ + 6e¯

4) Add:

3H₂O + I¯ + 6MnO₄¯ ---> 6MnO₄²¯ + IO₃¯ + 6H⁺

5) Change to basic by adding six hydroxides to each side:

6OH¯ + 3H₂O + I¯ + 6MnO₄¯ ---> 6MnO₄²¯ + IO₃¯ + 6H₂O

6) Remove three H₂O:

6OH¯ + I¯ + 6MnO₄¯ ---> 6MnO₄²¯ + IO₃¯ + 3H₂O

Problem #25: Cu²⁺ + SO₃²¯ ---> SO₄²¯ + Cu

Solution:

1) Half-reactions:

SO₃²¯ ---> SO₄²¯
Cu²⁺ ---> Cu

2) Balance them:

H₂O + SO₃²¯ ---> SO₄²¯ + 2H⁺ + 2e¯
2e¯ + Cu²⁺ ---> Cu

3) Note that I balanced the first half-reaction in acidic solution. I'll change it to basic in a moment. Since there are already an equal number of electrons on each side, we may add the two half-reactions:

H₂O + SO₃²¯ + Cu²⁺ ---> SO₄²¯ + Cu + 2H⁺

4) Add two hydroxide ions to each side to make the conversion from acidic to basic:

2OH¯ + H₂O + SO₃²¯ + Cu²⁺ ---> SO₄²¯ + Cu + 2H₂O

Note that the hydrogen ions and hydroxide ions form water.

5) Cancel out one water:

2OH¯ + SO₃²¯ + Cu²⁺ ---> SO₄²¯ + Cu + H₂O