### Balancing redox reactions in basic solutionFifteen Examples

 Problems 1-10 Problems 26-50 Balancing in acidic solution Problems 11-25 Only the examples and problems Return to Redox menu

Points to remember:

1) Electrons NEVER appear in a correct, final answer. In order to get the electrons in each half-reaction equal, one or both of the balanced half-reactions will be multiplied by a factor.

2) Duplicate items are always removed. These items are usually the electrons, water and hydroxide ion.

3) The technique below is almost always balance the half-reactions as if they were acidic. You would then add hydroxide at the end to convert it to basic. Sometimes (see example #5), you can balance using hydroxide directly.

Example #1: NH3 + ClO¯ ---> N2H4 + Cl¯

Solution:

1) The two half-reactions, balanced as if in acidic solution:

2NH3 ---> N2H4 + 2H+ + 2e¯
2e¯ + 2H+ + ClO¯ ---> Cl¯ + H2O

2) Electrons already equal, convert to basic solution:

2OH¯ + 2NH3 ---> N2H4 + 2H2O + 2e¯
2e¯ + 2H2O + ClO¯ ---> Cl¯ + H2O + 2OH¯

Comment: that's 2 OH¯, not 20 H¯. Misreading the O in OH as a zero is a common mistake.

3) The final answer:

2HN3 + ClO¯ ---> N2H4 + Cl¯ + H2O

Notice that no hydroxide appears in the final answer. That means this is a base-catalyzed reaction. For the reaction to occur, the solution must be basic and hydroxide IS consumed. It is just regenerated in the exact same amount, so it cancels out in the final answer.

Example #2: Au + O2 + CN¯ ---> Au(CN)2¯ + H2O2

Solution:

1) The two half-reactions, balanced as if in acidic solution:

2CN¯ + Au ---> Au(CN)2¯ + e¯
2e¯ + 2H+ + O2 ---> H2O2

2) Make electrons equal, convert to basic solution:

4CN¯ + 2Au ---> 2Au(CN)2¯ + 2e¯ <--- multiplied by a factor of 2
2e¯ + 2H2O + O2 ---> H2O2 + 2OH¯

3) The final answer:

4CN¯ + 2Au + 2H2O + O2 ---> 2Au(CN)2¯ + H2O2 + 2OH¯

Comment: the CN¯ is neither reduced nor oxidized, but it is necessary for the reaction. For example, you might see this way of writing the problem:

Au + O2 ---> Au(CN)2¯ + H2O2

Notice that CN¯ does not appear on the left side, but does so on the right. Since you MUST balance the equation, that means you are allowed to use CN¯ in your balancing. An important point here is that you know the cyanide polyatomic ion has a negative one charge.

Example #3: Br¯ + MnO4¯ ---> MnO2 + BrO3¯

Solution:

1) The two half-reactions, balanced as if in acidic solution:

3H2O + Br¯ ---> BrO3¯ + 6H+ + 6e¯
3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O

2) Make the number of electrons equal:

3H2O + Br¯ ---> BrO3¯ + 6H+ + 6e¯
6e¯ + 8H+ + 2MnO4¯ ---> 2MnO2 + 4H2O <--- multiplied by a factor of 2

3) Convert to basic solution, by adding 6OH¯ to the first half-reaction and 8OH¯ to the second:

6OH¯ + Br¯ ---> BrO3¯ + 3H2O + 6e¯
6e¯ + 4H2O + 2MnO4¯ ---> 2MnO2 + 8OH¯

4) The final answer:

H2O + 2MnO4¯ + Br¯ ---> 2MnO2 + BrO3¯ + 2OH¯

5) What happens if you add the two half-reactions without converting them to basic?

You get this:
2H+ + 2MnO4¯ + Br¯ ---> 2MnO2 + BrO3¯ + H2O

Then, add 2OH¯ to each side:

2H2O + 2MnO4¯ + Br¯ ---> 2MnO2 + BrO3¯ + H2O + 2OH¯

Eliminate one water for the final answer:

H2O + 2MnO4¯ + Br¯ ---> 2MnO2 + BrO3¯ + 2OH¯

The answer to the question? Nothing happens. You get the right answer if convert before adding the half-reactions or after. There will even be cases where balancing one half-reaction using hydroxide can easily be done while the other half-reaction gets balanced in acidic solution before converting. You can add the two half-reactions while one is basic and one is acidic, then convert after the adding (see example #5 and example #8 below for examples of this).

Example #4: AlH4¯ + H2CO ---> Al3+ + CH3OH

Solution:

1) The two half-reactions, balanced as if in acidic solution:

AlH4¯ ---> Al3+ + 4H+ + 8e¯
2e¯ + 2H+ + H2CO ---> CH3OH

2) Converted to basic by addition of hydroxide, second half-reaction multiplied by 4 (note that the hydrogen is oxidized from -1 to +1):

4OH¯ + AlH4¯ ---> Al3+ + 4H2O + 8e¯
8e¯ + 8H2O + 4H2CO ---> 4CH3OH + 8OH¯

3) The final answer:

AlH4¯ + 4H2O + 4H2CO ---> Al3+ + 4CH3OH + 4OH¯

Example #5: Se + Cr(OH)3 ---> Cr + SeO32¯

Solution:

1) The unbalanced half-reactions:

Se ---> SeO32¯
Cr(OH)3 ---> Cr

2) Note that only the first half-reaction is balanced using the balance-first-in-acid technique, the second is balanced using hydroxide:

Se + 3H2O ---> SeO32¯ + 6H+ + 4e¯
3e¯ + Cr(OH)3 ---> Cr + 3OH¯

3) Convert the first half-reaction by adding 6 hydroxide to each side, eliminate duplicate waters, then make the electrons equal (factor of 3 for the first half-reaction and a factor of 4 for the second). The final answer:

6OH¯ + 3Se + 4Cr(OH)3 ---> 4Cr + 3SeO32¯ + 9H2O

4) What would happen if we didn't make the first half-reaction basic and just added them?

Make the electrons equal:
3Se + 9H2O ---> 3SeO32¯ + 18H+ + 12e¯
12e¯ + 4Cr(OH)3 ---> 4Cr + 12OH¯

Add:

3Se + 4Cr(OH)3 + 9H2O ---> 4Cr + 3SeO32¯ + 18H+ + 12OH¯

Combine hydrogen ion and hydroxide ion on the right-hand side:

3Se + 4Cr(OH)3 + 9H2O ---> 4Cr + 3SeO32¯ + 6H+ + 12H2O

Eliminate water:

3Se + 4Cr(OH)3 ---> 4Cr + 3SeO32¯ + 6H+ + 3H2O

Add six hydroxides:

6OH¯ + 3Se + 4Cr(OH)3 ---> 4Cr + 3SeO32¯ + 9H2O

Note that I combined the H+ and the OH¯ to make six waters and then added it to the three waters that were already there.

Also, I could have added the six hydroxides before eliminating water. You may try that out, if you wish.

Example #6: Au + NaCN + O2 + H2O ---> NaAu(CN)2 + NaOH

Solution:

1) Net-ionic:

Au + CN¯ + O2 ---> Au(CN)2¯ + OH¯

I decided to treat the Au(CN)2¯ as a polyatomic ion. I could have eliminated the cyanide and added it back in after balancing the net-ionic.

2) Half-reactions:

Au + CN¯ ---> Au(CN)2¯
O2 ---> OH¯

3) Balance:

Au + 2CN¯ ---> Au(CN)2¯ + e¯
4e¯ + 2H+ + O2 ---> 2OH¯ <--- note hydrogen ion on one side and hydroxide ion on the other

4) Equalize electrons:

4Au + 8CN¯ ---> 4Au(CN)2¯ + 4e¯
4e¯ + 2H+ + O2 ---> 2OH¯

5) Add:

4Au + 8CN¯ + 2H+ + O2 ---> 4Au(CN)2¯ + 2OH¯

Only the electrons cancel.

6) Convert to basic solution:

4Au + 8CN¯ + 2H2O + O2 ---> 4Au(CN)2¯ + 4OH¯

Example #7: Ag2S + CN¯ + O2 ---> Ag(CN)2¯ + S8 + OH¯

Solution:

1) Examination shows that the sulfide is oxidized and the oxygen is reduced. However, the sulfide is attached to a silver. What to do? Let's keep it in the half-reaction:

Ag2S ---> S8
O2 ---> OH¯

Notice that there isn't any cyanide ion present. I'll add that in during the balancing.

2) Balance the silver sulfide half-reaction only:

8Ag2S ---> S8 <--- balances the sulfur

8Ag2S ---> S8 + 16Ag(CN)2¯ <--- balances the silver

8Ag2S + 32CN¯ ---> S8 + 16Ag(CN)2¯ <--- balances the cyanide

8Ag2S + 32CN¯ ---> S8 + 16Ag(CN)2¯ + 16e¯ <--- balances the electrons

3) Balance the oxygen half-reaction only:

O2 ---> 2OH¯ <--- balances the oxygen

2H+ + O2 ---> 2OH¯ <--- balance the hydrogens

2H2O + O2 ---> 4OH¯ <--- change over to basic

4e¯ + 2H2O + O2 ---> 4OH¯ <--- balance charge with electrons

4) Write both balanced half-reactions:

8Ag2S + 32CN¯ ---> S8 + 16Ag(CN)2¯ + 16e¯
4e¯ + 2H2O + O2 ---> 4OH¯

5) Equalize electrons:

8Ag2S + 32CN¯ ---> S8 + 16Ag(CN)2¯ + 16e¯
16e¯ + 8H2O + 4O2 ---> 16OH¯

6) Add:

8Ag2S + 32CN¯ + 4O2 + 8H2O ---> S8 + 16Ag(CN)2¯ + 16OH¯

Example #8: N2H4 + Cu(OH)2 ---> N2 + Cu

Solution:

1) Half-reactions:

N2H4 ---> N2
Cu(OH)2 ---> Cu

2) Balance:

N2H4 ---> N2 + 4H+ + 4e¯
2e¯ + Cu(OH)2 ---> Cu + 2OH¯

Note how one half-reaction is balanced in acidic and the other in basic. That will not create a problem.

3) Equalize electrons:

N2H4 ---> N2 + 4H+ + 4e¯
4e¯ + 2Cu(OH)2 ---> 2Cu + 4OH¯

4) Add

N2H4 + 2Cu(OH)2 ---> N2 + 4H+ + 2Cu + 4OH¯

5) Combine hydrogen ion and hydroxide ion to make water:

N2H4 + 2Cu(OH)2 ---> N2 + 2Cu + 4H2O

Example #9: MnO4¯ + C2O42¯ ---> MnO2 + CO32¯

Solution:

1) Half-reactions:

MnO4¯ ---> MnO2
C2O42¯ ---> CO32¯

2) Balance as if in acidic solution:

3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O
2H2O + C2O42¯ ---> 2CO32¯ + 4H+ + 2e¯

3) Equalize electrons:

6e¯ + 8H+ + 2MnO4¯ ---> 2MnO2 + 4H2O
6H2O + 3C2O42¯ ---> 6CO32¯ + 12H+ + 6e¯

4) Add:

2H2O + 3C2O42¯ + 2MnO4¯ ---> 6CO32¯ + 2MnO2 + 4H+

5) Convert to basic by adding four hydroxides to each side (and then eliminating two waters from each side):

4OH¯ + 3C2O42¯ + 2MnO4¯ ---> 6CO32¯ + 2MnO2 + 2H2O

Example #10: Zn + NO3¯ ---> Zn(OH)42¯ + NH3

Solution:

1) Half-reactions:

Zn ---> Zn(OH)42¯
NO3¯ ---> NH3

2) Balance:

Zn + 4OH¯ ---> Zn(OH)42¯ + 2e¯
8e¯ + 9H+ + NO3¯ ---> NH3 + 3H2O

3) Equalize electrons:

4Zn + 16OH¯ ---> 4Zn(OH)42¯ + 8e¯
8e¯ + 9H+ + NO3¯ ---> NH3 + 3H2O

4) Add:

4Zn + 9H+ + 16OH¯ + NO3¯ ---> 4Zn(OH)42¯ + NH3 + 3H2O

5) Allow nine hydrogen ions and nine hydroxide ions to react (and then eliminate three duplicate waters):

4Zn + 6H2O + 7OH¯ + NO3¯ ---> 4Zn(OH)42¯ + NH3

Notice how water and the hydroxide ion wind up on the same side. Usually, they are on opposite sides.

Example #11: Balance the equation for the reaction of stannous ion with pertechnetate in basic solution. Products are stannic ion, Sn4+ and technetium(IV), Tc4+ ions.

Solution:

1) Net ionic:

TcO4¯ + Sn2+ ---> Tc4+ + Sn4+

2) Half-reactions:

TcO4¯ ---> Tc4+
Sn2+ ---> Sn4+

3) Balance in acidic:

3e¯ + 8H+ + TcO4¯ ---> Tc4+ + 4H2O
Sn2+ ---> Sn4+ + 2e¯

4) Equalize electrons:

6e¯ + 16H+ + 2TcO4¯ ---> 2Tc4+ + 8H2O
3Sn2+ ---> 3Sn4+ + 6e¯

5) Add:

16H+ + 2TcO4¯ + 3Sn2+ ---> 2Tc4+ + 3Sn4+ + 8H2O

6) Convert to basic by adding 16 hydroxides to each side:

8H2O + 2TcO4¯ + 3Sn2+ ---> 2Tc4+ + 3Sn4+ + 16OH¯

Example #12: Cr2O72¯ + I2 ---> Cr3+ + IO3¯

Solution:

1) Separate into half-reactions:

Cr2O72¯ ---> Cr3+
I2 ---> IO3¯

2) Balance in acidic solution:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
6H2O + I2 ---> 2IO3¯ + 12H+ + 10e¯

3) Equalize electrons:

30e¯ + 70H+ + 5Cr2O72¯ ---> 10Cr3+ + 35H2O
18H2O + 3I2 ---> 6IO3¯ + 36H+ + 30e¯

4) Add and eliminate duplicates:

34H+ + 5Cr2O72¯ + 3I2 ---> 6IO3¯ + 10Cr3+ + 17H2O

5) Add 34 OH¯ to each side and eliminate duplicates:

17H2O + 5Cr2O72¯ + 3I2 ---> 6IO3¯ + 10Cr3+ + 34OH¯

Example #13: Bi3+ + MnO4¯ ---> MnO2 + BiO3¯

Solution:

1) Half-reactions:

Bi3+ ---> BiO3¯
MnO4¯ ---> MnO2

2) Balance in acidic solution:

3H2O + Bi3+ ---> BiO3¯ + 6H+ + 2e¯
3e¯ + 4H+ + MnO4¯ ---> MnO2 + 2H2O

3) Equalize electrons:

9H2O + 3Bi3+ ---> 3BiO3¯ + 18H+ + 6e¯
6e¯ + 8H++ 2MnO4¯ ---> 2MnO2 + 4H2O

4) Add and eliminate like items:

5H2O + 3Bi3+ + 2MnO4¯ ---> 3BiO3¯ + 2MnO2 + 10H+

5) Change to basic solution:

10OH¯ + 5H2O + 3Bi3+ + 2MnO4¯ ---> 3BiO3¯ + 2MnO2 + 10H2O

6) Remove water:

10OH¯ + 3Bi3+ + 2MnO4¯ ---> 3BiO3¯ + 2MnO2 + 5H2O

Example #14: Co(OH)2(s) + SO32¯(aq) ---> SO42¯(aq) + Co(s)

Solution:

1) Half-reactions:

Co(OH)2(s) ---> Co(s)
SO32¯(aq) ---> SO42¯(aq)

3) Balance:

Co(OH)2(s) ---> Co(s) + 2OH¯
H2O + SO32¯(aq) ---> SO42¯(aq) + 2H+ + 2e¯

3) Electrons equal, so add:

H2O + SO32¯(aq) + Co(OH)2(s) ---> Co(s) + 2OH¯ + SO42¯(aq) + 2H+

4) The H+ and the OH¯ on the right-hand side unite to form water. Write the final answer:

SO32¯(aq) + Co(OH)2(s) ---> Co(s) + SO42¯(aq) + H2O

Example #15: PtO42¯ + Be ---> BeO32¯ + Pt(OH)62¯

Solution:

1) Half-reactions:

PtO42¯ ---> Pt(OH)62¯
Be ---> BeO32¯

2) Balance:

2e¯ + 2H+ + 2H2O + PtO42¯ ---> Pt(OH)62¯
3H2O + Be ---> BeO32¯ + 6H+ + 4e¯

3) Equalize electrons:

4e¯ + 4H+ + 4H2O + 2PtO42¯ ---> 2Pt(OH)62¯
3H2O + Be ---> BeO32¯ + 6H+ + 4e¯

4) Add:

7H2O + 2PtO42¯ + Be ---> 2Pt(OH)62¯ + BeO32¯ + 2H+

5) Convert to basic:

2OH¯ + 5H2O + 2PtO42¯ + Be ---> 2Pt(OH)62¯ + BeO32¯

Bonus Example: CuS + HNO3 ---> Cu(NO3)2 + NO + H2SO4 + H2O

Solution:

This is a bit of an odd duck. That's because this equation is always seen on the acidic side. In my nearly 40 years of classroom teaching, I have never seen this equation balanced in basic solution. So, here we gooooo . . . . . .

1) Write the equation in net-ionic form:

S2¯ + NO3¯ ---> NO + SO42¯

2) Half-reactions:

S2¯ ---> SO42¯
NO3¯ ---> NO

3) Balance in basic:

8OH¯ + S2¯ ---> SO42¯ + 4H2O + 8e¯
3e¯ + 2H2O + NO3¯ ---> NO + 4OH¯

I used an old school technique to equalize the oxygens. It's based on the idea that two OH (ignore the negative charge) equals one water plus one "left over" oxygen. This was the technique in the days before the "balance in acid first" technique took over.

4) Equalize electrons:

24OH¯ + 3S2¯ ---> 3SO42¯ + 12H2O + 24e¯
24e¯ + 16H2O + 8NO3¯ ---> 8NO + 32OH¯

5) Add:

4H2O + 3S2¯ + 8NO3¯ ---> 3SO42¯ + 8NO + 8OH¯

6) Start to recover the molecular equation by adding in three Cu2+ ions:

4H2O + 3CuS + 8NO3¯ ---> 3Cu2+ + 3SO42¯ + 8NO + 8OH¯

7) Add six nitrates:

4H2O + 3CuS + 14NO3¯ ---> 3Cu(NO3)2 + 3SO42¯ + 8NO + 8OH¯

8) Add 14 hydrogen ions:

4H2O + 3CuS + 14HNO3 ---> 3Cu(NO3)2 + 3H2SO4 + 8NO + 8H2O

On the right, six H+ made sulfuric acid and eight reacted with the 8 hydroxide.

9) Eliminate water for the final answer:

3CuS + 14HNO3 ---> 3Cu(NO3)2 + 3H2SO4 + 8NO + 4H2O

 Problems 1-10 Problems 26-50 Balancing in acidic solution Problems 11-25 Only the examples and problems Return to Redox menu