Balancing redox reactions in acidic solutionFifteen Examples

 Problems 1-10 Problems 26-50 Balancing in basic 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 hydrogen ion.

After Example #5c, I have a suggestion for searching for help on balancing a specific equation.

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯

Solution:

1) Split into unbalanced half-reactions:

ClO3¯ ---> Cl¯
SO2 ---> SO42¯

2) Balance the half-reactions:

6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O
2H2O + SO2 ---> SO42¯ + 4H+ + 2e¯

3) Make the number of electrons equal:

6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O
6H2O + 3SO2 ---> 3SO42¯ + 12H+ + 6e¯ <--- multiplied through by a factor of 3

ClO3¯ + 3H2O + 3SO2 ---> 3SO42¯ + Cl¯ + 6H+

Note that items duplicated on each side were cancelled out. The duplicates are 6e¯, 3H2O, and 6H+

Example #2a: H2S + NO3¯ ---> S8 + NO

Solution:

1) The unbalanced half-reactions:

H2S ---> S8
NO3¯ ---> NO

2) balance each half-reaction:

8H2S ---> S8 + 16H+ + 16e¯
3e¯ + 4H+ + NO3¯ ---> NO + 2H2O

3) Make the number of electrons equal:

24H2S ---> 3S8 + 48H+ + 48e¯ <--- multiplied by a factor of 3
48e¯ + 64H+ + 16NO3¯ ---> 16NO + 32H2O <--- multiplied by a factor of 16

Note that 16 and 3 have no common factors except 1, so both 16 and 3 had to be used to obtain the lowest common multiple of 48 for the number of electrons.

24H2S + 16H+ + 16NO3¯ ---> 3S8 + 16NO + 32H2O

Comment: removing a factor of 8 does look tempting, doesn't it? However, the three in front of the S8 (or the five in the next example) makes it impossible. Also, note that duplicates of 48 electrons and 48 hydrogen ions were removed.

5) Sometimes, you will see the nitric acid in molecular form:

24H2S + 16HNO3 ---> 3S8 + 16NO + 32H2O

Example #2b: H2S + HNO3 ---> NO + S + H2O

Discussion:

Many times, teachers and textbooks will use S rather than S8. Notice that, in the answer, the S coefficient stays the same (but the subscript of 8 goes away) and the other coefficients are all reduced by a factor of 8. The answer to 2b is the exact same as 2a in terms of the stoichiometry of the reaction.

Solution:

1) Half-reactions:

NO3¯ ---> NO
H2S ---> S

2) Balance:

3e¯ + 4H+ + NO3¯ ---> NO + 2H2O
H2S ---> S + 2H+ + 2e¯

3) Equalize electrons:

6e¯ + 8H+ + 2NO3¯ ---> 2NO + 4H2O
3H2S ---> 3S + 6H+ + 6e¯

3H2S + 2HNO3 ---> 2NO + 3S + 4H2O

Example #3: MnO4¯ + H2S ---> Mn2+ + S8

Solution:

1) Half-reactions:

H2S ---> S8
MnO4¯ ---> Mn2+

2) Balance:

8H2S ---> S8 + 16H+ + 16e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

3) Make the number of electrons equal (note that there are no common factors between 5 and 16 except 1):

40H2S ---> 5S8 + 80H+ + 80e¯ <--- factor of 5
80e¯ + 128H+ + 16MnO4¯ ---> 16Mn2+ + 64H2O <--- factor of 16

40H2S + 48H+ + 16MnO4¯ ---> 5S8 + 16Mn2+ + 64H2O

Another possibility of removing a factor of 8 destroyed by an odd number, in this case, the 5 in front of the S8. Curses, foiled again!

Example #4: Cu + SO42¯ ---> Cu2+ + SO2

Solution:

1) The unbalanced half-reactions:

Cu ---> Cu2+
SO42¯ ---> SO2

2) The balanced half-reactions:

Cu ---> Cu2+ + 2e¯
2e¯ + 4H+ + SO42¯ ---> SO2 + 2H2O

Cu + 4H+ + SO42¯ ---> Cu2+ + SO2 + 2H2O

No need to equalize electrons since it turns out that, in the course of balancing the half-reactions, the electrons are equal in amount. Note how easy it was to balance the copper half-reaction. All you needed were the two electrons.

Example #5a: MnO4¯ + CH3OH ---> HCOOH + Mn2+

Solution:

1) The balanced half-reactions:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
H2O + CH3OH ---> HCOOH + 4H+ + 4e¯

2) Equalize electrons:

20e¯ + 32H+ + 4MnO4¯ ---> 4Mn2+ + 16H2O <--- factor of 4
5H2O + 5CH3OH ---> 5HCOOH + 20H+ + 20e¯ <--- factor of 5

12H+ + 5CH3OH + 4MnO4¯ ---> 5HCOOH + 4Mn2+ + 11H2O

Example #5b: MnO4¯ + CH3OH ---> CH3COOH + Mn2+

Solution:

1) Half-reactions:

MnO4¯ ---> Mn2+
CH3OH ---> CH3COOH

2) Balance:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
2CH3OH ---> CH3COOH + 4H+ + 4e¯

3) Equalize electrons:

20e¯ + 32H+ + 4MnO4¯ ---> 4Mn2+ + 16H2O
10CH3OH ---> 5CH3COOH + 20H+ + 20e¯

12H+ + 4MnO4¯ + 10CH3OH ---> 4Mn2+ + 5CH3COOH + 16H2O

Example #5c: MnO4¯ + H2O2 ---> Mn2+ + O2

Solution:

1) Half reactions:

MnO4¯ ---> Mn2+
H2O2 ---> O2

2) Balance:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
H2O2 ---> O2 + 2H+ + 2e¯

3) Equalize electrons:

10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O
5H2O2 ---> 5O2 + 10H+ + 10e¯

6H+ + 2MnO4¯ + 5H2O2 ---> 2Mn2+ + 5O2 + 8H2O

The search suggestion is to type the reactants plus an arrow, like this:

mno4- + h2o2 --->

Notice that I completely ignore capitals in the formulas. In the hits from that search, there will be a number of websites that will help you figure out the answer.

However, be careful. Take a look at this website. You will see 14H2O rather than 8H2O. What happened?

Answer: the writer of that page represented hydrogen ion as H3O+ rather than H+, thus adding six H2O to each side.

Example #6: VO2+ + MnO4¯ ---> V(OH)4+ + Mn2+

Solution:

1) Half reactions:

VO2+ ---> V(OH)4+
MnO4¯ ---> Mn2+

2) Balance:

3H2O + VO2+ ---> V(OH)4+ + 2H+ + e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

3) Equalize electrons:

15H2O + 5VO2+ ---> 5V(OH)4+ + 10H+ + 5e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

11H2O + 5VO2+ + MnO4¯ ---> 5V(OH)4+ + Mn2+ + 2H+

Example #7: Cr2O72¯ + Cl¯ ---> Cr3+ + Cl2

Solution:

1) Half-reactions:

Cr2O72¯ ---> Cr3+
Cl¯ ---> Cl2

2) Balance:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
2Cl¯ ---> Cl2 + 2e¯

3) Equalize electrons:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
6Cl¯ ---> 3Cl2 + 6e¯

14H+ + Cr2O72¯ + 6Cl¯ ---> 2Cr3+ + 3Cl2 + 7H2O

5) A more detailed discussion about balancing this equation can be found here.

6) I once saw an unusual method to balancing this particular example equation. It winds up with the equation balanced in basic solution. Here it is, in all its glory:

Cr2O72¯ + Cl¯ ---> Cr3+ + Cl2 + O2¯
there is a minimum of 2 Cr's
2Cr6+ + 6e¯ ---> 2Cr3+
and a minimum of 2 Cl's
2Cl¯ ---> Cl2 + 2e¯
we need to triple this to get 6e¯
6Cl¯ ---> 3Cl2 + 6e¯
Cr2O72¯ + 6Cl¯ ---> 2Cr3+ + 3Cl2 + 7O2¯
but oxide ions would immediately react with water
Cr2O72¯ + 6Cl¯ + 7H2O ---> 2Cr3+ + 3Cl2 + 14OH¯

Balancing with oxide ions!! You don't see that one every day.

7) And then, since are in acidic solution, we use 14H+ to react with the hydroxide:

Cr2O72¯ + 6Cl¯ + 7H2O + 14H+ ---> 2Cr3+ + 3Cl2 + 14H2O

8) And then remove seven waters from each side to arrive at the answer given in step 4.

Example #8: MnO4¯ + S2¯ ---> MnS + S

1) Half-reactions:

MnO4¯ ---> Mn2+
S2¯ ---> S

Note that I eliminated the sulfide from the MnS. I'll add it back in at the end.

2) Balance half-reactions:

5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O
S2¯ ---> S + 2e¯

3) Equalize electrons:

10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O
5S2¯ ---> 5S + 10e¯

16H+ + 2MnO4¯ + 5S2¯ ---> 2Mn2+ + 5S + 8H2O

5) Add two sulfides on each side to make MnS:

16H+ + 2MnO4¯ + 7S2¯ ---> 2MnS + 5S + 8H2O

6) This document balances the equation in basic solution. This is an easy transformation from the answer in step 5, just add 16 hydroxides to each side:

8H2O + 2MnO4¯ + 7S2¯ ---> 2MnS + 5S + 16OH¯

7) The linked document also keeps the MnS in the half-reaction and balances it with a sulfide on the left-hand side of the half-reaction.

Example #9: As2S5(s) + NO3¯(aq) ---> H3AsO4(aq) + HSO4¯(aq) + NO2(g)

Solution:

1) Half-reactions:

8H2O + As210+ ---> 2H3AsO4(aq) + 10H+ <--- note: neither reduction nor oxidation
20H2O + S510¯ ---> 5HSO4¯(aq) + 35H+ + 40e¯
e¯ + 2H+ + NO3¯(aq) ---> NO2(g) + H2O

Notice how I have separated the arsenic and sulfur. Often, both the arsenic and the associated anion are either oxidized or reduced. In this particular example, only the sulfur gets oxidized.

I deliberately wrote As210+ and S510¯. I did it so as to make it easy to recombine them to make As2S5.

2) Combine the first two half-reactions:

28H2O + As2S5 ---> 2H3AsO4(aq) + 5HSO4¯(aq) + 45H+ + 40e¯

3) Add in the second half-reaction and equalize for electrons:

28H2O + As2S5 ---> 2H3AsO4(aq) + 5HSO4¯(aq) + 45H+ + 40e¯
40e¯ + 80H+ + 40NO3¯(aq) ---> 40NO2(g) + 40H2O

35H+ + As2S5 + 40NO3¯(aq) ---> 2H3AsO4(aq) + 5HSO4¯(aq) + 40NO2(g) + 12H2O

Example #10: H3AsO3 + I2 ---> H3AsO4 + I¯

Solution:

1) Half-reactions:

H3AsO3 ---> H3AsO4
I2 ---> I¯

2) Balance:

H2O + H3AsO3 ---> H3AsO4 + 2H+ + 2e¯
2e¯ + I2 ---> 2I¯

H2O + H3AsO3 + I2 ---> H3AsO4 + 2H+ + 2I¯

4) If so needed, you could report this as fully molecular (instead of showing the HI - a strong acid - as fully ionized:

H2O + H3AsO3 + I2 ---> H3AsO4 + 2HI

Example #11: Balance the equation for the reaction of stannous ion with pertechnetate in acidic 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:

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

4) Equalize electrons:

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

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

Example #12: H3AsO4 + Zn + HNO3 --> AsH3 + Zn(NO3)2

Solution:

1) Net-ionic form:

AsO43¯ + Zn ---> AsH3 + Zn2+

2) Half-reactions:

AsO43¯ ---> AsH3
Zn ---> Zn2+

3) Balance:

8e¯ + 11H+ + AsO43¯ ---> AsH3 + 4H2O
Zn ---> Zn2+ + 2e¯

4) Equalize electrons:

8e¯ + 8H+ + H3AsO4 ---> AsH3 + 4H2O <--- I used 3 H+ to recreate the H3AsO4
4Zn ---> 4Zn2+ + 8e¯

8H+ + H3AsO4 + 4Zn ---> AsH3 + 4Zn2+ + 4H2O

8HNO3 + H3AsO4 + 4Zn ---> AsH3 + 4Zn(NO3)2 + 4H2O

Example #13: O3 + Cl¯ ---> H2O + ClO3¯

Solution:

1) Half-reactions:

O3 ---> H2O
Cl¯ ---> ClO3¯

2) Balance:

6e¯ + 6H+ + O3 ---> 3H2O
3H2O + Cl¯ ---> ClO3¯ + 6H+ + 6e¯

O3 + Cl¯ ---> ClO3¯

4) Or, you can notice that dropping the water right at the start results in an equation balanced for atoms and for charge.

Example #14: H2SO5 is named peroxymonosulfuric acid. One of its salts, KHSO5 (potassium peroxymonosulfate) is widely used as an oxidizing agent. Balance the following reaction in acidic solution:

HSO5¯ + ClO2¯ ---> ClO3¯ + SO42¯

Solution:

Comment: look to see if this one can be balanced for atoms and charge by sight. Hint: it can. The half-reaction method follows.

1) Separate into half-reactions:

HSO5¯ ---> SO42¯
ClO2¯ ---> ClO3¯

2) Balance:

2e¯ + H+ + HSO5¯ ---> SO42¯ + H2O
H2O + ClO2¯ ---> ClO3¯ + 2H+ + 2e¯

HSO5¯ + ClO2¯ ---> ClO3¯ + SO42¯ + H+

Example #15: As ---> H3AsO4 + AsH3

Solution:

1) Half-reactions:

As ---> H3AsO4
As ---> AsH3

2) Balance:

4H2O + As ---> H3AsO4 + 5H+ + 5e¯
3e¯ + 3H+ + As ---> AsH3

3) Equalize electrons:

12H2O + 3As ---> 3H3AsO4 + 15H+ + 15e¯
15e¯ + 15H+ + 5As ---> 5AsH3

12H2O + 8As ---> 3H3AsO4 + 5AsH3

Bonus Example: Cr2O72¯ + SO2 + H+ ---> Cr3+ + HSO4¯ + H2O

Solution:

1) Half-reactions:

Cr2O72¯ ---> Cr3+
SO2 ---> HSO4¯

2) Balance in acidic solution:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
2H2O + SO2 ---> HSO4¯ + 3H+ + 2e¯

3) Equalize electrons:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
6H2O + 3SO2 ---> 3HSO4¯ + 9H+ + 6e¯

5H+ + Cr2O72¯ + 3SO2 ---> 2Cr3+ + 3HSO4¯ + H2O

Sometimes you are given a net-ionic equation and asked to take it back to a full molecular equation. Sometimes, no context is added, so you have to make some informed predictions. Here's what I mean:

Since the equation is in acidic solution, you can use HCl or HNO3. I'll use HCl. The most common dichromate that is soluble is potassium dichromate, so we will use that. Using those, we find this:

5HCl + K2Cr2O7 + 3SO2 ---> 2CrCl3 + 3KHSO4 + H2O

However, there is a problem. One too many K and Cl on the right-hand side. The solution is to add one KCl to the left-hand side:

KCl + 5HCl + K2Cr2O7 + 3SO2 ---> 2CrCl3 + 3KHSO4 + H2O

You can write the equation using HNO3 and the nitrate would simply replace the chloride.

Using sulfuric acid can be done but (and this is part of the informed prediction) probably should not. The chromium(III) ion is presented as an ion, meaning it's soluble. Chromium(III) sulfate is not soluble, which means you would have to write the full formula. Since that was not done, we conclude that the chromium ion was part of a soluble compound.

This whole balance-a-redox-reaction-in-molecular-form is a thing and it's not covered very much in most textbooks. Here are some examples.

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