Balancing redox reactions in acidic solutionProblems #1-10

Problem #1: Cr2O72¯ + Fe2+ ---> Cr3+ + Fe3+

Solution:

1) Balanced half-reactions:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
Fe2+ ---> Fe3+ + e¯

2) Equalize the electrons:

6e¯ + 14H+ + Cr2O72¯ ---> 2Cr3+ + 7H2O
6Fe2+ ---> 6Fe3+ + 6e¯ <--- multiplied by a factor of 6

3) Add (and cancel) for the final answer:

14H+ + Cr2O72¯ + 6Fe2+ ---> 2Cr3+ + 7H2O + 6Fe3+

Note that the only thing that cancels are the six electrons.

Problem #2: HNO2 ---> NO + NO2

Solution:

1) The balanced half-reactions:

e¯ + H+ + HNO2 ---> NO + H2O
HNO2 ---> NO2 + H+ + e¯

2HNO2 ---> NO + NO2 + H2O

Note that the electrons were already balanced, so no need to multiply one or both half-reactions by a factor.

Comment #1: notice that this is no H+ in the final answer, but please keep in mind that its presence is necessary for the reaction to proceed. In cases like this, the H+ is acting in a catalytic manner; it is used up in one reaction and regenerated in another (in equal amount), consequently it does not appear in the final answer.

Comment #2: this type of a reaction is called a disproportionation. It is often found in redox situations, although not always. An important disproportionation reaction which does not involve redox is 2H2O ---> H3O+ + OH¯. This reaction is of central importance in aqueous acid-base chemistry.

Problem #3: H2C2O4 + MnO4¯ ---> CO2 + Mn2+

Solution:

1) The balanced half-reactions:

H2C2O4 ---> 2CO2 + 2H+ + 2e¯
5e¯ + 8H+ + MnO4¯ ---> Mn2+ + 4H2O

2) Equalize the electrons:

5H2C2O4 ---> 10CO2 + 10H+ + 10e¯ <--- factor of 5
10e¯ + 16H+ + 2MnO4¯ ---> 2Mn2+ + 8H2O <--- factor of 2

3) The final answer (electrons and some hydrogen ion get cancelled):

5H2C2O4 + 6H+ + 2MnO4¯ ---> 10CO2 + 2Mn2+ + 8H2O

Problem #4: O2 + As ---> HAsO2 + H2O

Solution:

1) First a bit of discussion before the correct answer. The H2O on the right side in the problem turns out to be a hint. This is because you need TWO half-reactions. For example, suppose the water wasn't in the equation and you saw this:

O2 + As ---> HAsO2

You'd think "Oh, that's easy" and procede to balance it like this:

H+ + O2 + As ---> HAsO2

Then, you'd "balance" the charge like this:

e¯ + H+ + O2 + As ---> HAsO2

And that is wrong because there is an electron in the final answer. You cannot have electrons appear in the final answer of a redox reaction. (You can in a half-reaction, but remember half-reactions do not occur alone, they occur in reduction-oxidation pairs.)

2) Here are the correct half-reactions:

4e¯ + 4H+ + O2 ---> 2H2O
2H2O + As ---> HAsO2 + 3H+ + 3e¯

3) In order to equalize the electrons, the first half-reaction is multiplied by a factor of 3 and the second by a factor of 4:

12e¯ + 12H+ + 3O2 ---> 6H2O
8H2O + 4As ---> 4HAsO2 + 12H+ + 12e¯

4) The final answer:

3O2 + 2H2O + 4As ---> 4HAsO2

Notice that the H2O winds up on the right-hand side of the equation.

By the way, try to balance

O2 + As ---> HAsO2

using H2O on the left rather than H+. That way leads to the correct answer without having to use half-reactions. There are some redox reactions where using half-reactions turns out to be "more" work, but there aren't that many.

Problem #5: NO3¯ + I2 ---> IO3¯ + NO2

Solution:

1) These are the balanced half-reactions:

6H2O + I2 ---> 2IO3¯ + 12H+ + 10e¯
e¯ + 2H+ + NO3¯ ---> NO2 + H2O

2) Only the second half-reaction needs to be multiplied through by a factor:

6H2O + I2 ---> 2IO3¯ + 12H+ + 10e¯
10e¯ + 20H+ + 10NO3¯ ---> 10NO2 + 10H2O

3) Adding the two half-reactions, but not eliminating anything except electrons:

6H2O + 20H+ + I2 + 10NO3¯ ---> 2IO3¯ + 12H+ + 10NO2 + 10H2O

4) Remove some water and hydrogen ion for the final answer:

8H+ + I2 + 10NO3¯ ---> 2IO3¯ + 10NO2 + 4H2O

Problem #6: HBr + SO42¯ ---> SO2 + Br2

Solution:

1) balanced half-reactions:

2e¯ + 4H+ + SO42¯ ---> SO2 + 2H2O
2HBr ---> Br2 + 2H+ + 2e¯

2) The final answer (note that electrons were already equal):

2H+ + 2HBr + SO42¯ ---> SO2 + Br2 + 2H2O

Problem #7: H5IO6 + Cr ---> IO3¯ + Cr3+

Solution:

1) The two half-reactions:

2e¯ + H+ + H5IO6 ---> IO3¯ + 3H2O
Cr ---> Cr3+ + 3e¯

2) Multiply top half-reaction by 3, bottom by 2; the final answer:

3H+ + 3H5IO6 + 2Cr ---> 2Cr3+ + 3IO3¯ + 9H2O

Problem #8: Fe + HCl ---> HFeCl4 + H2

Solution:

1) This problem poses interesting problems, especially with the Cl. The key to solving ths problem is to eliminate everything not directly involved in the redox. That means the H in HFeCl4 as well as the Cl in it and HCl. When we do that, this is the unbalanced, ionic form we wind up with:

Fe + H+ ---> Fe3+ + H2

2) The half-reactions (already balanced) are as follows:

Fe ---> Fe3+ + 3e¯
2e¯ + 2H+ ---> H2

3) The final answer:

2Fe + 6H+ ---> 2Fe3+ + 3H2

We will go back to the molecular equation with 8HCl. Six of the HCl molecules supply the 6H+ going to 3H2. The 7th and 8th HCl molecules supply the two H present in 2HFeCl4, as well as the 8 Cl needed in the two HFeCl4 molecules.

2Fe + 8HCl ---> 2HFeCl4 + 3H2

Problem #9: NO3¯ + H2O2 ---> NO + O2

Solution:

1) The half-reactions (already balanced) are as follows:

3e¯ + 4H+ + NO3¯ ---> NO + 2H2O
H2O2 ---> O2 + 2H+ + 2e¯

2) Multiply the top half-reaction by 2 and the bottom one by 3, add them and eliminate 6H+:

2H+ + 2NO3¯ + 3H2O2 ---> 2NO + 4H2O + 3O2

3) You can combine the hydrogen ion and the nitrate ion like this:

2HNO3 + 3H2O2 ---> 2NO + 4H2O + 3O2

This creates a what is called a molecular equation.

Problem #10: BrO3¯ + Fe2+ ---> Br¯ + Fe3+

Solution:

1) These are the half-reactions:

6e¯ + 6H+ + BrO3¯ ---> Br¯ + 3H2O
Fe2+ ---> Fe3+ + e¯

2) Only the second half-reaction needs to be multiplied through by a factor, then we add the two half-reactions for the final answer:

6H+ + BrO3¯ + 6Fe2+ ---> 6Fe3+ + Br¯ + 3H2O