Balancing redox reactions in acidic solution
Problems #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¯

2) Add for the final answer:

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 #3a: 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 #3b: C₂O₄²¯ + MnO₂ ---> CO₂ + Mn²⁺

Solution:

1) The unbalanced half-reactions:

C₂O₄²¯ ---> CO₂
MnO₂ ---> Mn²⁺

2) The balanced half-reactions:

C₂O₄²¯ ---> 2CO₂ + 2e¯
2e¯ + 4H⁺ + MnO₂ ---> Mn²⁺ + 2H₂O

3) Electrons already balanced, so add:

4H⁺ + C₂O₄²¯ + MnO₂ ---> 2CO₂ + Mn²⁺ + 2H₂O

4) Make oxalic acid, then add two chlorides to make it molecular:

2H⁺ + H₂C₂O₄ + MnO₂ ---> 2CO₂ + Mn²⁺ + 2H₂O

2HCl + H₂C₂O₄ + MnO₂ ---> 2CO₂ + MnCl₂ + 2H₂O

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 do show electrons in a half-reaction, but remember, half-reactions do not occur alone. They occur together, with at least one reduction and at least one oxidation involved.)

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