Redox reactions can be balanced with the half-reaction method. The half-reaction shows the oxidation and reduction processes separately.

**For example,**

** **Zn + Ag^{+} → Zn^{2+} + Ag

Follow these steps to balance this reaction:

**1)** Add the oxidation states and **write the half-reactions**. The oxidations states are only added to help identify the oxidation and reduction processes:

**2)** **Balance the elements.** In this case, they are balanced, so proceed to the next step.

3) **Balance the charges** by adding electrons:

Zn → Zn^{2+} + 2e^{–}1e^{–} + Ag^{+} → Ag

**4) Balance the number of electrons** by multiplying the equation(s) with a whole number. The first equation has two electrons, and therefore, we multiply the second equation by two:

2 x [1e^{–} + Ag^{+} → Ag] = 2e^{–} + 2Ag^{+} → 2Ag

5) Once everything is balanced, add the half-reactions together and check if all the atoms and charges are balanced again.

Zn → Zn^{2+} + 2e^{–}2e^{–} + 2Ag^{+} → 2Ag

__________________

Zn + 2e^{–} + 2Ag^{+} → 2Ag + Zn^{2+} + 2e^{–}

**6)** Cancel any species that appear in equal quantitates on both sides of the equation:

Zn + 2e^{–} + 2Ag^{+} → 2Ag + Zn^{2+} + 2e^{–}

Zn + 2Ag^{+} → 2Ag + Zn^{2+}

**Balancing Redox Equations in Acidic Solution**

In the example above, there are only two elements which simplifies balancing the reaction. There is an additional part to balancing redox reactions in **acidic or basic conditions** which results from the presence of oxygen and hydrogen atoms.

For example,

Fe^{2+} + Cr_{2}O_{7}^{2- }→ Fe^{3+} + Cr^{3+}

Do as we did in the previous example;

**1)** Add the oxidation states and **write the half-reactions**:

**2)** Balance the atoms. *Add H _{2}O to balance the O atoms, and H^{+} to balance the H atoms.*

The atoms in the first half-reaction are balanced. For the second reaction, we need to equalize the Cr atoms, and add 7 H_{2}O on the right side to balance the oxygen atoms:

Cr_{2}O_{7}^{2- }→ 2Cr^{3+ }+ 7H_{2}O

Now, to balance the H atoms, we need to **add 14 H ^{+} **on the left side:

14H^{+} + Cr_{2}O_{7}^{2- }→ 2Cr^{3+ }+ 7H_{2}O

**3)** Balance the charges by adding electrons.

In the first equation, we add one electron on the right side:

Fe^{2+} → Fe^{3+}+ 1e^{–}

^{ }

In the second equation the charge is zero on the left side (+14 + (-7 x 2) = 0), we have +6 (+3 x 2) on the right side:

^{ }

14H^{+} + Cr_{2}O_{7}^{2- }→ 2Cr^{3+ }+ 7H_{2}O

Therefore, **add 6e ^{– }**to balance the charges in the second equation:

6e^{– }+ 14H^{+} + Cr_{2}O_{7}^{2- }→ 2Cr^{3+ }+ 7H_{2}O

**4)** Balance the number of electrons by multiplying the equation(s) with a whole number. There are 6 e^{–} in the second equation, so multiply the first equation by 6:

6 x [Fe^{2+} → Fe^{3+}+ 1e^{–}] = 6Fe^{2+} → 6Fe^{3+}+ 6e^{–}

^{ }

**5)** Once everything is balanced, add the half-reactions together and cancel the electrons:

6Fe^{2+} → 6Fe^{3+}+ 6e^{–}6e^{–} + 14H^{+} + Cr_{2}O_{7}^{2- }→ 2Cr^{3+ }+ 7H_{2}O

______________________________

6Fe^{2+}+ 14H^{+} + Cr_{2}O_{7}^{2- }→ 6Fe^{3+ }+ 2Cr^{3+ }+ 7H_{2}O

Check if everything is balanced. In this case, it is!

**Balancing Redox Equations in Basic Solution**

For reactions in basic solutions, we proceed all the way to balancing the equation as if it is in an acidic solution. At the end, **for every H+, we add an equal number of OH ^{–} on both sides of the equation**. A pair of H

^{+}and OH

^{–}ions is then represented as a water molecule.

For example,

MnO_{4}^{–} + I^{– }→^{ }MnO_{2} + I_{2}

^{ }

**1)** Write the half-reactions:

MnO_{4}^{–} → MnO_{2}I^{–} → I_{2}

_{ }

**2)** Balance the atoms. *Add H _{2}O to balance the O atoms, and H^{+} to balance the H atoms.*

The atoms in the first half-reaction are balanced.

For the first reaction, we need to equalize the oxygens, and therefore, we **add 2 H _{2}O** on the right side:

MnO_{4}^{–} → MnO_{2 }+ 2H_{2}O

And now, **add 4 H+** on the left side to balance the hydrogens:

4H^{+ }+ MnO_{4}^{–} → MnO_{2 }+ 2H_{2}O

For the second reaction, **add 2 in front of I ^{–}**:

2I^{–} → I_{2}

**3)** Balance the charges by adding electrons.

To balance the charges in the first equation, we add 3e^{–} on the left side because there are three extra positive charges (+4 from the protons and -1 from the permanganate ion):

3e^{– }+ 4H^{+ }+ MnO_{4}^{–} → MnO_{2 }+ 2H_{2}O

For the second equation, we need to add 2e^{– }on the right side:

2I^{–} → I_{2 }+ 2e^{–}

**4)** Balance the number of electrons by multiplying the equation with whole numbers. There are 3 electrons in the first equation, and 2 electrons in the second, so ta balance this, we multiply the first equation by 2, and the second equation by 3:

2 x [3e^{– }+ 4H^{+ }+ MnO_{4}^{–} → MnO_{2 }+ 2H_{2}O] = 6e^{– }+ 8H^{+ }+ 2MnO_{4}^{–} → 2MnO_{2 }+ 4H_{2}O

3 x [2I^{–} → I_{2 }+ 2e^{–}] = 6I^{–} → 3I_{2 }+ 6e^{–}

_{ }

**5)** Once everything is balanced, add the half-reactions together and cancel the electrons:

6e^{–} + 8H^{+ }+ 2MnO_{4}^{–} → 2MnO_{2 }+ 4H_{2}O

6I^{–} → 3I_{2 }+ 6e^{–}

^{__________________________________________________}

8H^{+ }+ 2MnO_{4}^{– }+ 6I^{–} → 2MnO_{2 }+ 4H_{2}O + 3I_{2 }

_{ }

**6)** This is the balanced equation in acidic media, and one more step to make it in a basic media, is to add 8 OH^{–} on both sides of the equation to get rid of the 8H^{+} ions on the left side. This works on the principle of combining H^{+} and OH^{–} ions and representing it as water:

8OH^{–} + 8H^{+ }+ 2MnO_{4}^{– }+ 6I^{–} → 2MnO_{2 }+ 4H_{2}O + 3I_{2 }+ 8OH^{–}^{ }

8H_{2}O + 2MnO_{4}^{– }+ 6I^{–} → 2MnO_{2 }+ 4H_{2}O + 3I_{2 }+ 8OH^{–}

_{ }

Notice that we now have 8 H_{2}O on the left side, and only for on the right side. 4 H_{2}O molecules on both sides cancel out, and we are left with the balanced equation in a basic media:

2MnO_{4}^{– }+ 6I^{–} + 4H_{2}O → 2MnO_{2 }+ 3I_{2 }+ 8OH^{–}

## To **summarize the balancing of redox reactions**, remember that:

**1)** For reactions with only two elements, balance the atoms and then the charges by adding electrons where needed. Once the half-reactions are balanced, add them up and cancel any identical species that appear on both sides of the equation.

**2)** Balancing redox reactions in acidic media: Follow the same steps and remember, we *add H _{2}O to balance the O atoms, and H^{+} to balance the H atoms.*

* *

**3)** Balancing redox reactions in basic media: Follow the same steps as far balancing in acidic media and, in the end, add as many OH- ions as there are on either side of the equation. Convert an equal number of these ions to water, and cancel the water molecules on both sides of the equation.^{ }

**Check Also**

- Solutions
- Strong and Weak Electrolytes
- Dissociation of Ionic Compounds
- Molecular, Ionic, and Net Ionic Equations
- Molarity
- Dilution
- Ion Concentration
- Precipitation Reactions
- Definitions of Acids and Bases
- Acid-Base Reactions
- Stoichiometry of Reactions in Aqueous Solutions
- Acid-Base Titrations
- Oxidation State
- Oxidation-Reduction (Redox) Reactions
**Reactions in Aqueous Solutions Practice Problems**

#### Practice

Balance the following redox reactions occurring in **acidic** aqueous solution:

a) Al(*s*) + Fe^{2}^{+}(*aq*) → Al^{3}^{+}(*aq*) + Fe(*s*)

b) SO_{3}^{2}^{–}(*aq*) + MnO_{4}^{–}(*aq*) → SO_{4}^{2}^{–}(*aq*) + Mn^{2}^{+}(*aq*)

c) Cr_{2}O_{7}^{2-} + C_{2}O_{2}^{2-} → Cr^{3+} + CO_{2}

d) PbO_{2} + Mn^{2}^{+} + SO_{4}^{2}^{–} → PbSO_{4} + MnO_{4}^{–}

e) MnO_{4}^{−}(*aq*) + H_{2}O_{2}(*aq*) → Mn^{2+}(*aq*) + O_{2}(*g*)

Balance the following redox reactions occurring in **basic** aqueous solution:

a) Mn^{2}^{+} + ClO_{3}^{–} → MnO_{2} + ClO_{2}

b) MnO_{4}^{–}(*aq*) + Br^{–}(*aq*) → MnO_{2} + BrO_{3}^{–}(*aq*)

c) MnO_{4}^{–}(*aq*) ^{ }+ CN^{–}(*aq*) → CNO-(*aq*) + MnO_{2}(*s*)

d) H_{2}O_{2}(*aq*) + ClO_{2}(*aq*) → ClO_{2}^{–}(*aq*) + O_{2}(*g*)

e) MnO_{4}^{–}(*aq*) + Fe(OH)_{2}(*s*) → MnO_{2}(*s*) + Fe(OH)_{3}(*s*)

I like your teaching method

thank you for giving so many practice problems, they’ve been very helpful

Glad to hear that. Doing my best to make this as helpful as possible.