# Balancing Redox Reactions - Examples

Oxidation-Reduction or "redox" reactions occur when elements in a chemical reaction gain or lose electrons, causing an increase or decrease in oxidation numbers. The Half Equation Method is used to balance these reactions.

In a redox reaction, one or more element becomes oxidized, and one or more element becomes reduced. Oxidation is the loss of electrons whereas reduction is the gain of electrons. An easy way to remember this is to think of the charges: an element's charge is reduced if it gains electrons (an acronym to remember the difference is LEO = Lose Electron Oxidation & GER = Gain Electron Reduction). Redox reactions usually occur in one of two environments: acidic or basic. In order to balance redox equations, understanding oxidation states is necessary.

Some points to remember when balancing redox reactions:

• The equation is separated into two half-equations, one for oxidation, and one for reduction.
• The equation is balanced by adjusting coefficients and adding H2O, H+, and e- in this order:
1. Balance the atoms in the equation, apart from O and H.
2. To balance the Oxygen atoms, add the appropriate number of water (H2O) molecules to the other side.
3. To balance the Hydrogen atoms (including those added in step 2), add H+ ions.
4. Add up the charges on each side. They must be made equal by adding enough electrons (e-) to the more positive side.
• The e- on each side must be made equal; if they are not equal, they must be multiplied by appropriate integers to be made the same.
• The half-equations are added together, cancelling out the electrons to form one balanced equation. Cancel out as much as possible.
• (If the equation is being balanced in a basic solution, the appropriate number of OH- must be added to turn the remaining H+ into water molecules)
• The equation can now be checked to make sure it is balanced.

Next, these steps will be shown in another example:

Example $$\PageIndex{1A}$$: In Acidic Aqueous Solution

Balance this reaction

$\ce{MnO4^{-} + I^{-} -> I2 + Mn^{2+}} \nonumber$

Solution

Steps to balance:

Step 1: Separate the half-reactions that undergo oxidation and reduction.

Oxidation: $\ce{ I^{-} -> I2} \nonumber$

This is the oxidation half because the oxidation state changes from -1 on the left side to 0 on the right side. This indicates a gain in electrons.

Reduction: $\ce{ MnO4^{-} -> Mn^{2+}} \nonumber$

This is the reduction half because the oxidation state changes from +7 on the left side to +2 on the right side. This indicates a reduction in electrons.

Step 2: In order to balance this half reaction we must start by balancing all atoms other than any Hydrogen or Oxygen atoms.

Oxidation: $\ce{2I^{-} -> I2} \nonumber$

In order to balance the oxidation half of the reaction you must first add a 2 in front of the $$\ce{I}$$ on the left hand side so there is an equal number of atoms on both sides.

Reduction: $\ce{MnO4^{-} -> Mn^{2+}} \nonumber$

For the reduction half of the reaction, you can notice that all atoms other than Hydrogen and Oxygen are already balanced because there is one manganese atom on both sides of the half reaction.

Step 3: Balance Oxygen atoms by adding $$\ce{H2O}$$ to the side of the equation that needs Oxygen. Once you have completed this step add $$\ce{H^{+}}$$ to the side of the equation that lacks $$\ce{H}$$ atoms necessary to be balanced.

Oxidation: $\ce{2 I^{-} -> I2} \nonumber$

Because there are no Oxygen or Hydrogen atoms in this half of the reaction, it is not required to perform any kind of balancing.

Reduction: $\ce{ MnO4^{-} -> Mn^{2+} + 4 H2O} \nonumber$

The first step in balancing this reaction using step 3 is to add4 H2O atoms in order to balance the Oxygen atoms with the 4 on the other side of MnO4-

Reduction: $\ce{MnO4^{-} + 8 H^{+} -> Mn^{2+} + 4 H2O} \nonumber$

Now that the Oxygen atoms have been balanced you can see that there are 8 $$\ce{H}$$ atoms on the right hand side of the equation and none on the left. Therefore, you must add 8 $$\ce{H^{+}}$$ atoms to the left hand side of the equation to make it balanced.

Step 4: Now that the two half reactions have been balanced correctly one must balance the charges in each half reaction so that both the reduction and oxidation halves of the reaction consume the same number of electrons.

Oxidation: $\ce{2 I^{-} -> I2 + 2e^{-} } \nonumber$

Because of the fact that there are two $$\ce{I}$$'s on the left hand side of the equation which a charge of -1 we can state that the left hand side has an overall charge of -2. The I on the left side of the equation has an overall charge of 0. Therefore to balance the charges of this reaction we must add 2 electrons to the right side of the equation so that both sides of the equation have equal charges of -2.

Reduction: $\ce{5 e^{-} + 8 H^{+} + MnO4^{-} -> Mn^{2+} + 4 H2O} \nonumber$

Looking at the left hand side of the equation you can notice that there are 8 Hydrogen atoms with a +1 charge. There is also a $$\ce{MnO4^{-}}$$ ion that has a charge of -1. When we add these two charges up we can calculate that the left hand side of the equation has an overall charge of +7. The right hand side has an $$\ce{Mn}$$ atom with a charge of +2 and then 4 water molecules that have charges of 0. Therefore, the overall charge of the right side is +2. We must add 5 electrons to the left side of the equation to make sure that both sides of the equation have equal charges of +2.

Step 5: Multiply both sides of both reactions by the least common multiple that will allow the half-reactions to have the same number of electrons and cancel each other out.

Oxidation: $$10I^- \rightarrow 5I_2 +10e^-$$

We multiply this half reaction by 5 to come up with the following result above.

Reduction: $$10e^- + 16H^+ + 2MnO_4^- \rightarrow 2Mn^{2+} + 8H_2O$$

We multiply the reduction half of the reaction by 2 and arrive at the answer above.

By multiplying the oxidation half by 5 and the reduction half by 2 we are able to observe that both half-reactions have 10 electrons and are therefore are able to cancel each other out.

Step 6: Add the two half reactions in order to obtain the overall equation by canceling out the electrons and any $$\ce{H2O}$$ and $$\ce{H^{+}}$$ ions that exist on both sides of the equation.

Overall: $\ce{10 I^{-} + 16 H^{+} + 2 MnO4^{-} -> 5 I2 + 2 Mn^{2+} + 8 H_2O} \nonumber$

In this problem, there is not anything that exists on both halves of the equation that can be cancelled out other than the electrons. Finally, double check your work to make sure that the mass and charge are both balanced. To double check this equation you can notice that everything is balanced because both sides of the equation have an overall charge of +4.

Example $$\PageIndex{1B}$$: In Basic Aqueous Solutions

The balancing procedure in basic solution differs slightly because $$\ce{OH^{-}}$$ ions must be used instead of $$\ce{H^{+}}$$ ions when balancing hydrogen atoms. To give the previous reaction under basic conditions, sixteen $$\ce{OH^{-}}$$ ions can be added to both sides. on the left the $$\ce{OH^{-}}$$ and the $$\ce{H^{+}}$$ ions will react to form water, which will cancel out with some of the $$\ce{H2O}$$ on the right.

$\ce{10I^{-} (aq) + 2MnO4^{-} (aq) + 16H^{+} (aq) + 16OH^{-} (aq) -> 5I2 (s) + 2Mn^{2+} (aq) + 8H2O (l) + 16OH^{-} (aq)} \nonumber$

On the left side the OH- and the $$\ce{H^{+}}$$ ions will react to form water, which will cancel out with some of the $$\ce{H2O}$$ on the right:

$\ce{10I^{-} (aq) + 2MnO4^{-} (aq) + 16H2O (l) -> 5I2 (s) + 2Mn^{2+} (aq) + 8H2O (l) + 16OH^{-} (aq)} \nonumber$

Eight water molecules can be canceled, leaving eight on the reactant side:

$\ce{10I^{-} (aq) + 2MnO4^{-} (aq) + 8H2O (l) -> 5I2 (s) + 2Mn^{2+} (aq) + 16OH^{-} (aq)} \nonumber$

This is the balanced reaction in basic solution.

Example $$\PageIndex{2}$$

Balance the following in an acidic solution.

$\ce{SO3^{2-} (aq) + MnO4^{-} (aq) \rightarrow SO4^{2-} (aq) + Mn^{2+} (aq)} \nonumber$

Solution

To balance a redox reaction, first take an equation and separate into two half reaction equations specifically oxidation and reduction, and balance them.

Step 1: Split into two half reaction equations: Oxidation and Reduction

• Oxidation: $\ce{SO3^{2-} (aq) -> SO4^{2-} (aq)} \nonumber$ [ oxidation because oxidation state of sulfur increase from +4 to +6]
• Reduction: $\ce{MnO4^{+} (aq) -> Mn^{2+ }(aq)} \nonumber$ [ Reduction because oxidation state of $$\ce{Mn}$$ decreases from +7 to +2]

Step 2: Balance each of the half equations in this order:

• Atoms other than $$\ce{H}$$ and $$\ce{O}$$
• O atoms by adding $$\ce{H2O}$$ molecules with proper coefficients
• H atoms by adding $$\ce{H^{+}}$$ with proper coefficients

The $$\ce{S}$$ and $$\ce{Mn}$$ atoms are already balanced,

Balancing $$\ce{O}$$ atoms

\begin{align*} &\text{Oxidation}: \quad \ce{SO3^{2-}(aq)} + \ce{H2O(l)} \rightarrow \ce{SO4^{2-} (aq)} \\[4pt] &\text{Reduction}: \quad \ce{MnO4^{-} (aq)} \rightarrow \ce{Mn^{2+}(aq)} + \ce{4H2O(l)} \end{align*}

Then balance out $$\ce{H}$$ atoms on each side

\begin{align*} &\text{Oxidation}: \quad \ce{SO3^{2-}(aq)} + \ce{H2O(l)} \rightarrow \ce{SO4^{2-} (aq)} + \ce{2H^{+}(aq)}\\[4pt] &\text{Reduction}: \quad \ce{MnO4^{-} (aq)} + 8 \ce{H^{+}} \rightarrow \ce{Mn^{2+}(aq)} + \ce{4H2O(l)} \end{align*}

Step 3: Balance the charges of the half reactions by adding electrons

\begin{align*} &\text{Oxidation}: \quad \ce{SO3^{2-}(aq)} + \ce{H2O(l)} \rightarrow \ce{SO4^{2-} (aq)} + \ce{2H^{+}(aq)} + \ce{2e^{-}}\\[4pt] &\text{Reduction}: \quad \ce{MnO4^{-} (aq)} + 8 \ce{H^{+}} + \ce{5e^{-}} \rightarrow \ce{Mn^{2+}(aq)} + \ce{4H2O(l)} \end{align*}

Step 4: Obtain the overall redox equation by combining the half reaction, but multiply entire equation by number of electrons in oxidation with reduction equation, and number of electrons in reduction with oxidation equation.

\begin{align*} &\text{Oxidation}: \quad 5 \times \left[\ce{SO3^{2-}(aq)} + \ce{H2O(l)} \rightarrow \ce{SO4^{2-} (aq)} + \ce{2H^{+}(aq)} + \ce{2e^{-}} \right] \\[4pt] &\text{Reduction}: \quad 2 \times \left[ \ce{MnO4^{-} (aq)} + 8 \ce{H^{+}} + \ce{5e^{-}} \rightarrow \ce{Mn^{2+}(aq)} + \ce{4H2O(l)}\right] \end{align*}

Overall Reaction:

\begin{align*} &\text{Oxidation}: \quad \ce{5SO3^{2-}(aq)} + \ce{5H2O(l)} \rightarrow \ce{5 SO4^{2-} (aq)} + \ce{10 H^{+}(aq)} + \ce{10e^{-}} \\[4pt] &\text{Reduction}: \quad \ce{2 MnO4^{-} (aq)} + \ce{16H^{+}} + \ce{10e^{-}} \rightarrow \ce{2Mn^{2+}(aq)} + \ce{8H2O(l)} \\[4pt] \hline &\text{total}: \quad \ce{5SO3^{2-}(aq)} + \ce{5H2O(l)} + \ce{2 MnO4^{-} (aq)} + \ce{16H^{+}} + \ce{10e^{-}} \rightarrow \ce{5 SO4^{2-} (aq)} + \ce{10 H^{+}(aq)} + \ce{2Mn^{2+}(aq)} + \ce{8H2O(l)} + \ce{10e^{-}} \end{align*}

Step 5: Simplify and cancel out similar terms on both sides

$\ce{5SO3^{2-}(aq)} + \cancel{\ce{5H2O(l)}} + \ce{2 MnO4^{-} (aq)} + \ce{\cancelto{6}{16}H^{+}} + \cancel{\ce{10e^{-}}} \rightarrow \ce{5 SO4^{2-} (aq)} + \cancel{\ce{10 H^{+}(aq)}} + \ce{2Mn^{2+}(aq)} + \ce{\cancelto{3}{8}H2O(l)} + \cancel{\ce{10e^{-}}} \nonumber$

To get

$\ce{5SO3^{2-}(aq)} + \ce{2 MnO4^{-} (aq)} + \ce{6H^{+}} \rightarrow \ce{5 SO4^{2-} (aq)} + \ce{2Mn^{2+}(aq)} + \ce{3H2O(l)} \nonumber$

Example $$\PageIndex{3}$$:

Balance this reaction in both acidic and basic aqueous solutions

$\ce{MnO4^{-}(aq) + SO3^{2-}(aq) -> MnO2(s) + SO4^{2-}(aq)} \nonumber$

Solution

First, they are separated into the half-equations:

$\ce{MnO4^{-}(aq) -> MnO2(s)} \nonumber$

This is the reduction half-reaction because oxygen is LOST)

and

$\ce{SO3^{2-}(aq) -> SO4^{2-}(aq)} \nonumber$

(the oxidation, because oxygen is GAINED)

Now, to balance the oxygen atoms, we must add two water molecules to the right side of the first equation, and one water molecule to the left side of the second equation:

$\ce{MnO4^{-}(aq) -> MnO2(s) + 2H2O(l)} \nonumber$

$\ce{H2O(l) + SO3^{2-}(aq) -> SO4^{2-}(aq)} \nonumber$

To balance the hydrogen atoms (those the original equation as well as those added in the last step), we must add four H+ ions to the left side of the first equation, and two H+ ions to the right side of the second equation.

$\ce{4H^{+} + MnO4^{-}(aq) -> MnO2(s) + 2H2O(l)} \nonumber$

$\ce{H2O(l) + SO3^{2-}(aq) -> SO4^{2-}(aq) + 2H^{+}} \nonumber$

Now we must balance the charges. In the first equation, the charge is +3 on the left and 0 on the right, so we must add three electrons to the left side to make the charges the same. In the second equation, the charge is -2 on the left and 0 on the right, so we must add two electrons to the right.

$\ce{3e- + 4H^{+} + MnO4^{-}(aq) -> MnO2(s) + 2H2O(l)} \nonumber$

$\ce{H2O(l) + SO3^{2-}(aq) --> SO4^{2-}(aq) + 2H^{+} + 2e^{-}} \nonumber$

Now we must make the electrons equal each other, so we multiply each equation by the appropriate number to get the common multiple (in this case, by 2 for the first equation, and by 3 for the second).

$\ce{2(3e^{-} + 4H^{+} + MnO4^{-}(aq) -> MnO2(s) + 2H2O(l))} \nonumber$

$\ce{3(H2O(l) + SO3^{2-}(aq) -> SO4^{2-}(aq) + 2H^{+} + 2e^{-})} \nonumber$

With the result:

$\ce{6e^{-} + 8H^{+} + 2MnO4^{-}(aq) -> 2MnO2(s) + 4H2O(l)} \nonumber$

$\ce{3H2O(l) + 3SO3^{2-}(aq) -> 3SO4^{2-}(aq) + 6H^{+} + 6e^{-}} \nonumber$

Now we cancel and add the equations together. We can cancel the 6e- because they are on both sides. We can get rid of the 6H+ on both sides as well, turning the 8H+ in the first equation to $$\ce{2H^{+}}$$. The same method gets rid of the $$\ce{3H2O(l)}$$ on the bottom, leaving us with just one $$\ce{H2O(l)}$$ on the top. In the end, the overall reaction should have no electrons remaining. Now we can write one balanced equation:

$\ce{2MnO4^{-}(aq) + 2H^{+} + 3SO3^{2-}(aq) -> H2O(l) + 2MnO2(s) + 3SO4^{2-}(aq)} \nonumber$

The equation is now balanced in an acidic environment.

To balance in a basic environment add $$\ce{OH^{-}}$$ to each side to neutralize the $$\ce{H^{+}}$$ into water molecules:

$\ce{2MnO4^{-}(aq) + 2H2O + 3SO3^{2-}(aq) -> H2O(l) + 2MnO2(s) + 3SO4^{2-}(aq) + 2OH^{-}} \nonumber$

and then cancel the water molecules

$\ce{2MnO4^{-}(aq) + H2O + 3SO3^{2-}(aq) -> + 2MnO2(s) + 3SO4^{2-}(aq) + 2OH^{-}} \nonumber$

The equation is now balanced in a basic environment.

Example $$\PageIndex{4}$$

Balance this reaction in acidic solution

$\ce{Fe(OH)3 + OCl^{-} \rightarrow FeO4^{2-} + Cl^{-}} \nonumber$

Solution

Step 1:

• Reduction: $\ce{OCl^{-} -> Cl^{-} }\nonumber$
• Oxidation: $\ce{ Fe(OH)3 -> FeO4^{2-} }\nonumber$

Steps 2 and 3:

• Reduction: $\ce{2H^{+} + OCl^{-} + 2e^{-} -> Cl^{-} + H2O }\nonumber$
• Oxidation: $\ce{Fe(OH)3 + H2O -> FeO4^{2-} + 3e^{-} + 5H^{+} }\nonumber$

Step 4:

Overall Equation:

\begin{align*} 3 \times \big[ \ce{ 2H^{+} + OCl^{-} + 2e^{-}} &\ce{ -> Cl^{-} + H2O} \big] \\[4pt] \ce{ 6H^{+} + 3OCl^{-} + 6e^{-}} &\ce{ -> 3Cl^{-} + 3H2O} \end{align*}

and

\begin{align*} 2 \times \big[ \ce{Fe(OH)3 + H2O} & \ce{-> FeO4^{2-} + 3e- + 5H^{+}} \big] \\[4pt] \ce{2Fe(OH)3 + 2H2O} & \ce{-> 2FeO4^{2-} + 6e- + 10H^{+}} \end{align*}

$\ce{6H^{+} + 3OCl^{-} + 2e^{-} + 2Fe(OH)3 +2 H2O -> 3Cl^{-} +3 H2O + 2FeO4^{2-} + 6e^{-} + 10H^{+} }\nonumber$

Step 5:

Simplify:

$\ce{3OCl^{-} + 2Fe(OH)3 \rightarrow 3Cl^{-} + H2O + 2FeO4^{2-} + 4H^{+}} \nonumber$

Example $$\PageIndex{5}$$

Balance this equation in acidic aqueous solution

$\ce{VO4^{3-} + Fe^{2+} -> VO^{2+} + Fe^{3+}} \nonumber$

Solution

Step 1:

• Oxidation: $\ce{Fe^{2+} -> Fe^{3+}} \nonumber$
• Reduction: $\ce{VO4^{3-} -> VO^{2+}} \nonumber$

Step 2/3:

• Oxidation: $\ce{Fe^{2+} -> Fe^{3+} + e^{-}} \nonumber$
• Reduction: $\ce{6H^{+} + VO4^{3-} + e^{-} -> VO^{2+} + 3H2O} \nonumber$

Step 4:

Overall Reaction:

$\ce{Fe^{2+} -> Fe^{3+} + e^{-}} \nonumber$

+

$\ce{6H^{+} + VO4^{3-} + e^{-} -> VO^{2+} + 3H2O} \nonumber$

____________________________

$\ce{Fe^{2+} + 6H^{+} + VO4^{3-}} + \cancel{e^{-}} \ce{ \rightarrow Fe^{3+}} + \cancel{e^{-}} \ce{ + VO^{2+} + 3H2O} \nonumber$

Step 5:

Simplify:

$\ce{Fe^{2+} + 6H^{+} + VO4^{3-} \rightarrow Fe^{3+} + VO^{2+} + 3H2O} \nonumber$

Exercise $$\PageIndex{1}$$

Balance the following equation in both acidic and basic environments:

$\ce{Cr2O7^{2-}(aq) + C2H5OH(l) -> Cr^{3+}(aq) + CO2(g)} \nonumber$

In acidic aqueous solution: $\ce{2Cr2O7^{-}(aq) + 16H^{+}(aq) + C2H5OH(l) -> 4Cr^{3+}(aq) + 2CO2(g) + 11H2O(l)} \nonumber$

In basic aqueous solution: $\ce{2Cr2O7^{-}(aq) + 5H2O(l) + C2H5OH(l) -> 4Cr^{3+}(aq) + 2CO2(g) + 16OH^{-}(aq) } \nonumber$

Exercise $$\PageIndex{2}$$

Balance the following equation in both acidic and basic environments:

$\ce{Fe^{2+}(aq) + MnO4^{-}(aq) -> Fe^{3+}(aq) + Mn^{2+}(aq)} \nonumber$

In acidic aqueous solution: $\ce{ MnO4^{-}(aq) + 5Fe^{2+}(aq) + 8H^{+}(aq) -> Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H2O(l)} \nonumber$
In basic aqueous solution: $\ce{MnO4^{-}(aq) + 5Fe^{2+}(aq) + 4H2O(l) -> Mn^{2+}(aq) + 5Fe^{3+}(aq) + 8OH^{-}(aq) } \nonumber$