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# 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 1a: In Acidic Aqueous Solution

For example looking at the following reaction in an acidic solution

$MnO_4^- (aq) + I^- (aq) \rightarrow Mn^{2+} (aq) + I_2(s)$

The first step is to identify which element(s) is being oxidized and which element(s) is being reduced. Manganese(Mn) goes from a charge of +7 to a charge of +2. Since it gains five electrons, it is reduced. Because it is reduced, it is considered the oxidizing agent of the reaction.

Iodine (I) goes from a charge of -1 to 0. Thus it loses electrons, and is oxidized. Iodine (I) is therefore considered to be the reducing agent of this reaction.

The next step is to break the equation down into two half equations, one each for the reduction and oxidation reactions. At a later stage, the equations will be recombined into a balanced overall equation.

The first half reaction is for the reduction: MnO4- (aq) $$\rightarrow$$ Mn2(aq)

The other describes the oxidation:

$I^- (aq) \rightarrow I_2 (s)$

The overall equation is balanced as follows. For the reduction half, oxygen is the only element that requires balancing, which is accomplished by adding H2O to the product side. Since there are four oxygen atoms in MnO4(aq), four water molecules must be added so that the oxygen of the MnO4(aq) balances with the oxygen in the H2O. However, the added water introduces hydrogen into the reaction, which must be balanced as well. Under acidic conditions, this is accomplished by adding hydrogen ions to the reactant side. Electrons must also be added to balance the charges. Because on the left there is a charge of +7 and on the right a charge of +2, five electrons must be added to the +7 side:

MnO4- (aq) + 8H+ (aq) + 5e- $$\rightarrow$$ Mn2(aq) +4H2O (l)

The reduction half reaction is now sufficiently balanced.

The only element in the oxidation half reaction is iodine. This is easily balanced by doubling the iodide ions on the reactant side and adding electrons to the product side to balance the negative charge:

2I(aq) $$\rightarrow$$ I2 (s) + 2e-

(Note that the electrons were also multiplied because another iodide, with another negative charge, was added. Also,  a general rule of thumb is that the electrons should appear on opposite sides in each half reaction so they will cancel out when the reactions are combined.)

Now that both parts of the reaction are balanced, it is time to reform the overall reaction equation. Consider the electrons in each reaction. There are currently five electrons on the left in the reduction half reaction, and two on the right in the oxidation half reaction. The least common multiple of five and two is ten, so the reduction reaction is multiplied by two and the oxidation reaction is multiplied by five:

2MnO4(aq) + 16H(aq) + 10e-$$\rightarrow$$ 2Mn2(aq) + 8H2O (l)

10I(aq) $$\rightarrow$$ 5I2 (s) + 10e-.

The two reactions are added together, and common terms on either side of the reaction arrow are eliminated (in this case, only the electrons) to add the two parts together and cancel out anything that is on both sides, in this case, only the electrons can be cancelled out:

2MnO4(aq) + 16H(aq) + 10e-$$\rightarrow$$ 2Mn2(aq) + 8H2O (l)

10I(aq) $$\rightarrow$$ 5I2 (s) + 10e-

_________________________________________________

10I(aq) + 2MnO4(aq) + 16H(aq) $$\rightarrow$$ 5I2 (s) + 2Mn2(aq) + 8H2O (l)

This is the balanced reaction equation in acidic solution.

Example 1b: In Basic Aqueous Solution

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

10I(aq) + 2MnO4(aq) + 16H(aq) + 16OH- (aq) $$\rightarrow$$ 5I2 (s) + 2Mn2(aq) + 8H2O (l) + 16OH- (aq)

On the left sidethe OHand the Hions will react to form water, which will cancel out with some of the H2O on the right:

10I(aq) + 2MnO4(aq) + 16H2O (l) $$\rightarrow$$ 5I2 (s) + 2Mn2(aq) + 8H2O (l) + 16OH- (aq)

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

10I(aq) + 2MnO4(aq) + 8H2O (l) $$\rightarrow$$ 5I2 (s) + 2Mn2(aq) + 16OH(aq)

This is the balanced reaction in basic solution.

Example 2
To balance a redox reaction, first take an equation and separate into two half reaction equations specifically oxidation and reduction, and balance them. Balance the following in an acidic solution.
SO32(aq) + MnO4- (aq) $$\rightarrow$$ SO42- (aq) + Mn2+ (aq)

Step 1: Split into two half reaction equations: Oxidation and Reduction
Oxidation: SO32- (aq) $$\rightarrow$$ SO42- (aq)  [ oxidation because oxidation state of sulfur increase from +4 to +6]
Reduction: MnO4+ (aq) $$\rightarrow$$ Mn2+ (aq) [ Reduction because oxidation state of Mn decreases from +7 to +2]

Step 2: Balance each of the half equations in this order:
• Atoms other than H and O
• O atoms by adding H2Os with proper coefficient
• H atoms by adding H+ with proper coefficient
The Sulfur atoms and Mn atoms are already balanced,

Balancing O atoms
Oxidation: SO32- (aq) + H2O (l) $$\rightarrow$$ SO4- (aq)
Reduction: MnO4- (aq) $$\rightarrow$$ Mn2+ (aq) + 4H2O (l)
Then balance out H atoms on each side
Oxidation: SO32- (aq) + 4H2O (l) $$\rightarrow$$ SO42- (aq) + 2H+ (aq)
Reduction: MnO4- (aq) + 8H+ (aq) $$\rightarrow$$ Mn2+ (aq) + 4H2O (l)

Step 3: Balance the charges of the half reactions by adding electrons
Oxidation: SO32- (aq) + H2O (l) $$\rightarrow$$ SO4- (aq) + 2H+  (aq) + 2e-
Reduction: MnO4- (aq) + 8H+ + 5e- $$\rightarrow$$ Mn2+ (aq) + 4H2O (l)

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.

Oxidation:[ SO32- (aq) + H2O (l) $$\rightarrow$$ SO4- (aq) + 2H+  (aq) + 2e-] x 5
Reduction: [ MnO4- (aq) + 8H+ + 5e-$$\rightarrow$$ Mn2+ (aq) + 4H2O (l) ] x 2

Overall Reaction:
Oxidation: 5 SO32- (aq) + 5H2O (l) $$\rightarrow$$ 5SO42- (aq) + 10H+ (aq) + 10e-
+
Reduction: 2 MnO4- (aq) + 16H+ (aq) +10e- $$\rightarrow$$ 2 Mn2+ (aq) + 8H2O (l)
5 SO32- (aq) + 5H2O (l) + 2 MnO4- (aq) + 16H+ (aq) +10e- $$\rightarrow$$ 5SO42- (aq) + 10H+ (aq) + 10e- +2 Mn2+ (aq) + 8H2O (l)

Step 5: Simplify and cancel out similar terms on both sides, like the 10e- and waters.
5 SO32- (aq) + 2 MnO4- (aq) + 6H+ (aq)  $$\rightarrow$$ 5SO42- (aq) + 2Mn2+ (aq) + 3H2O (l)
Example 3

MnO4-(aq) + SO32-(aq) --> MnO2(s) + SO42-(aq)

First, they are separated into the half-equations:

MnO4-(aq) --> MnO2(s) (the reduction, because oxygen is LOST) and

SO32-(aq) --> SO42-(aq) (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:

MnO4-(aq) --> MnO2(s) + 2H2O(l)

H2O(l) + SO32-(aq) --> SO42-(aq)

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.

4H+ + MnO4-(aq) --> MnO2(s) + 2H2O(l)

H2O(l) + SO32-(aq) --> SO42-(aq) + 2H+

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.

3e- + 4H+ + MnO4-(aq) --> MnO2(s) + 2H2O(l)

H2O(l) + SO32-(aq) --> SO42-(aq) + 2H+ + 2e-

Now we must make the electrons equal eachother, 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).

2(3e- + 4H+ + MnO4-(aq) --> MnO2(s) + 2H2O(l))

3(H2O(l) + SO32-(aq) --> SO42-(aq) + 2H+ + 2e-)

With the result:

6e- + 8H+ + 2MnO4-(aq) --> 2MnO2(s) + 4H2O(l)

3H2O(l) + 3SO32-(aq) --> 3SO42-(aq) + 6H+ + 6e-

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 2H+. The same method gets rid of the 3H2O(l) on the bottom, leaving us with just one H2O(l) on the top. In the end, the overall reaction should have no electrons remaining. Now we can write one balanced equation:

2MnO4-(aq) + 2H+ + 3SO32-(aq) --> H2O(l) + 2MnO2(s) + 3SO42-(aq)

The equation is now balanced in an acidic environment. If necessary, we can balance in a basic environment by adding OH- to turn the H+ into water molecules as follows:

2MnO4-(aq) + H2O + 3SO32-(aq) --> H2O(l) + 2MnO2(s) + 3SO42-(aq) + 2OH-

The equation is now balanced in a basic environment.

Example 4

Fe(OH)3 + OCl- $$\rightarrow$$ FeO42+ Cl- in acidic solution

Step 1:
Reduction: OCl-  $$\rightarrow$$ Cl-
Oxidation:   Fe(OH)3 $$\rightarrow$$ FeO42-
Step 2/3:
Reduction:  2H+  OCl- + 2e- $$\rightarrow$$ Cl- + H2O
Oxidation:  Fe(OH)3 + H2O  $$\rightarrow$$ FeO42- + 3e- + 5H+
Step 4:
Overall Equation:
[ 2H+  OCl- + 2e- $$\rightarrow$$ Cl- + H2O ] x 3
[ Fe(OH)3 + H2O  $$\rightarrow$$ FeO42- + 3e- + 5H] x 2
=
6H+ 3OCl- + 6e- $$\rightarrow$$ 3Cl- +3 H2O
+
2Fe(OH)3 +2 H2O  $$\rightarrow$$ 2FeO42- + 6e- + 10H+
6H+ 3OCl- + 2e+ 2Fe(OH)3 +2 H2O  $$\rightarrow$$ 3Cl- +3 H2O + 2FeO42- + 6e- + 10H+

Step 5:
Simplify:
3OCl- + 2Fe(OH)3  $$\rightarrow$$ 3Cl- + H2O + 2FeO42- + 4H+
Example 2VO43+ Fe2$$\rightarrow$$ VO2+ Fe3in acidic solution
Step 1:
Oxidation:    Fe2$$\rightarrow$$ Fe3+
Reduction:   VO43$$\rightarrow$$ VO2+

Step 2/3:
Oxidation: Fe2$$\rightarrow$$ Fe3+ + e-
Reduction: 6H+ + VO43+ e$$\rightarrow$$ VO2+ + 3H2O
Step 4:
Overall Reaction:
Fe2$$\rightarrow$$ Fe3+ + e-
+
6H+ + VO43e-$$\rightarrow$$ VO2+ + 3H2O
____________________________

Fe2 6H+ + VO43+ e- $$\rightarrow$$ Fe3+ + e- + VO2+ + 3H2O
Step 5:
Simplify:
Fe2 6H+ + VO43-  $$\rightarrow$$ Fe3+  + VO2+ + 3H2O

### Practice Problems

Balance the following equations in both acidic and basic environments:

1) Cr2O72-(aq) + C2H5OH(l) --> Cn3+(aq) + CO2(g)

2) Fe2+(aq) + MnO4-(aq) --> Fe3+(aq) + Mn2+(aq)

Solutions:

1. (Acidic Answer: 2Cr2O7-(aq) + 16H+(aq) + C2H5OH(l) --> 4Cr3+(aq) + 2CO2(g) + 11H2O(l))

(Basic Answer: 2Cr2O7-(aq) + 5H2O(l) + C2H5OH(l) --> 4Cr3+(aq) + 2CO2(g) + 16OH-(aq))

2. (Acidic Answer: MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) --> Mn2+(aq) + 5Fe3+(aq) + 4H2O(l))

(Basic Answer: MnO4-(aq) + 5Fe2+(aq) + 4H2O(l) --> Mn2+(aq) + 5Fe3+(aq) + 8OH-(aq))

In a redox reaction, also known as an oxidation-reduction reaction, it is a must for oxidation and reduction to occur simultaneously. In the oxidation half of the reaction, an element gains electrons. A species loses electrons in the reduction half of the reaction. These reactions can take place in either acidic or basic solutions.

Example 1: Balance in Acid Solution

Problem : $$MnO_4^- + I^- \rightarrow I_2 + Mn^{2+}$$

Steps to balance :

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

Oxidation: $$I^- \rightarrow I_2$$

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: $$MnO_4^- \rightarrow Mn^{2+}$$

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.

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

Oxidation: $$2I^- \rightarrow I_2$$

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

Reduction: $$MnO_4^- \rightarrow Mn^{2+}$$

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.

3) Balance Oxygen atoms by adding H2O to the side of the equation that needs Oxygen. Once you have completed this step add H+ to the side of the equation that lacks H atoms necessary to be balanced.

Oxidation: $$2 I^- \rightarrow I_2$$

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: $$MnO_4^- \rightarrow Mn^{2+} + 4 H_2O$$

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: $$MnO_4^- + 8 H^+ \rightarrow Mn^{2+} + 4 H_2O$$

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

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: $$2 I^- \rightarrow I_2 + 2e^-$$

Because of the fact that there are two 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: $$5 e^- + 8 H^+ + MnO_4^- \rightarrow Mn^{2+} + 4 H_2O$$

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 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 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.

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.

6) Add the two half reactions in order to obtain the overall equation by canceling out the electrons and any H2O and H+ ions that exist on both sides of the equation.

Overall: $$10 I^- + 16 H^+ + 2 MnO_4^- \rightarrow 5 I_2 + 2 Mn^{2+} + 8 H_2O$$

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 1b: Balance in Basic Solution

The remaining question lies as to how to balance the reaction is basic solution. The simplest way to do this is to realize that OH- and H+ ions will combine to form water. Therefore, we must add OH- ions to both sides of the equation. Chances are that after completing this step you will also need to subtract water molecules from one side.

$10I^- + 2MnO_4^- + 8H_2O \rightarrow 5I_2 + 2Mn{2+} + 16OH^-$

As you can see, there are 16H+ and 16OH- ions on the left side of equation which will combine to form the 16 water molecules on the left side. Because there are 8 water molecules on the right side of the equation and 16 on the left we can subtract the 8 from the right side over to the left and obtain the overall equation above. Now you have balanced the reaction is basic solution.

### References

1. Petrucci, Ralph, William Harwood, Geoffrey Herring, and Jeffry Madura. General Chemistry: Principles & Modern Applications. 9th edition. Upper Saddle River, New Jersey: Pearson Prentince Hall, 2007.