BALANCING OF REDOX REACTIONS

Balancing by Ion Electron or Half Reaction Method

We are familiar with the balancing of chemical equations by inspection method. However, inspection method may not be useful for balancing the redox equations because in these equations, we have to keep in mind the conservation of charge as well as conservation of mass. The redox equations, are therefore, are balanced by using the concept of half equations and following certain set of rules. One of the methods used for balancing redox reactions is called ion-electron method.

Table of Contents

BALANCING BY ION ELECTRON OR HALF REACTION METHOD

We know that during redox reactions there is a change in oxidation number of the elements due to the transference of electrons. The basic principle involved in balancing the redox equation is that the number of electrons lost during oxidation is equal to the number of electrons gained during reduction.

STEPS INVOLVED IN BALANCING REDOX EQUATIONS

The various steps involved in the balancing of redox equations ion-electron method are as follows:

Indicate the oxidation number of each atom involved in the reaction. Identify the elements which undergo a change in the oxidation number.
Divide the skeleton redox equation into two half reactions; oxidation half and reduction half. In each half reaction balance the atoms which undergo the change in oxidation number.
In order to make up for the difference in O.N. add electrons to left hand side or right hand side of the arrow in each half reaction.
Balance oxygen atoms by addition of proper number of H₂O molecules to the side which is falling short of 0 atoms in each half reaction.
This step is meant for only ionic equations. It involves the balancing of H atoms in each half reaction as follows:

(i) For acidic medium Add proper number of H+ ions to the side falling short of H atoms.

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( ii) For basic medium. Add proper number of H₂O molecules to the side falling short of H atoms and equal number of OH- ions to the other side.

Equalise the number of electrons lost and gained by multiplying the half reactions with suitable integer and add them to get the final equation. The application of various steps described above has been illustrated as follows by balancing the redox equation representing the reaction between iodine and nitric acid.

HNO₃ + I₂ à HIO₃ + NO₂ + H₂O

Step 1. Indication of oxidation numbers of each atom

Thus, only nitrogen and iodine undergo change in oxidation number.

Step 2. Division into two half reactions and balancing the atoms undergoing change in O.N.

Step 3. Addition of electrons to make up the difference in O.N.

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(Each I atom loses 5e- therefore, two iodine atoms would lose 10e-)

(Each N atom gains 1 electron).

Step 4. Balancing of 0 by adding proper number of H₂O molecule to the side falling short of 0 atoms.

Step 5. Not required because the equation is not ionic.

Step 6. To equalise the electrons multiply reduction half reaction by 10 and add the two.

Let us now proceed to balance the ionic equations in acidic and basic mediums.

Consider the chemical reaction between zinc and copper(II) tetraoxosulphate(V) solution.

Zn(s) + CuSO₄(aq) à ZnSO(aq) + Cu(s)

Writing ionic form of each species we have

Zn(s) + Cu₂+(aq) + SO₄(aq) à Zn₂+(aq) + SO₄(aq) + Cu(s)

Now, it is evident that Cu atoms and Zn²⁺ ions undergo oxidation and reduction respectively and are active ions. SO₄ lion, on the other hand, do not participate in the reaction and remain as such. They are inactive towards the reaction and are called spectator ions. In the similar way K+ ions are spectator ions in the following reaction.

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)→Mn2+(aq)+I2(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) → Mn²⁺(aq)

The other describes the oxidation:

I−(aq)→I2(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^– → 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) → 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) → 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^–→ 2Mn²⁺(aq) + 8H₂O (l)

10I^–(aq) → 5I₂(s) + 10e^–

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10I^–(aq) + 2MnO₄^–(aq) + 16H⁺(aq) → 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) → 5I₂(s) + 2Mn²⁺(aq) + 8H₂O (l) + 16OH^–(aq)

On the left sidethe 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) + 16H₂O (l) → 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) → 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.

SO₃^2-(aq) + MnO₄^–(aq) → SO₄^2-(aq) + Mn²⁺(aq)

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

Oxidation: SO₃^2-(aq) → SO₄^2-(aq) [ oxidation because oxidation state of sulfur increase from +4 to +6]

Reduction: MnO₄⁺(aq) → 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: SO₃^2- (aq) + H₂O (l) → SO₄^–(aq)

Reduction: MnO₄^–(aq) → Mn²⁺(aq) + 4H₂O (l)

Then balance out H atoms on each side

Oxidation: SO₃^2-(aq) + 4H₂O (l) → SO₄^2-(aq) + 2H⁺(aq)

Reduction: MnO₄^–(aq) + 8H⁺(aq) → Mn²⁺ (aq) + 4H₂O (l)

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

Oxidation: SO₃^2-(aq) + H₂O (l) → SO₄^–(aq) + 2H⁺ (aq) + 2e^–

Reduction: MnO₄^– (aq) + 8H⁺+ 5e^– → 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:[ SO₃^2-(aq) + H₂O (l) → SO₄^–(aq) + 2H⁺ (aq) + 2e^–] x 5

Reduction: [ MnO₄^– (aq) + 8H⁺+ 5e^–→ Mn²⁺(aq) + 4H₂O (l) ] x 2

Overall Reaction:

Oxidation: 5 SO₃^2-(aq) + 5H₂O (l) → 5SO₄^2- (aq) + 10H⁺(aq) + 10e^–

Reduction: 2 MnO₄^–(aq) + 16H⁺(aq) +10e^–→ 2 Mn²⁺ (aq) + 8H₂O (l)

5 SO₃^2-(aq) + 5H₂O (l) + 2 MnO₄^–(aq) + 16H⁺(aq) +10e^–→ 5SO₄^2- (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 SO₃^2-(aq) + 2 MnO₄^–(aq) + 6H⁺(aq) → 5SO₄^2- (aq) + 2Mn²⁺ (aq) + 3H₂O (l)

Example 3

MnO₄^–(aq) + SO₃^2-(aq) –> MnO₂(s) + SO₄^2-(aq)

First, they are separated into the half-equations:

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

SO₃^2-(aq) –> SO₄^2-(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) + SO₃^2-(aq) –> SO₄^2-(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) + SO₃^2-(aq) –> SO₄^2-(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) + SO₃^2-(aq) –> SO₄^2-(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) + SO₃^2-(aq) –> SO₄^2-(aq) + 2H⁺ + 2e^–)

With the result:

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

3H₂O(l) + 3SO₃^2-(aq) –> 3SO₄^2-(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⁺ + 3SO₃^2-(aq) –> H₂O(l) + 2MnO₂(s) + 3SO₄^2-(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 + 3SO₃^2-(aq) –> H₂O(l) + 2MnO₂(s) + 3SO₄^2-(aq) + 2OH^–

The equation is now balanced in a basic environment.

Example 4

Fe(OH)₃ + OCl^– → FeO₄^2-+ Cl^–in acidic solution

Step 1:

Reduction: OCl^– → Cl^–

Oxidation: Fe(OH)₃ → FeO₄^2-

Step 2/3:

Reduction: 2H⁺+ OCl^– + 2e^– → Cl^– + H₂O

Oxidation: Fe(OH)₃ + H₂O → FeO₄^2- + 3e^– + 5H⁺

Step 4:

Overall Equation:

[ 2H⁺+ OCl^– + 2e^– → Cl^– + H₂O ] x 3

[ Fe(OH)₃ + H₂O → FeO₄^2- + 3e^– + 5H⁺] x 2

6H⁺+ 3OCl^– + 6e^– → 3Cl^– +3 H₂O

2Fe(OH)₃ +2 H₂O → 2FeO₄^2- + 6e^– + 10H⁺

6H⁺+ 3OCl^– + 2e^–+ 2Fe(OH)₃ +2 H₂O → 3Cl^– +3 H₂O + 2FeO₄^2- + 6e^– + 10H⁺

Step 5:

Simplify:

3OCl^– + 2Fe(OH)₃ → 3Cl^– + H₂O + 2FeO₄^2- + 4H⁺

Example 2: VO₄^3-+ Fe²⁺→ VO²⁺+ Fe³⁺in acidic solution

Step 1:

Oxidation: Fe²⁺→ Fe³⁺

Reduction: VO₄^3-→ VO²⁺

Step 2/3:

Oxidation: Fe²⁺→ Fe³⁺ + e^–

Reduction: 6H⁺ + VO₄^3-+ e^–→ VO²⁺ + 3H₂O

Step 4:

Overall Reaction:

Fe²⁺→ Fe³⁺ + e^–

6H⁺ + VO₄^3-+ e^–→ VO²⁺ + 3H₂O

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Fe²⁺+ 6H⁺ + VO₄^3-+ e^– → Fe³⁺ + e^– + VO²⁺ + 3H₂O

Step 5:

Simplify:

Fe²⁺+ 6H⁺ + VO₄^3-→ Fe³⁺ + VO²⁺ + 3H₂O

Practice Problems

Balance the following equations in both acidic and basic environments:

1) Cr₂O₇^2-(aq) + C₂H₅OH(l) –> Cn³⁺(aq) + CO₂(g)

2) Fe²⁺(aq) + MnO₄^–(aq) –> Fe³⁺(aq) + Mn²⁺(aq)

Solutions:

(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))

(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−→I2+Mn2+

Steps to balance :

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

Oxidation: I−→I2

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→Mn2+

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−→I2

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→Mn2+

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: 2I−→I2

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→Mn2++4H2O

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+8H+→Mn2++4H2O

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: 2I−→I2+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: 5e−+8H++MnO−4→Mn2++4H2O

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−→5I2+10e−

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

Reduction: 10e−+16H++2MnO−4→2Mn2++8H2O

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: 10I−+16H++2MnO−4→5I2+2Mn2++8H2O

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+8H2O→5I2+2Mn2++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.

Neutral Conditions

The first step to balance any redox reaction is to separate the reaction into half-reactions. The substance being reduced will have electrons as reactants, and the oxidized substance will have electrons as products. (Usually all reactions are written as reduction reactions in half-reaction tables. To switch to oxidation, the whole equation is reversed and the voltage is multiplied by -1.) Sometimes it is necessary to determine which half-reaction will be oxidized and which will be reduced. In this case, whichever half-reaction has a higher reduction potential will by reduced and the other oxidized.

Example 1.1: Balancing in a Neutral Solution

Balance the following reaction

Cu+(aq)+Fe(s)→Fe3+(aq)+Cu(s)(1.4)

Solution

Step 1: Separate the half-reactions. By searching for the reduction potential, one can find two separate reactions:

Cu+(aq)+e−→Cu(s)(1.5)

Fe3+(aq)+3e−→Fe(s)(1.6)

The copper reaction has a higher potential and thus is being reduced. Iron is being oxidized so the half-reaction should be flipped. This yields:

Cu+(aq)+e−→Cu(s)(1.7)

Fe(s)→Fe3+(aq)+3e−(1.8)

Step 2: Balance the electrons in the equations. In this case, the electrons are simply balanced by multiplying the entire Cu+(aq)+e−→Cu(s) half-reaction by 3 and leaving the other half reaction as it is. This gives:

3Cu+(aq)+3e−→3Cu(s)(1.9)

Fe(s)→Fe3+(aq)+3e−(1.10)

Step 3: Adding the equations give:

3Cu+(aq)+3e−+Fe(s)→3Cu(s)+Fe3+(aq)+3e−(1.11)

The electrons cancel out and the balanced equation is left.

3Cu+(aq)+Fe(s)→3Cu(s)+Fe3+(aq)(1.12)

Acidic Conditions

Acidic conditions usually implies a solution with an excess of H⁺ concentration, hence making the solution acidic. The balancing starts by separating the reaction into half-reactions. However, instead of immediately balancing the electrons, balance all the elements in the half-reactions that are not hydrogen and oxygen. Then, add H₂O molecules to balance any oxygen atoms. Next, balance the hydrogen atoms by adding protons (H⁺). Now, balance the charge by adding electrons and scale the electrons (multiply by the lowest common multiple) so that they will cancel out when added together. Finally, add the two half-reactions and cancel out common terms.

Example 1.2: Balancing in a Acid Solution

Balance the following redox reaction in acidic conditions.

Cr2O2−7(aq)+HNO2(aq)→Cr3+(aq)+NO−3(aq)(1.13)

Solution

Step 1: Separate the half-reactions. The table provided does not have acidic or basic half-reactions, so just write out what is known.

Cr2O2−7(aq)→Cr3+(aq)(1.14)

HNO2(aq)→NO−3(aq)(1.15)

Step 2: Balance elements other than O and H. In this example, only chromium needs to be balanced. This gives:

Cr2O2−7(aq)→2Cr3+(aq)(1.16)

HNO2(aq)→NO−3(aq)(1.17)

Step 3: Add H₂O to balance oxygen. The chromium reaction needs to be balanced by adding 7 H₂O molecules. The other reaction also needs to be balanced by adding one water molecule. This yields:

Cr2O2−7(aq)→2Cr3+(aq)+7H2O(l)(1.18)

HNO2(aq)+H2O(l)→NO−3(aq)(1.19)

Step 4: Balance hydrogen by adding protons (H⁺). 14 protons need to be added to the left side of the chromium reaction to balance the 14 (2 per water molecule * 7 water molecules) hydrogens. 3 protons need to be added to the right side of the other reaction.

14H+(aq)+Cr2O2−7(aq)→2Cr3+(aq)+7H2O(l)(1.20)

HNO2(aq)+H2O(l)→3H+(aq)+NO−3(aq)(1.21)

Step 5: Balance the charge of each equation with electrons. The chromium reaction has (14+) + (2-) = 12+ on the left side and (2 * 3+) = 6+ on the right side. To balance, add 6 electrons (each with a charge of -1) to the left side:

6e−+14H+(aq)+Cr2O2−7(aq)→2Cr3+(aq)+7H2O(l)(1.22)

For the other reaction, there is no charge on the left and a (3+) + (-1) = 2+ charge on the right. So add 2 electrons to the right side:

HNO2(aq)+H2O(l)→3H+(aq)+NO−3(aq)+2e−(1.23)

Step 6: Scale the reactions so that the electrons are equal. The chromium reaction has 6e^– and the other reaction has 2e^–, so it should be multiplied by 3. This gives:

3∗[HNO2(aq)+H2O(l)→3H+(aq)+NO−3(aq)+2e−]⇒(1.24)

3HNO2(aq)+3H2O(l)→9H+(aq)+3NO−3(aq)+6e−(1.25)

6e−+14H+(aq)+Cr2O2−7(aq)→2Cr3+(aq)+7H2O(l).(1.26)

Step 7: Add the reactions and cancel out common terms.

[3HNO2(aq)+3H2O(l)→9H+(aq)+3NO−3(aq)+6e−]+(1.27)

[6e−+14H+(aq)+Cr2O2−7(aq)→2Cr3+(aq)+7H2O(l)]=(1.28)

3HNO2(aq)+3H2O(l)+6e−+14H+(aq)+Cr2O2−7(aq)→9H+(aq)+3NO−3(aq)+6e−+2Cr3+(aq)+7H2O(l)(1.29)

The electrons cancel out as well as 3 water molecules and 9 protons. This leaves the balanced net reaction of:

3HNO2(aq)+5H+(aq)+Cr2O2−7(aq)→3NO−3(aq)+2Cr3+(aq)+4H2O(l)(1.30)

Basic Conditions

Bases dissolve into OH^– ions in solution; hence, balancing redox reactions in basic conditions requires OH^–. Follow the same steps as for acidic conditions. The only difference is adding hydroxide ions (OH^–) to each side of the net reaction to balance any H⁺. OH^– and H⁺ ions on the same side of a reaction should be added together to form water. Again, any common terms can be canceled out.

Example 1.1: Balancing in Basic Solution

Balance the following redox reaction in basic conditions.

Ag(s)+Zn2+(aq)→Ag2O(aq)+Zn(s)(1.31)

Solution

Go through all the same steps as if it was in acidic conditions.

Step 1: Separate the half-reactions.

Ag(s)→Ag2O(aq)(1.32)

Zn2+(aq)→Zn(s)(1.33)

Step 2: Balance elements other than O and H.

2Ag(s)→Ag2O(aq)(1.34)

Zn2+(aq)→Zn(s)(1.35)

Step 3: Add H₂O to balance oxygen.

H2O(l)+2Ag(s)→Ag2O(aq)(1.36)

Zn2+(aq)→Zn(s)(1.37)

Step 4: Balance hydrogen with protons.

H2O(l)+2Ag(s)→Ag2O(aq)+2H+(aq)(1.38)

Zn2+(aq)→Zn(s)(1.39)

Step 5: Balance the charge with e^–.

H2O(l)+2Ag(s)→Ag2O(aq)+2H+(aq)+2e−(1.40)

Zn2+(aq)+2e−→Zn(s)(1.41)

Step 6: Scale the reactions so that they have an equal amount of electrons. In this case, it is already done.

Step 7: Add the reactions and cancel the electrons.

H2O(l)+2Ag(s)+Zn2+(aq)→Zn(s)+Ag2O(aq)+2H+(aq).(1.42)

Step 8: Add OH^– to balance H⁺. There are 2 net protons in this equation, so add 2 OH^– ions to each side.

H2O(l)+2Ag(s)+Zn2+(aq)+2OH−(aq)→Zn(s)+Ag2O(aq)+2H+(aq)+2OH−(aq).(1.43)

Step 9: Combine OH^– ions and H⁺ ions that are present on the same side to form water.

H2O(l)+2Ag(s)+Zn2+(aq)+2OH−(aq)→Zn(s)+Ag2O(aq)+2H2O(l)(1.44)

Step 10: Cancel common terms.

2Ag(s)+Zn2+(aq)+2OH−(aq)→Zn(s)+Ag2O(aq)+H2O(l)(1.45)

IDENTIFICATION OF OXIDIZING AND REDUCING AGENTS

The main function of oxidising agents is to bring about the oxidation of desired material. Similarly, reducing agents are required to bring about the reduction of desired material in the laboratory. The choice of the oxidising agent/reducing agents to be used depends upon the nature of the material to be oxidised or reduced. The reduction half reactions of some important oxidising agents are given in Table 30.2. Similarly, the oxidation half reactions of some important reducing agents are given in Table 30.3.

(a). Iron(II) chloride, FeCl2 – the solution of this substance (Fe2+) is green in color. When reacted with a substance suspected to be an oxidizing agent, and its color changes to reddish brown, then the substance is confirmed to be an oxidizing agent. The Fe2+ ion is oxidized to Fe3+ ion.
(b). When a strip of moist starch – iodide paper is dipped into a substance suspected to be an oxidizing agent, and a blue black color appears, then the substance is confirmed to be an oxidizing agent.
The oxidizing agent oxidizes the iodide to free iodine. The solution of the free iodine then reacts with the starch to give the blue black color (notice that it is only the solution of the free iodine which gives the blue black color with the starch).
(c). When hydrogen sulphide (H2S) is reacted with a substance, and some yellow deposits (sulphur) are settled on the bottom of the container, then the substance is an oxidizing agent – H2S is oxidized to sulphur.

TEST FOR REDUCING AGENTS
(a). Reacting the suspected reducing agent with potassium tetraoxomanganate(VII) solution, KMnO4 (purple color), and the color is observed to become colorless (Mn2+), the substance is confirmed to be a reducing agent. KMnO4 is reduced to Mn2+.
(b). Also, a solution of potassium heptaoxodichromate(VI), K2Cr2O7 (orange in color) would become greenish (Cr3+) when reacted with a reducing agent.

EVALUATION

2

Which of the following process involves the use of an oxidising agent?

Acidified potassium permanganate being used to form ethanoic acid from ethanol

Using carbon monoxide to obtain iron from iron ore in a blast furnace

Lithium aluminium hydride being used to change propanone into propan-2-ol

3

Which of the following reactions is an example of a redox process?

Black carbon powder burning in oxygen to form colourless carbon dioxide gas

Ethanol and a butanoic acid reacting together to form an ester

Magnesium metal displacing zinc from a solution of zinc sulfate

4

What is the equation for the redox reaction involving the following two ion-electron equations?

5

What is added to complex ion-electron equations to balance the electrical charge?

Water molecules (H₂O)

Hydrogen ions (H⁺)

Electrons (e^–)

6

Which of the following is the correctly balanced ion-electron equation for the change ?

7

Which of the following is the correctly balanced ion-electron equation for the change Br₂ (l) giving BrO^–(aq)?

8

Which of the following is the correctly balanced ion-electron equation for the change ClO₃^– (aq) giving Cl₂ (g)?

9

Which of the following is the correctly balanced ion-electron equation for the change XeO₃ (s) giving Xe (g)?

10

Which of the following is the correctly balanced ion-electron equation for the change CrO₄^2- (aq) giving Cr₂O₇^2- (aq)?

Balancing by Ion Electron or Half Reaction Method

BALANCING BY ION ELECTRON OR HALF REACTION METHOD

STEPS INVOLVED IN BALANCING REDOX EQUATIONS

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Balancing Redox Reactions: Examples

Practice Problems

Neutral Conditions

Acidic Conditions

Basic Conditions

IDENTIFICATION OF OXIDIZING AND REDUCING AGENTS

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