# Balancing Redox Reactions I

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.

### Introduction

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.

### Balancing

The first step is to identify which element(s) is being oxidized and which element(s) is being reduced.

For example looking at the following reaction in an acidic solution

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

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.

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.

### Step-by-step Examples

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.
Original Equation:  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]
>THESE ARE THE SKELETAL HALF EQUATIONS
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, hence we can go onto 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. You can 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)

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:

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.

### More Examples

Example 1Fe(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

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))

### 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.
2. Madsen, Dorte. Chapter 5 Lecture. 6 February 2009.
3. Petrucci, Ralph. General Chemistry. Ninth. Upper Saddle River: Pearson Education, 2007. 157. Print
4. Petrucci, Ralph. General Chemistry. Ninth. Upper Saddle River: Pearson Education, 2007. 159. Print
5. Helmenstine, Anne Marie. "How to Balance Redox Reactions - Balancing Redox Reactions." Balancing Redox Reactions - Half-Reaction Method (2009): n. pag. Web. 1 Dec 2009. http://chemistry.about.com/od/generalchemistry/ss/redoxbal.htm
6. Stanitski, Conrad L. "Chemical Equations." Chemistry Explained Foundations and Applications. 1st. Chemistry Encyclopedia, 2009. Print.
7. "How to Balance Redox Equations." Youtube. Web. 1 Dec 2009. <http://www.youtube.com/results?searc...balance+redox+>.

Online Practice:

13:56, 4 Dec 2013

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