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Balancing Redox Equations

Iron (Fe) rusts (is oxidized) when it reacts with oxygen (O2) in the air.

2Fe + 3O2 Fe2O3 [iron(III) oxide]

A new iron bolt (Fe) A rusted (oxidized) bolt(rust is Fe2O3)

Reduction/Oxidation = Redox

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Redox EquationsRusting is a an example of a redox reaction.

2Fe + 3O2 Fe2O3 [iron(III) oxide]

In redox reactions, one chemical is oxidized (e.g., Fe is oxidized) and another chemical is reduced (e.g., O2 is reduced).

Oxidation and reduction always occur together. One cannot occur without the other.In this reaction, when iron is oxidized, it gains oxygen.This definition of oxidation (‘gain of oxygen’), while correct in this case, is inadequate, because many redox reactions do not involve oxygen. A better definition follows, but do not disregard the example of rusting. It is something we all have observed and will help you understand and remember the broader definitions.

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Redox Equations

2Fe + 3O2 Fe2O3 [iron(III) oxide] 0

lose 3 e’s, LEO

0 +3 -2

gain 2 e’s, GER

What are the oxidation numbers of iron and oxygen in the rust equation?Definitions to Memorize: Loss of Electrons = OxidationGain of Electrons = ReductionTo do redox chemistry, you must be able to determine which atoms have lost e’s (have been oxidized) and which atoms have gained e’s (have been reduced).Thus, iron has been oxidized by oxygen. Oxygen is an ‘oxidizing agent’. and oxygen has been reduced by iron. Iron is a ‘reducing agent’.

the Lion saysLEO GER

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Balancing Redox Equations by the Half-Reaction Method

The following rules describe the most versatile method for balancing redox equations ...1. Break the eqn. into two ½-reactions, one for oxidation (oxid.) and one for reduction (red.)2. Balance all atoms in each ½-reaction other than H and O3. Balance O and H as follows ...

a) first balance O by adding H2O, b) then balance H’s by adding H+

c) finally, balance the charges by adding sufficient electronse-'s are always added to the right side of the oxid. ½-reaction because oxid. = loss of e-'se-'s are always added to the left side of the red. ½-reaction because red. = gain of e-'s4. Multiply each ½-reaction by a coefficient so that the same number of e-'s are transferred in

each ½-reaction. The number of e’s gained must equal the number of e’s lost.5. Add the ½-reactions and cancel common terms (e.g., H2O and/or H+) from each side.6. If the reaction is carried out in neutral or basic solution, add enough OH- to neutralize all H+

(combine them to make H2O) and be sure to add the same quantity of OH- to both sides to keep mass and charge balanced.

7. Cancel common terms (H2O and/or OH-) from each side and check for mass and charge balance

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Bromate and sulfite react as shown : BrO3- + SO3

-2 Br2 + SO4-2

Oxidation ½-reaction:1. 3a) balance O w. H2O3b) balance H w. H+

3c) balance charge w e’s4.

Reduction ½-reaction:1.

2. balance Br

3a) balance O w. H2O3b) balance H w. H+

3c) balance charge w. e’s

+5 +4 0 +6

lose 2 e’s, LEO

gain 5 e’s, GER

4. Multiply both ½-reactions by the appropriate coefficient so that both reactions involve the same number of e’s.Find the lowest common multiple (LCM) of the e’s transferred (10 e’s is the LCM of 2e’s and 10e’s).Combine the left sides of the ½-reactions and combine the right sides of the ½-reactions and reduce them.

5H2O + 5SO3-2 + 10e- + 12H+ + 2BrO3

- 5SO4-2 + 10H+ + Br2 + 6H2O + 10e-

5SO3-2 + 2H+ + 2BrO3

- 5SO4-2 + Br2 + H2O

SO3-2 SO4

-2 H2O + SO3

-2 SO4-2

H2O + SO3-2 SO4

-2 + 2H+

H2O + SO3-2 SO4

-2 + 2H+ + 2e-

(H2O + SO3-2 SO4

-2 + 2H+ + 2e-) × 5

BrO3- Br2

2BrO3- Br2

2BrO3- Br2 + 6H2O

12H+ + 2BrO3- Br2 + 6H2O

10e- + 12H+ + 2BrO3- Br2 + 6H2O

BrO3-, the oxidizer is reduced

SO3-2, the reducer is oxidized

O can only be balanced w. H2OH can only be balanced w. H+

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H2O2 O2 H2O2 O2 + 2H+

H2O2 O2 + 2H+ + 2e-

2OH- + H2O2 O2 + 2H2O + 2e-

(2OH- + H2O2 O2 + 2H2O + 2e-) × 3

MnO4- MnO2

MnO4- MnO2 + 2H2O

4H+ + MnO4- MnO2 + 2H2O

3e- + 4H+ + MnO4- MnO2 + 2H2O

3e- + 4H2O + MnO4- MnO2 + 2H2O + 4OH-

3e- + 2H2O + MnO4- MnO2 + 4OH-

(3e- + 2H2O + MnO4- MnO2 + 4OH-) × 2

Permanganate reacts w. hydrogen peroxide: MnO4- + H2O2 MnO2 + O2

Oxidation ½-reaction:1. 3b) balance H w. H+

3c) balance charge w e’s6. neutralize H+ w. OH-

Reduction ½-reaction:1.

3a) balance O w. H2O3b) balance H w. H+

3c) balance charge w. e’s6. neutralize H+ w. OH-

7. cancel H2O

lose 1 e’s, LEOgain 3 e’s, GER

+7 -1 +4 0

Multiply both ½-reactions by the appropriate coefficient so that both reactions involve the same number of e’s.Find the lowest common multiple (LCM) of the e’s transferred (6 e’s is the LCM of 2e’s and 3e’s)

6OH- + 3H2O2 + 6e- + 4H2O + 2MnO4- 2MnO2 + 8OH- + 3O2 + 6H2O + 6e-

3H2O2 + 2MnO4- 2MnO2 + 2OH- + 3O2 + 2H2O

MnO4-, the oxidizer is reduced

H2O2, the reducer is oxidized

e’s are always on the right side of an oxidation ½-reaction

e’s are always on the left side of a reduction ½-reaction

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lose 7 e’s, LEO

C2H6 CO2 C2H6 2CO2

4H2O + C2H6 2CO2

4H2O + C2H6 2CO2 + 14H+ 4H2O + C2H6 2CO2 + 14H+ + 14e-

4H2O + C2H6 + 14OH- 2CO2 + 14H2O + 14e-

C2H6 + 14OH- 2CO2 + 10H2O + 14e-

Cr2O7-2 Cr+3

Cr2O7-2 2Cr+3

Cr2O7-2 2Cr+3 + 7H2O

14H+ + Cr2O7-2 2Cr+3 + 7H2O

6e- + 14H+ + Cr2O7-2 2Cr+3 + 7H2O

6e- + 14H2O + Cr2O7-2 2Cr+3 + 7H2O + 14OH-

6e- + 7H2O + Cr2O7-2 2Cr+3 + 14OH-

Dichromate and methane react as shown : Cr2O7-2 + C2H6 CO2 + Cr+3

Oxidation ½-reaction:1. 2. balance C’s3a) balance O w. H2O3b) balance H w. H+ 3c) balance charge w e’s6) neutralize H+ w. OH-

7) cancel H2O

Reduction ½-reaction:1. 2. balance Cr3a) balance O w. H2O3b) balance H w. H+

3c) balance charge w. e’s6. neutralize H+ w. OH-

7. cancel H2O

gain 3 e’s, GER

+6 -3 +4 +3

Cr2O7-2, the oxidizer is reduced

C2H6, the reducer is oxidized

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lose 7 e’s, LEO Cr2O7

-2 + C2H6 CO2 + Cr+3 continued:

Oxidation ½-reaction: Reduction ½-reaction:C2H6 + 14OH- 2CO2 + 10H2O + 14e- 7H2O + Cr2O7

-2 + 6e- 2Cr+3 + 14OH-

Multiply both ½-reactions by the appropriate coefficient so that both reactions involve the same number of e’s.Find the lowest common multiple (LCM) of the e’s transferred ( 42 is the LCM’s of 14e’s and 6e’s)

(C2H6 + 14OH- 2CO2 + 10H2O + 14e-) × 3 (7H2O + Cr2O7-2 + 6e- 2Cr+3 + 14OH-) × 7

Combine the two ½-reactions:3C2H6 + 42OH- 49H2O + 7Cr2O7

-2 + 42e- 14Cr+3 + 98OH- + 6CO2 + 30H2O + 42e-

3C2H6 + 19H2O + 7Cr2O7-2 14Cr+3 + 56OH- + 6CO2

gain 3 e’s, GER

+6 -3 +4 +3

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Cr2O7-2 + C2H6 CO2 + Cr+3

lose 7 e’s, LEOgain 3 e’s, GER

+6 -3 +4 +3

MnO4- + H2O2 MnO2 + O2

lose 1 e’s, LEOgain 3 e’s, GER

+7 -1 +4 0

BrO3- + SO3

-2 Br2 + SO4-2

+5 +4 0 +6

lose 2 e’s, LEO

gain 5 e’s, GER

each Br atom in BrO3- gains 5 e’s.

each S atom in SO3-2 loses 2 e’s, 10e’s is the LCM

5 e’s × 2BrO3- = 10e’s gained

2 e’s × 5SO3-2 = 10e’s lost

2BrO3- + 5SO3

-2 Br2 + 5SO4-2

each Mn atom in MnO4- gains 3 e’s.

each O atom in H2O2 loses 1 e’s oreach H2O2 lose 2 e’s, 6e’s is the LCM3 e’s × 2MnO4

- = 6e’s gained2 e’s × 3H2O2 = 6e’s lost2MnO4

- + 3H2O2 2MnO2 + 3O2

each Cr atom in Cr2O7-2 gains 3 e’s or

each Cr2O7-2 gains 6 e’s

each C atom in C2H6 loses 7 e’s oreach C2H6 lose 14 e’s, 42e’s is the LCM6 e’s × 7Cr2O7

-2 = 42e’s gained14e’s × 3C2H6 = 42e’s lost7Cr2O7

-2+ 3C2H6 14Cr+3 + 6CO2

Oxidation Number Method for Balancing Redox Equations

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MnO4- + C5H8 MnO2 + CO2

lose 8/5 e’s, LEOgain 3 e’s, GER

+7 -8/5 +4 +4

Equations like the one above are not uncommon in organic chemistry, where carbon may have fractional oxidation numbers. Although this can be balanced by the oxidation number method (with a bit of mental arithmetic),most students will likely have more success using the Half-Reaction method.The balanced equation is shown below. Can you obtain this answer?

A Difficult Redox Equation to Balance

28MnO4- + 3C5H8 + 2H2O 28MnO2 + 15CO2 + 28

OH-

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Practice is the best teacher

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metal metal cation + e’s Potential (V)Li Li+ + 1e +3.05K K++ 1e +2.924Na Na+ + 1e +2.71Mg Mg+2 + 2e +2.36Al Al+3 +3e +1.7Zn Zn+2 + 2e +0.76Fe Fe+2 + 2e +0.41Ni Ni+2 + 2e +0.23Sn Sn+2 + 2e +0.14Pb Pb+2 + 2e +0.13½ H2 H+ + e 0.00Cu Cu+2 + 2e -0.34Hg Hg+2 + 2e -0.85Ag Ag+ + e -0.80Pt Pt+2 + 2e -1.20Au Au+3 + 3e -1.42

Activity Series of Metals and Oxidation Potential (V = emf)

These very reactive metals will be oxidized by H2O and liberate H2 gas.

These moderately reactive metals are not oxidized by H2O, but are oxidized by dilute acids (H2SO4, HCl, etc.) and liberate H2 gas.

These inert (‘noble’) metals are unreactive. They are not oxidized by dilute acids.

A positive oxidation potential means that a reaction is favorable (spontaneous) compared to hydrogen and will produce the emf shown (under standard conditions).

A negative oxidation potential means that a reaction is unfavorable compared to hydrogen and will require that a emf greater than the value shown be applied to force the reaction to occur as written.

The oxidation potential of hydrogen is set to zero and used as a reference for other redox reactions.Reversing an oxidation reaction yields a reduction reaction with the same magnitude of emf, but with opposite sign.