balancing redox equations. electron transfer method (change in oxidation number method) works best...

Balancing

Redox

Equations

Electron Transfer Method

(Change in Oxidation Number Method)

works best for formula equations (no ions present)

Steps:

1. Write formulas for reactants and products (skeletal equation) – usually given

2. Assign oxidation numbers for every atom

3. Write "electronic" equations of atoms that changed oxidation number

Electron Transfer Method (Change in Oxidation Number Method)Steps:

4. Adjust coefficients to get equal number of electrons lost (ox.) and gained (red.)

5. Insert coefficients into skeletal equation. CAUTION: the coefficients in the electronic equations reflect ONLY those atoms oxidized and reduced – sometimes the OVERALL equation is not balanced yet

H2SO3 + I2 + H2O H2SO4 + HI+1 +4-2 0 +1 -2 +1 +6 -2 +1-1

+4 +6

oxidation rxn: S S + 2 e-1

0 -1

reduction rxn: I2 + 2 e-1 2 I

H2SO3 + I2 + H2O H2SO4 + HI+1 +4-2 0 +1 -2 +1 +6 -2 +1-1

+4 +6

oxidation rxn: S S + 2 e-1

0 -1

reduction rxn: I2 + 2 e-1 2 I

put in coefficients from above

H2SO3 + I2 + H2O H2SO4 + 2 HI (balanced)

KClO3 + FeSO4 + H2SO4 KCl + Fe2(SO4)3 + H2O+1+5-2 +2+6-2 +1+6-2 +1-1 +3 +6-2 +1 -2

+2 +3

ox. rxn: 2 Fe+2 Fe2+3 + 2 e-1

+2 +3

6 Fe+2 3 Fe2+3 + 6 e-1

+5 -1

red. rxn: Cl + 6 e-1 Cl-1

x 3


+2 +3

ox. rxn: 2 Fe+2 Fe2+3 + 2 e-1 x 3

+2 +3

6 Fe+2 3 Fe2+3 + 6 e-1

+5 -1


put in the coefficients from above

KClO3 + 6 FeSO4 + H2SO4 KCl + 3 Fe2(SO4)3 + H2O


+2 +3

ox. rxn: 2 Fe+2 Fe2+3 + 2 e-1 x 3

+2 +3

6 Fe+2 3 Fe2+3 + 6 e-1

+5 -1




Finish (O not balanced)


+2 +3

ox. rxn: 2 Fe+2 Fe2+3 + 2 e-1 x 3

+2 +3

6 Fe+2 3 Fe2+3 + 6 e-1

+5 -1




Finish (O not balanced)

KClO3 + 6 FeSO4 + 3 H2SO4 KCl + 3 Fe2(SO4)3 + 3 H2O

Ag2S + HNO3 AgNO3 + NO + S + H2O

+1 -2 +1+5-2 +1+5-2 +2-2 0 +1 -2


+1 -2 +1+5-2 +1+5-2 +2-2 0 +1 -2

-2 0

ox. rxn: S-2 S + 2 e-1

+5 +2

red. rxn: N + 3 e-1 N


+1 -2 +1+5-2 +1+5-2 +2-2 0 +1 -2

-2 0

ox. rxn: S-2 S + 2 e-1 x 3 -2 0

3 S-2 3 S + 6 e-1 +5 +2

red. rxn: N + 3 e-1 N x 2 +5 +2

2 N + 6 e-1 2 N


+1 -2 +1+5-2 +1+5-2 +2-2 0 +1 -2

-2 0

ox. rxn: S-2 S + 2 e-1 x 3 -2 0

3 S-2 3 S + 6 e-1 +5 +2

red. rxn: N + 3 e-1 N x 2 +5 +2

2 N + 6 e-1 2 N


3 Ag2S + 2 HNO3 AgNO3 + 2 NO + 3 S + H2O

Finish (Ag not balanced, fixing the Ag changes other coefficients)


+1 -2 +1+5-2 +1+5-2 +2-2 0 +1 -2

-2 0

ox. rxn: S-2 S + 2 e-1 x 3 -2 0

3 S-2 3 S + 6 e-1 +5 +2

red. rxn: N + 3 e-1 N x 2 +5 +2

2 N + 6 e-1 2 N


3 Ag2S + 2 HNO3 AgNO3 + 2 NO + 3 S + H2O

Finish (Ag not balanced, fixing the Ag changes other coefficients)

3 Ag2S + 8 HNO3 6 AgNO3 + 2 NO + 3 S + 4 H2O

Pt + HCl + KNO3 + KCl K2PtCl6 + NO + H2O

0 +1-1 +1+5-2 +1-1 +1+4-1 +2-2 +1-2

0 +4

ox. rxn: Pt Pt + 4 e-1 0 +4

3 Pt 3 Pt + 12 e-1

+5 +2

red. rxn: N + 3 e-1 N +5 +2

4 N + 12 e-1 4 N

put in the coefficients

3 Pt + HCl + 4 KNO3 + KCl 3 K2PtCl6 + 4 NO + H2O

Finish 3 Pt + 16 HCl + 4 KNO3 + 2 KCl 3 K2PtCl6 + 4 NO + 8 H2O

X 3

X 4

Half Reaction Method (see Chapter 20)works best for ionic reactions (ions present)reactions occur in acidic or basic solutionSteps:1. Break skeletal ionic equation into an

oxidation equation and a reduction equation (called half reactions)

2. Balance both reactions for all atoms except O and H

3. Balance both reactions for O by adding H2O to the side with less O

Balance both reactions for H by adding H+1 to the side with less H

Half Reaction Method (continued)

Steps:4. Balance both reactions for charge by

adding electrons to the side with the greater positive charge

5. Adjust coefficients in the balanced half reactions to get the number of electrons lost equal to the number of electrons gained

6. Add the two half reactions and cancel electrons and other species that

appear on both sides of the equation

Cr2O7-2(aq) + Cl-1(aq) Cr+3(aq) + Cl2(aq)

in acid solution

Step 1 – write half reactions

ox. rxn: Cl-1(aq) Cl2(aq) (check ox #’s if

necessary)

red. rxn: Cr2O7-2(aq) Cr+3(aq)

Step 2 - balance except for O and H

ox 2 Cl-1(aq) Cl2(aq)

red Cr2O7-2(aq) 2 Cr+3(aq)

Step 3 - balance for O by adding H2O, balance H by adding H+1

ox 2 Cl-1(aq) Cl2(aq)

red Cr2O7-2(aq) + 14 H+1(aq) 2 Cr+3(aq) + 7 H2O(l)

Step 4 - balance for charge by adding electrons

ox 2 Cl-1(aq) Cl2(aq) + 2 e-1

red Cr2O7-2(aq) + 14 H+1(aq) + 6 e-1 2 Cr+3(aq) + 7 H2O(l)

Step 5 - make TOTAL electrons lost and gained equal

ox 3 [ 2 Cl-1(aq) Cl2(aq) + 2 e-1 ]

6 Cl-1(aq) 3 Cl2(aq) + 6 e-1


Step 6 - add half reactions and cancel species that appear on both sides

ox 6 Cl-1(aq) 3 Cl2(aq) + 6 e-1


-------------------------------------------------------------------------------------

Cr2O7-2(aq) + 14 H+1(aq) + 6 Cl-1(aq) 3 Cl2(aq) + 2 Cr+3(aq) + 7 H2O(l)

ClO-1(aq) + Cr(OH)4-1(aq) CrO4

-2(aq) + Cl-1(aq) basic solution

Balance like an acidic reaction, then add OH-1 ions to neutralize the H+1

Step 1 – half reactions

ox Cr(OH)4-1(aq) CrO4

-2(aq)

red ClO-1(aq) Cl-1(aq)

Step 2 – they are already balanced for atoms other than O and H

Step 3 – balance O (add H2O) and THEN H (add H+)


-2(aq) + 4 H+1(aq)

red ClO-1(aq) + 2 H+1(aq) Cl-1(aq) + H2O(l)

Step 4 – balance charge


-2(aq) + 4 H+1(aq) + 3 e-1

red ClO-1(aq) + 2 H+1(aq) + 2 e-1 Cl-1(aq) + H2O(l)

Step 5 – balance TOTAL electrons

ox 2 x [Cr(OH)4-1(aq) CrO4

-2(aq) + 4 H+1(aq) + 3 e-1]

red 3 x [ClO-1(aq) + 2 H+1(aq) + 2 e-1 Cl-1(aq) + H2O(l)]

ox 2 Cr(OH)4-1(aq) 2 CrO4

-2(aq) + 8 H+1(aq) + 6 e-1

red 3 ClO-1(aq) + 6 H+1(aq) + 6 e-1 3 Cl-1(aq) + 3 H2O(l)

Step 6 – add and cancel

2 Cr(OH)4-1(aq) 2 CrO4

-2(aq) + 8 H+1(aq) + 6 e-1

3 ClO-1(aq) + 6 H+1(aq) + 6 e-1 3 Cl-1(aq) + 3 H2O(l)

------------------------------------------------------------------------------------

2Cr(OH)4-1(aq) + 3ClO-1(aq) 2CrO4

-2(aq) + 2H+1(aq) + 3Cl-1(aq) + 3H2O(l)

because this is in basic solution, we add 2 OH-1 ions to each side to convert the 2 H+1 on the right side into H2O

final equation balanced for atoms and charge is2Cr(OH)4

-1(aq) + 3ClO-1(aq) + 2OH-1(aq)2CrO4-2(aq) + 3Cl-1(aq) + 5H2O(l)

the 2 H+1 (on the right) combine with the 2 OH-1 to make 2 H2O and then the 2 H2O combine with the 3 H2O already there to make 5 H2O total on the right

Becomes2 H2O

1. Fe+2 + MnO4-1 Fe+3 + Mn+2 (in acidic

solution)

1&2) ox Fe+2 Fe+3

red MnO4-1 Mn+2

3) ox Fe+2 Fe+3

red MnO4-1 + 8 H+1 Mn+2 + 4 H2O

4) ox Fe+2 Fe+3 + e-1

red MnO4-1 + 8 H+1 + 5 e-1 Mn+2 + 4 H2O

5) ox 5 [ Fe+2 Fe+3 + e-1 ] 5 Fe+2 5 Fe+3 + 5 e-1

red MnO4-1 + 8 H+1 + 5 e-1 Mn+2 + 4 H2O

6) 5 Fe+2 + MnO4-1 + 8 H+1 5 Fe+3 + Mn+2 + 4 H2O

6. MnO4-1 + NO2

-1 MnO2 (s) + NO3-1 (in

basic solution)

1&2) NO2-1 NO3

-1

MnO4-1 MnO2

3) NO2-1 + H2O NO3

-1 + 2 H+1 MnO4

-1 + 4 H+1 MnO2 + 2 H2O

4) NO2-1 + H2O NO3

-1 + 2 H+1 + 2 e-1 MnO4

-1 + 4 H+1 + 3 e-1 MnO2 + 2 H2O

5) ox 3 [ NO2-1 + H2O NO3

-1 + 2 H+1 + 2 e-1 ] 3 NO2

-1 + 3 H2O 3 NO3-1 + 6 H+1

+ 6 e-1 red 2 [ MnO4

-1 + 4 H+1 + 3 e-1 MnO2 + 2 H2O ] 2 MnO4

-1 + 8 H+1 + 6 e-1 2 MnO2 + 4 H2O

6) 3 NO2-1 + 2 MnO4

-1 + 2 H+1 3 NO3-1 + 2 MnO2 + H2O

basic solution – have to add 2 OH-1

3 NO2-1 + 2 MnO4

-1 + 2 H2O 3 NO3-1 + 2 MnO2 + H2O

+ 2 OH-1

3 NO2-1 + 2 MnO4

-1 + H2O 3 NO3-1 + 2 MnO2 + 2 OH-1

oxred

2. Cr2O7-2 + I-1 Cr+3 + I2 (s) (in acidic solution)

1) ox I-1 I2 red Cr2O7

-2 Cr+3

2) ox 2 I-1 I2 red Cr2O7

-2 2 Cr+3

3) ox 2 I-1 I2 red Cr2O7

-2 + 14 H+1 2 Cr+3 + 7 H2O

4) ox 2 I-1 I2 + 2 e-1

red Cr2O7-2 + 14 H+1 + 6 e-1 2 Cr+3 + 7 H2O

5) ox 3 [ 2 I-1 I2 + 2 e-1 ] 6 I-1 3 I2 + 6 e-1 red Cr2O7

-2 + 14 H+1 + 6 e-1 2 Cr+3 + 7 H2O

6) Cr2O7-2 + 14 H+1 + 6 I-1 2 Cr+3 + 7 H2O + 3 I2

balancing redox equations. electron transfer method (change in oxidation number method) works best...

Documents

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