balancing redox equations handout

OFB Chapter 12 111/13/09

Chapter 4.11Balancing Redox Equations

This section is intended to be a supplemental resources for students.

A stepwise process to balance oxidation-reduction (redox) reactions is introduced.

Student’s should use these slides to practice balancing redox reactions.

The number of electrons transferred will be used for the Electrochemical cell reactions in Chapter 16


• This section 4.11deals with Oxidation, Reduction, and electron transfer

• Balancing Redox Reactions• All reactions take place in either

Acid (H+) or Basic (OH-) conditions• Very important to remember the

problem you are working. Is it acidic or basic conditions?

• We will be adding H2O and H+ for acid reactions or H2O and OH- for basic reaction in order to properly balance the reactions

• Practice, Practice, Practice


Gains ElectronsDecreaseSubstance Reduced

Loses ElectronsIncreaseSubstance Oxidized

Supplies ElectronsIncrease

Reducing Agent, does the reducing

Picks Up electronsDecrease

Oxidizing Agent, does the oxidizing

Gain of ElectronsDecreaseReduction

Loss of ElectronsIncreaseOxidation

Electron Change

Oxidation Number ChangeTerm

The Oxidation States Method

Cl2 Na NaCl+ →


• Step 1 Write two unbalanced half-equations, one for the species that is oxidized and its product and one for the species that is reduced and its product

• Step 2 Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-equation.

• Step 3 Balance oxygen by adding H 2O to the side deficient in O in each half-equation

• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on

to the side deficient in hydrogen.

– For a half-reaction in basic solution, add H2O to the side that is deficient in hydrogen and an equal amount of OH- to the other side


• Step 5 Balance charge by inserting e- (electrons) as a reactant or product in each half-reaction.

• Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication

• Step 7 Check for balance


Balancing Redox Reactions

• Dithionate ion reacting with chlorous acid in aqueous acidic conditions

S2O62-(aq) +HClO2(aq)

→ SO42-(aq) + Cl2(g)

Will have:

Oxidation half reaction

Reduction half reaction



S2O62- → SO4

2-

Dithionate

(we will see eventually that this is the oxidation

reaction)

HClO2 → Cl2

Chlorous acid

(we will see eventually that this is the reduction reaction)

S2O62-(aq) +HClO2(aq)

→ SO42-(aq) + Cl2(g)



S2O62- → SO4

2-

HClO2 → Cl2

2

2



S2O62- → 2SO4

2-

2HClO2 → Cl2

(8Os)(6Os)

2H2O +(Need 2 Os)

(4Os)

+ 4H2O(Need 4 Os)


• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on to the

side deficient in hydrogen.

– For a half-reaction in basic solution, add H2O to the

side that is deficient in hydrogen and an equal amount of OH- to the other side

2H2O + S2O62- → 2SO4

2-

2HClO2 → Cl2 + 4H2O

Dithionate ion reacting with chlorous acid in aqueous acidic conditions

Need 4 Hs

Need 6 Hs (2 Hs) (8 Hs)

+ 4H+

6H+ +

(4 Hs)


• Step 5 Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction.

2H2O + S2O62-

→ 2SO42- + 4H+

6H+ + 2HClO2

→ Cl2 + 4H2O

-2

-4 +4 Need -2

+ 2e-

+6

Zero charge

Need -6

+ 6e-

Oxidation reaction, electrons are lost on the product side

Reduction reaction, electrons are gained on the reactant side


• .Step 6 Multiply the two half-equations by numbers chosen to make the number of electrons given off by the oxidation equal to the number taken up by the reduction. Then add the two half-equations and cancel out the electrons. If H+ ion, OH- ion, or H2O appears on both sides of the final equation, cancel out the duplication

Oxidation Reaction has 2 electrons

Reduction reaction has 6 electrons

Multiply the oxidation reaction by 3. This will give 6 electrons on both sides of the reaction

2H2O + S2O62-

→ 2SO42- + 4H+ + 2e-

3 times



3 times

6H2O + 3S2O62-

→ 6SO42- + 12H+ + 6e-

2H2O + S2O62-

→ 2SO42- + 4H+ + 2e-

equals

Now add both Re-Ox reactions



6H2O + 3S2O62-

→ 6SO42- + 12H+ + 6e-

6H+ + 2HClO2 + 6e-

→ Cl2 + 4H2O

6H2O + 3S2O62- + 6H+ +

2HClO2 + 6e-

→ 6SO42- + 12H+ + Cl2 +

4H2O + 6e-

2

6



2H2O + 3S2O62- + 2HClO2

→ 6SO42- + 6H+ + Cl2

Left side

H 6

O 24

S 6

Cl 2

Right side

H 6

O 24

S 6

Cl 2

If it doesn’t balance, look for simple mistakes first

Omitted subscripts

Omitted superscripts


Find the (integer) stoichiometry coefficients for the correctly balanced version of the following reaction

Cr2O72- + H+ + I-

→ Cr+3 + H2O + I3-



Cr2O72- + H+ + I-

→ Cr+3 + H2O + I3-

Cr2O72- → Cr+3

I- → I3-



2

3

Cr2O72- → Cr+3

I- → I3-



(7Os)

+ 7H2O(Need 7 Os)

(No Oxygen, okay)

Cr2O72- → 2Cr+3

3I- → I3-


• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on to the

side deficient in hydrogen.

– For a half-reaction in basic solution, add H2O to the

side that is deficient in hydrogen and an equal amount of OH- to the other side

Dichromate ion reacting with iodide ion in aqueous acidic conditions

Need 14 Hs

14H+ +

(14 Hs)

Cr2O72-

→ 2Cr+3 + 7H2O

3I- → I3-

(No Hydrogen, okay)



-2+14

+6

Need -6

+ 2e-

-3 Need -2

6e- +

Oxidation reaction

Reduction reaction

14H+ + Cr2O72-

→ 2Cr+3 + 7H2O

3I- → I3-

-1





Multiply the oxidation reaction by 3 this will give 6 electrons on both sides of the reaction

3 times

3I- → I3- + 2e-



equals


3 times

3I- → I3- + 2e-

9I- → 3 I3- + 6 e-



6e- + 14H+ + Cr2O72-

→ 2Cr+3 + 7H2O

9I- → 3 I3- + 6 e-

6e- + 14H+ + Cr2O72- + 9I-

→ 2Cr+3 + 7H2O + 3 I3- + 6 e-

equals



Left side

H 14

Cr 2

O 7

I 9

Right side

H 14

Cr 2

O 7

I 9

14H+ + Cr2O72- + 9I-

→ 2Cr+3 + 7H2O + 3 I3-


Basic SolutionMethod 1

• At Step 4– Instead of adding H+ on the

deficient side as did with the acid solution method

– Add H2O on the H deficient side and add an equal amount of OH- on the other side.


Find the (integer) stoichiometry coefficients for the correctly balanced version of the following reaction conducted in a basic solution

Ag2S + Cr(OH)3

→ Ag0 + HS- + CrO42-



Ag2S + Cr(OH)3

→ Ag0 + HS- + CrO42-

Ag2S → Ag + HS-

Cr(OH)3 → CrO42-

Reduction reaction

Oxidation State of Ag

+1

0

Oxidation reactionOxidation State of Cr

+3

+6



2Ag2S → Ag + HS-

Cr(OH)3 → CrO42-

Already Balanced



(4Os)

H2O +

(Oxygen not needed)

Ag2S → 2Ag + HS-

Cr(OH)3 → CrO42-

(3Os)(need 1O)


• Step 4 Balance hydrogen. – For half-reaction in acidic solution, add H+ on

to the side deficient in hydrogen.

– For a half-reaction in basic solution, add H2O

to the side that is deficient in hydrogen and an equal amount of OH- to the other side

This problem is in aqueous basic conditions

Needs H

(1 H)

Ag2S

→ 2Ag + HS-

H2O +

+ OH-

(need O)

H2O + Cr(OH)3

→ CrO42-

(3 H)(2 H)

+ 5H2O

5OH- +

Needs H

Add OH-



-2

Need -2

+ 3e-

-5

Need -3

H2O + Ag2S

→ 2Ag + HS- + OH-

5OH- + H2O + Cr(OH)3

→ CrO42- + 5H2O

-1 -1

2e- +

Reduction reaction

Oxidation reaction





2e x 3 = 6e

3e x 2 = 6e



6e- + 3H2O + 3Ag2S

→ 6Ag + 3HS- + 3OH-

Reduction reaction

3 times Reduction Reaction =

10OH- + 2H2O + 2Cr(OH)3

→ 2CrO42- + 10H2O + 6e-

2 times Oxidation Reaction =

Oxidation reaction




6e- + 3H2O + 3Ag2S

→ 6Ag + 3HS- + 3OH-

10OH- + 2H2O + 2Cr(OH)3

→ 2CrO42- + 10H2O + 6e-


equals


6e- + 3H2O + 3Ag2S

→ 6Ag + 3HS- + 3OH-

10OH- + 2H2O + 2Cr(OH)3

→ 2CrO42- + 10H2O + 6e-

6e- + 3H2O + 3Ag2S +

10OH- + 2H2O + 2Cr(OH)3

→ 6Ag + 3HS- + 3OH- +

2CrO42- + 10H2O + 6e-


simplify

6e- + 3H2O + 3Ag2S +

10OH- + 2H2O + 2Cr(OH)3

→ 6Ag + 3HS- + 3OH- +

2CrO42- + 10H2O + 6e-

5

7

3Ag2S + 7OH- + 2Cr(OH)3

→

6Ag + 3HS- + 2CrO42- + 5H2O



Left side

Ag 6

S 3

O 13

H 13

Cr 2

3Ag2S + 7OH- + 2Cr(OH)3

→ 6Ag + 3HS- + 2CrO4

2- + 5H2O

Right side

Ag 6

S 3

O 13

H 13

Cr 2


Find the (integer) stoichiometry

coefficients for the correctly balanced version of the following reaction conducted in a basic solution

Solution

3 Ag2S + 2 Cr(OH)3+ 7 OH-

→ 6 Ag + 3 HS- + 2CrO4

2- + 5H2O


Chapter 4.11Redox reactions

• Practice

• We will use the number of electrons transferred again in Chapter 16


Balancing Disproportionation Reactions

• Same Steps

• For example

Cl2 → ClO3- + Cl-

• Step 1– Oxidation:

Cl2 → ClO3-

– Reduction:

Cl2 → Cl-

balancing redox equations handout

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