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Balancing redox reactions 2

Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

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Page 1: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

Balancing redox reactions 2

Page 2: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

• Balance oxidation-reduction reactions using redox methods

Include: oxidation number method, and half-reaction method

Additional KEY Terms

Page 3: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

Some redox reactions require an acidic or basic solution.

Half-reaction method

It depends on the reaction and substance needing to be oxidized – I will always tell you if it is in an acidic or basic solution

Half-reaction method is used for balancing redox reactions in the presence of acid or base.

• Acid / base not oxidized or reduced in reaction• Usually converted to water

Page 4: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

Cr2O72–

(aq) + SO32–

(aq) → Cr3+(aq) + SO4

2–(aq)

Step 1: Assign O#s and write half-reactions

Balancing in Acidic Solutions

+3+4-2+6 -2+6-2

Oxidation: SO32- → SO4

2- + 2e–

Reduction: Cr2O72- + 3e– → Cr3+

Aqueous ions cannot exist by themselves – the spectator ions must have already been removed

Page 5: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

Step 2: Balance all elements except H and O

Step 3: Balance oxygen atoms by adding H2O

Oxidation: SO32- → SO4

2- + 2e–

Reduction: Cr2O72- + e– → Cr3+2

Oxidation: SO32- + H2O → SO4

2- + 2e–

Reduction: Cr2O72- + 6e– → 2 Cr3+ + 7 H2O

Be sure to adjust the electrons to the number of atoms for each half-reaction Each Cr gains 3e- but there are 2 Cr atoms - so a total of 6e- gained

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Page 6: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

Step 4: Balance hydrogen atoms adding H+ ions

Step 5: Balance the number of electrons between half-reactions

Oxidation: SO32- + H2O → SO4

2- + 2e–

Reduction: Cr2O72- + 6e– → 2 Cr3+ + 7 H2O

+ 2 H+

14 H+ +

Oxidation: 3 x (SO32- + H2O → SO4

2- + 2e– + 2 H+)

Reduction: 14 H+ + Cr2O72- + 6e– → 2 Cr3+ + 7 H2O

3 SO32- + 3 H2O → 3 SO4

2- + 6e– + 6 H+

Multiply the whole equation by the common multiple needed to make the half-reaction electrons equal

Page 7: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

6. Add the two half-reactions

Oxidation: 3 SO32- + 3 H2O → 3 SO4

2- + 6e– + 6 H+

Reduction: 14 H+ + Cr2O7

2- + 6e– → 2 Cr3+ + 7 H2O

8 H+ + Cr2O72- + 3 SO3

2- → 2 Cr3+ + 3 SO42- + 4 H2O

Take care here - cancel out what you can and combine the half-reactions into a single equation

8 4

Page 8: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

MnO4– + I– → MnO2 + I2

8 H+ + 2 MnO4– + 6 I– → 2 MnO2 + 3 I2 + 4 H2O

Balance the following reaction in a acidic solution.

-1+7 0+4

Oxidation: I- → I2 + e–

Reduction: MnO4- + 3e– → MnO2

2

+ 2 H2O4 H+ +

3 x ()

2x (

)Oxidation: 6 I- → 3 I2 + 6e–Reduction: 8 H+ + 2 MnO4

- + 6e– → 2 MnO2 + 4 H2O

No spectator ion on the reactant side – keep compound together – so you have to keep the product together too

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Page 9: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

Balancing in Basic Solutions• Steps 1-4 are the same as in Acid solutions

MnO4– + C2O4

2– → CO2 + MnO2

+4+3+7 +4

Oxidation: C2O42- → CO2 + e–

Reduction: MnO41- + 3e– → MnO2

2

+ 2 H2O4 H+ +

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Tip: Take your time and do each step on a new line to avoid any mis-steps

Page 10: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

**5b. Eliminate H+ / OH- by forming water

Oxidation C2O42- → 2 CO2 + 2e–

Reduction: 4 H+ + MnO41- + 3e– → MnO2 + 2

H2O+ 4 OH-4 OH- +

**5a. Add the same number of OH- as H+ to BOTH sides of the equation

The point here is to cancel any water to simplify the half-reactions

Oxidation C2O42- → 2 CO2 + 2e–

Reduction: 4 OH- + 4 H+ + MnO4- + 3e– → MnO2 + 2 H2O + 4

OH-

4 H2O2

Page 11: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

4 H2O + 2 MnO4– + 3 C2O4

2– → 2 MnO2 + 6 CO2 + 8 OH–

Oxidation 3 C2O42-

→ 6 CO2 + 6e–

Reduction: 4 H2O + 2 MnO41- + 6e– → 2 MnO2 + 8

OH-

Step 7: Add the two half-reactions

Oxidation C2O42- → 2 CO2 + 2e–

Reduction: 2 H2O + MnO41- + 3e– → MnO2 + 4

OH-

3 x ()

2x (

)

Step 6: Balance the number of electrons between half-reactions

Page 12: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

N2O + ClO– → NO2– + Cl–

Balance the following reaction in a basic solution.

+3+1+1 -1

Oxidation: N2O → NO2- + e–

Reduction: ClO- + 2e– → Cl-

23 H2O +

2 H+ +

+ 6 H+

+ 1 H2O

• Steps 1-4

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Page 13: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

2 OH– + 2 ClO– + N2O → 2 Cl– + 2 NO2– + H2O

Oxidation: N2O → NO2- + 4e–

Reduction: ClO- + 2e– → Cl-

23 H2O +

2 H+ +

+ 6 H+

+ 1 H2O

+ 6 OH-

2 OH- + 2 H2O

3

+ 2 OH-

6 OH- + 6 H2O

1

Reduction: ClO- + 2e– → Cl-1 H2O + + 2 OH-

Oxidation: N2O → NO2- + 4e–2 + 3 H2O6 OH- +

2x ( )

2 H2O + 2 ClO- + 4e– → 2 Cl- + 4 OH-

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• Steps 5-7

Page 14: Balancing redox reactions 2. Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional

CAN YOU / HAVE YOU?

• Balance oxidation-reduction reactions using redox methods

Include: oxidation number method, and half-reaction method

Additional KEY Terms