Redox Reactions• Redox Reactions

Oxidation - Reduction reactions

• Terms

Oxidation

loss of electrons

electrons are a product

Na --> Na+ + e-

Reduction

gain of electrons

electrons are a reactant

Br2 + 2e- --> 2Br-

Oxidizing AgentReactant that causes the oxidation

(this reactant will be reduced)Reducing Agent

Reactant that causes the reduction(this reactant will be oxidized)

Oxidation Numbercharge on an atomcan be explicit or implicitoxidation number of elements = 0Na+ oxidation number = +1Br - oxidation number = -1

What is the oxidation number of iron in ferric chloride?

A) +1

B) +2

C) +3

D) -1

E) -2

What is the oxidation number of iron in ferric chloride?

Ferric chloride = ?

Chloride has a charge of?

So iron has to be?

FeCl3

-1

+3

What is the oxidation number on S in sulfur dioxide?

A) +2

B) +4

C) +6

D) -2

E) -4

What is the oxidation number on S in sulfur dioxide?

sulfur dioxide = ?

most of the time, oxygen in combination is the oxide oxygen. So each oxygen has an oxidation number of ?

the molecule has to have an overall charge of 0

S + 2(-2) = 0

S = +4

SO2

-2

What is the oxidation number of Cr in CrO4

-2?

A) +4

B) +6

C) +8

D) -4

E) -2

What is the oxidation number of Cr in CrO4

-2?

CrO4-2 is called?

ion has a charge of -2

Cr + 4(O) = -2

Cr + 4(-2) = -2

Cr = +6

Chromate ion

Na + Br2 --> NaBr2 2

2 Na -- > 2 Na+ + 2e- Oxidation half reaction

Br2 + 2e- -- > 2 Br- Reduction half reaction

2 Na + Br2 -- > 2 NaBr

Balancing Redox Reaction Ion Electron Method

Split the reaction into half reactions and balance each half reaction

1. Balance all elements other than H and O

2. Balance O by adding H2O3. Balance H by adding H+. If the reaction is in acid go to

4, if in base go to 3A.3A. If the reaction is in base, add OH- equal to the number of H+

to both sides of the equation. Combine the H+ and OH- that are on the same side to make H2O

• Balance the charge by adding e-

• Multiply the 2 half reactions so that the number of electrons is the same in each half reaction

• Add the half reactions back together and eliminate redundancies

Cr2O7-2 + SO3

-2 --> Cr+3 + SO4-2

Cr2O7-2 --> Cr+3

SO3-2 --> SO4

-2

acid

2 +7H2O+ 14H+

+12 +6+ 6e-

H2O + + 2H+

- 2 0

+ 2e-

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

SO3-2 --> SO4

-2H2O + + 6H+ + 6e-

3(

3 3 3

8 4

Cr2O7-2 +8H+ +3SO3

-2 -->2Cr+3 + 4H2O + 3SO4-2

MnO4- + C2O4

-2 --> MnO2 + CO2 base

MnO4- --> MnO2

2MnO4- +4H2O+3C2O4

-2-->2MnO2+ 8OH- + 6CO2

C2O4-2 --> + CO2

+ 2 H2O + 4H+ + 4OH- + 4OH-

4H2O

-1 -4

3e-+

2 0

2e- -2

2(

3(

2MnO4- --> 2MnO2 + 4 H2O + 8H2O + 8OH- 6e-+

3C2O4-2 --> + CO2 66e-

4

Br2 + OH- --> BrO3- + Br - + H2O

disproportionationReaction where the same reactant is both oxidized and reduced

Br2 --> BrO3-

Br2 --> Br -

2+ 6H2O-2

- 12

+10e-

2+ 2e-

+12H++ 12OH - +12OH- 12H2O

- 2

0

5(

Br2 --> BrO3- 2+ 6H2O +10e- + 12H2O+ 12OH -

5 Br2 + 10e- --> 10 Br-

Br2 + 5 Br2 + 12OH- --> 2 BrO3- + 6 H2O + 10 Br-

6

• Redox titration

a known concentration of oxidizing agent is used to find an unknown concentration of reducing agent (or vice versa)

37.5 mL of 0.658 M KMnO4 solution were required to completely react with 22.0 mL of a basic K2C2O4soluiton. What is the concentration of the oxalate solution?

A) 0.748 MB) 1.12 MC) 1.68 MD) 3.89 ME) 0.579 M

2MnO4- +4H2O+3C2O4

-2-->2MnO2+ 8OH- + 6CO2

37.5 mL of o 658 M KMnO4 solution were required to completely react with 22.0 mL of a basic K2C2O4soluiton. What is the concentration of the oxalate solution?

1. Balance the redox reaction. K+ is spectator.4H2O + 2MnO4

- + 3C2O4-2 --> 2MnO2 + 6CO2 + 8OH-

2. Concentration = M = moles C2O4-2/ L. So,

need to find moles C2O4-2. Use the moles of

MnO4-2 and the mole ratio from the balanced

reaction.

37.5 mL of 0. 658 M KMnO4 solution were required to completely react with 22.0 mL of a basic K2C2O4soluiton. What is the concentration of the oxalate solution?

4H2O + 2MnO4- + 3C2O4

-2 --> 2MnO2 + 6CO2 + 8OH-

Moles = ?Moles = VM

37.5 mL KMnO4 0.658 mmol KMnO4

mL KMnO4

1 mmol MnO4

1 mmol KMnO4

3 mmol C2O4 2

2 mmol MnO4

37.01 mmol C2O4

2

37.01 mmol C2O4 2

22.0 mL 1.68 M

Activity Seriesoxidation series that shows the relative reactivity of metals

• Li• K• Ba• Ca• Na• Mg• Al• Mn• Zn• Cr• Fe• Cd• Co• Ni• Sn• Pb• H2

• Cu• Hg• Ag• Au

Reactions are oxidation reactions.

--> Li+

--> Ca+2

--> Al+3

--> Zn+2

--> Cd+2

Metals reactwith ions that lie below them.

Li + Ca+2 --> Li+ + Ca

Al + Cd+2 --> Al+3 + Cd

Al + Ca+2 --> ? NR

Reactions occur top left with bottom right

• Li• K• Ba• Ca• Na• Mg• Al• Mn• Zn• Cr• Fe• Cd• Co• Ni• Sn• Pb• H2

• Cu• Hg• Ag• Au

--> Li+

--> Ca+2

--> Al+3

--> Zn+2

--> Cd+2

Which of the following combination will react?A) Li+ and CaB) Ca+2 and MgC) Ca and K+

D) Ni and H+

E) Ni and Sn