Chapter 17 Oxidation and Reduction

Notes One Unit Six

•Redox•Oxidation Numbers•Identifying what is oxidized and what is reduced

Oxidation Reduction Chemisty: Redox Chemistry

Oxidation and Reduction reactions always take place simultaneously.

Loss of electrons – oxidation (Increase in Oxidation Number)

Ex:Na ------> Na+1 + e-1

Gain of electrons - reduction ( Decrease in Oxidation Number)

Cl2 + 2 e-1 ------> 2 Cl-1

Redox: Reduction occurs when an atom gains one or more electrons. Ex: O + 2e-1 O2-

  Oxidation Foccurs when an atom or ion loses one or more electrons. Ex: Fe Fe+3 + 3e-1

  LEO goes GERCopper metal reacts with silver nitrate to form silver metal and copper nitrate:   Cu + 2 Ag(NO3) 2 Ag + Cu(NO3)2.

Redox reactions involve electron transfer:

Lose e - =Oxidation

Cu (s) + 2 Ag+1 (aq) Cu +2 (aq) + 2 Ag(s)

Gain e - =Reduction

Oxidation occurs when a molecule does any of the following:

   Loses electrons    Loses hydrogen    Gains oxygen

If a molecule undergoes oxidation, it has

been oxidized and it is the reducing agent

(aka reductant).

Reduction occurs when a molecule does any of the following:

Gains electrons Gains hydrogen Loses oxygen

If a molecule undergoes reduction, it has been reduced and it is the oxidizing agent (aka oxidant).

zinc is being oxidized while the copper is being reduced. Why?

RedoxBurning: C6H12O6 + 6O2 6CO2 + 6H2O+ HeatRusting Iron: 4Fe + 3O22Fe2O3 + HeatOxidation - Loss of e-1.Na(s)Na+1 +1e-1

Reduction - Gain of e-1.Cl2+ 2e-1 2Cl-1

Number line (Oxidation…Left or right?)

Oxidation Numbers• Rules for Assigning Oxidation States • The oxidation state of an atom in an uncombined element is 0.• The oxidation state of a monatomic ion is the same as its charge.• Oxygen is assigned an oxidation state of –2 in most of its covalent

compounds. Important exception: peroxides (compounds containing the O2 2- group), in which each oxygen is assigned an oxidation state of –1)

• In its covalent compounds with nonmetals, hydrogen is assigned an oxidation state of +1

• For a compound, sum total of Oxidation Numbers is zero.

• For an ionic species (like a polyatomic ion), the sum of the oxidation states must equal the overall charge on that ion.

Key Elements

• (99%) H+1 H-1

• (99%)O-2 O-1

• (Always) Li+1, Na+1, K+1, Rb+1, Cs+1, Fr+1

• (Always) Be+2, Mg+2, Ca+2, Ba+2, Sr+2, Ra+2

• (Always) Al+3

• (with only a metal) F-1, Cl-1, Br-1, I-1

• (NO3-1) ion is always +5

• (SO4-2) ion is always +6

Finding Oxidation Numbers

+1 -2H2O

2 (+1)+ 1(-2) = Zero

The sum of the oxidation numbers must be equal to _____ for a compound.Find Ox#’s for H2O?

zero

2 (H)+ 1(O) = Zero

+1 -2H3PO4

Find Ox#’s for H3PO4?

3 (H)+ 4(O) = Zero+5

1 (P)+3(+1)+ 4(-2) = Zero1(+5)+

Finding Oxidation #’s for Compounds+1 -2

+1+5-2HH33POPO44

H2O

HNO3

+1+5-2

H2SO4

+1 -2+6

Hg2SO4

+6+1 -2

Na2Cr2O7

+1 +6 -2

H2CO3

+1 -2+4

(NH4)2CO3

-3 +1 +4-2

Ca3(AsO4)2

+2 +5 -2

Fe2(SO4)3

+6+3 -2

Ba(ClO4)2

+2 +7-2

Al2(CO3)3

+3 +4 -2

Identifying OX, RD, SI Species

• Ca0 + 2 H+1Cl-1 Ca+2Cl-12 + H2

0

• Oxidation = loss of electrons. The species becomes

more positive in charge. For example, Ca0 Ca+2, so Ca0 is the species that is oxidized.

• Reduction = gain of electrons. The species becomes

more negative in charge. For example, H+1 H0, so the H+1 is the species that is reduced.

• Spectator Ion = no change in charge. The species does not gain or lose any electrons. For example, Cl-1 Cl-1, so the Cl-1 is the spectator ion.

Oxidizing Agent and Reducing Agent:

Oxidizing agent gets reduced itself and reducing agent gets oxidized itself, so a strong oxidizing agent should have a great tendency to accept e and a strong reducing agent should be willing to lose e easily. What are strong oxidizing agents- metals or non metals? Why?

Spectator Ions are ions that do NOT change their oxidation number from the reactant side of a RXN to the product side of a RXN. They are just “hanging out”.

Notes Two Unit Six

• Activity Series of Metals• Balancing by Redox• Electrolysis• Lab A-Electrolysis

The activity series of metals is an empirical tool used to predict products in displacement reactions and reactivity of metals with water and acids in replacement reactions and ore extraction. It can be used to predict the products in similar reactions involving a different metal.

The activity series is a chart of metals listed in order of declining relative reactivity. The top metals are more reactive than the metals on the bottom. For example, both magnesium and zinc can react with hydrogen ions to displace H2 from a solution by the reactions:

Mg(s) + 2 H+(aq) → H2(g) + Mg2+(aq)

Zn(s) + 2 H+(aq) → H2(g) + Zn2+(aq)

Both metals react with the hydrogen ions, but magnesium metal can also displace zinc ions in solution by the reaction:

Mg(s) + Zn2+ → Zn(s) + Mg2+

Reduction: Cu+2(aq) + 2 e- Cu(s)

The Cation becomes a solid metal (the + charge GAINS ELECTRONS to become a zero charge.

Oxidation: Cu(s) Cu+2(aq) + 2 e-

The metal becomes a cation (the zero charge metal LOSES ELECTRONSTo become a + charge.

Redox: Oxidation Reduction Reaction

3Cu+2 (aq) + 2Fe (s) 3 Cu (s) + 2Fe+3 (aq)

The paired reduction and oxidation

Electrons transfer from the metal to the cation if the metal

Is above (ie higher) on the Activity Series Chart in your packet.

6 e-

Cu+2 (aq) + Mg (s) Cu (s) + Mg+2 (aq)

Zn+2 (aq) + Ag (s) No RXN

Balancing By Redox Example One

H2O + P4+ H2SO4 H3PO4+ H2S

#1. Find the oxidation #’s.#2. ID the element (i) oxidized and (ii) reduced.#3. Find # of electrons lost or gained.#4. Cross-multiply.#5. Balance using whole # ratios.#6. Find whole number coefficients.

+1 -2 0 +1 -2+6 +1 -2+5 +1 -25e-1 lost

8e-1 Gained

X8

X52 85 512

Multiply by 1.12H2O + 2P4+ 5H2SO4 8H3PO4+ 5H2S

Balancing By Redox Example Two

K3PO4 + Cl2 P4+ K2O+ KClO2

#1. Find the oxidation #’s.#2. ID the element (i) oxidized and (ii) reduced.#3. Find # of electrons lost or gained.#4. Cross-multiply.#5. Balance using whole # ratios.#6. Find whole number coefficients.

+1 -20+1 -2+5 +1 -2+305e-1 Gained

3e-1 Lost

X3

X55/2 2 3/4 5 3

Multiply by 4.

12K3PO4 + 10Cl2 3P4+ 8K2O+ 20KClO2

Applications of Oxidation-Reduction

Reactions

Batteries

Alkaline Batteries

Hydrogen Fuel Cells

Corrosion and…

…Corrosion Prevention

How to use the ½ Reaction Resource:

Notice they are all written as Reductions (gaining of electrons)1.Find the highest one on the left hand side and write it in the forward direction.

2. Find the lowest one on the right and write it backwards as an oxidation ,along with changing the sign of the voltage.

3. Make sure to balance electrons lost and gained (you DO NOT MULTIPLY the voltages !!!)

4. So you will have two ½ cell reactions and you can cancel electrons and write the WHOLE CELL RXN.

Let’s Practice????? X

NO3-+2H++e-NO2(g)+H2O

Fe3++e-Fe2+

I2(s)+2e-2I-

Cu++e-Cu(s) Cu2++2e-Cu(s)

SO42-+4H++2e-SO2(g)

Sn4++2e-Sn(s) 2H++2e-H2(g)

Pb2++2e-Pb(s) Sn2++2e-Sn(s) Ni2++2e-Ni(s) Co2++2e-Co(s)

2H+(pH=7)+2e-H2(g) Fe2++2e-Fe(s) Cr3++3e-Cr(s) Zn2++2e- Zn(s)

2H2O+2e-2OH-+H2(g)

+0.78+0.77

+0.53+0.52 +0.34+0.17 +0.15 0.00 -0.13-0.14

-0.25-0.28 -0.41-0.44

-0.74

-0.76-0.84

Hg2++2e- Hg(l) +0.78Ag++e- Ag(l) +0.80

1/2O2(g)+2H+(pH=7)+2e-H2O +0.82NO3

-+4H++3e-NO(g)+2H2O +0.96Br2(l)+2e-2Br- +1.06

1/2O2(g)+2H+(pH=7)+2e-

H2O+1.23

ReducingAgent

StrongerReducing

Agent

Loses e-

Oxidizing Agent

Weaker Oxidizing

Agent

Gains e-

Stronger Weaker

Electrolysis

anode

Cathode

e-1Electrolysis

CathodeAnode

Electron flow?Mass Gain=?

Which is the…Cathode=? Anode=?

Cathode(spoon)Mass Loss=?Copper

-reduction-oxidation

-electric current produced chemical reaction

Electrolysis Lab- Demo KI (aq)What is available to react?

K+1 I-1 H2O

Anode Reaction lowest reaction on right.

Cathode Reaction highest reaction on left.

NO3-+2H++e-NO2(g)+H2O

Fe3++e-Fe2+

I2(s)+2e-2I-

Cu++e-Cu(s) Cu2++2e-Cu(s)

SO42-+4H++2e-SO2(g)

Sn4++2e-Sn(s) 2H++2e-H2(g)

Pb2++2e-Pb(s) Sn2++2e-Sn(s) Ni2++2e-Ni(s) Co2++2e-Co(s)

2H+(pH=7)+2e-H2(g) Fe2++2e-Fe(s) Cr3++3e-Cr(s) Zn2++2e-Zn(s)

2H2O+2e-2OH-+H2(g)

+0.78+0.77

+0.53+0.52 +0.34+0.17 +0.15 0.00 -0.13-0.14

-0.25-0.28 -0.41-0.44

-0.74

-0.76-0.84

Hg2++2e-Hg(l) +0.78Ag++e-Ag(l) +0.80

1/2O2(g)+2H+(pH=7)+2e-H2O +0.82NO3

-+4H++3e-NO(g)+2H2O +0.96Br2(l)+2e-2Br- +1.06

1/2O2(g)+2H+(pH=7)+2e-

H2O+1.23

K+1 I-1 H2O

K++e-K(s) -2.92

An: lowest on right.

Cath: highest on left.

2I-I2(s)+2e-

We see brown:I2(s)

We see pink.

We saw bubbles

KI(aq)

NO3-+2H++e-NO2(g)+H2O

Fe3++e-Fe2+

I2(s)+2e-2I-

Cu++e-Cu(s) Cu2++2e-Cu(s)

SO42-+4H++2e-SO2(g)

Sn4++2e-Sn(s) 2H++2e-H2(g)

Pb2++2e-Pb(s) Sn2++2e-Sn(s) Ni2++2e-Ni(s) Co2++2e-Co(s)

2H+(pH=7)+2e-H2(g) Fe2++2e-Fe(s) Cr3++3e-Cr(s) Zn2++2e-Zn(s)

2H2O+2e-2OH-+H2(g)

+0.78+0.77

+0.53+0.52 +0.34+0.17 +0.15 0.00 -0.13-0.14

-0.25-0.28 -0.41-0.44

-0.74

-0.76-0.84

Hg2++2e-Hg(l) +0.78Ag++e-Ag(l) +0.80

1/2O2(g)+2H+(pH=7)+2e-H2O +0.82NO3

-+4H++3e-NO(g)+2H2O +0.96Br2(l)+2e-2Br- +1.06

1/2O2(g)+2H+(pH=7)+2e-

H2O+1.23

Na++e-Na(s) -2.71

An: lowest(right)

Cath: highest(left)

H2O1/2O2(g)+2H+

(pH=7)+2e-

We saw bubbles!

We saw copper on the pencil tip!

Cu+2 SO4-2 H2O

CuSO4(aq)

NO3-+2H++e-NO2(g)+H2O

Fe3++e-Fe2+

I2(s)+2e-2I-

Cu++e-Cu(s) Cu2++2e-Cu(s)

SO42-+4H++2e-SO2(g)

Sn4++2e-Sn(s) 2H++2e-H2(g)

Pb2++2e-Pb(s) Sn2++2e-Sn(s) Ni2++2e-Ni(s) Co2++2e-Co(s)

2H+(pH=7)+2e-H2(g) Fe2++2e-Fe(s) Cr3++3e-Cr(s) Zn2++2e-Zn(s)

2H2O+2e-2OH-+H2(g)

+0.78+0.77

+0.53+0.52 +0.34+0.17 +0.15 0.00 -0.13-0.14

-0.25-0.28 -0.41-0.44

-0.74

-0.76-0.84

Hg2++2e-Hg(l) +0.78Ag++e-Ag(l) +0.80

1/2O2(g)+2H+(pH=7)+2e-H2O +0.82NO3

-+4H++3e-NO(g)+2H2O +0.96Br2(l)+2e-2Br- +1.06

1/2O2(g)+2H+(pH=7)+2e-

H2O+1.23

Na++e-Na(s) -2.71

An: lowest(right)

Cath: highest(left)

H2O1/2O2(g)+2H+

(pH=7)+2e-

We saw bubbles.

We saw pink.

We saw bubbles

Na+1 SO4-2 H2O

Na2SO4(aq)

Electrolysis lab A

Notes Three Unit Six

• Electrolysis Lab Results• Concept Check Assignment• Quiz-Balancing/Electrolysis

Na+1+e-Na(s)

NO3-+2H++e-NO2(g)+H2O

I2(s)+2e-2I-

Cu++e-Cu(s) Cu2++2e-Cu(s)

SO42-+4H++2e-SO2(g)

Sn4++2e-Sn(s) 2H++2e-H2(g)

Pb2++2e-Pb(s) Sn2++2e-Sn(s) Ni2++2e-Ni(s) Co2++2e-Co(s)

2H+(pH=7)+2e-H2(g) Fe2++2e-Fe(s) Cr3++3e-Cr(s) 2H2O+2e-2OH-+H2(g)

+0.78+0.53+0.52 +0.34+0.17 +0.15 0.00 -0.13-0.14

-0.25-0.28 -0.41-0.44

-0.74

-0.84

Hg2++2e-Hg(l) +0.78Ag++e-Ag(l) +0.80

1/2O2(g)+2H+(pH=7)+2e-H2O +0.82NO3

-+4H++3e-NO(g)+2H2O +0.96Br2(l)+2e-2Br- +1.06

1/2O2(g)+2H+(pH=7)+2e-

H2O+1.23

Mg+2+e-Mg(s) -2.37

An: lowest(right)

Cath: highest(left)

Mg(s)2e-1+Mg+2

Cu+2 SO4-2 H2O

CuSO4(aq)

Cl2(g)2e-2Cl-1 +1.36

+2.37

Na+1 SO4-2 H2O

Na2SO4(aq)

H+1 Cl-1 H2O

HCl(aq)

Mg(s) Cu(s)

-2.71Cu+2+Mg(s)Cu(s)+

Mg+2 +2.71

Lab C Voltaic Cell

Mg(s)2e-1+Mg+2Mg(s)

Cu(s)

Cathode

Anode

Cu2++2e-Cu(s)

e-1

SO4-2 H+1 Na+1

Spectator IonsSO4

-2Cu+2

H+1

Cl-1

Cl-1

Na+1SO4

-2

rxn

rxn

Current Flow

+

-

Notes Four Unit Six

• Faraday’s Law Lab B

• This as a chemical process that uses electricity to produce industrial quantities of specific chemicals.

Application of Faraday’s law

F = 96500 C/mol e-)

A x s = C

Faraday’s Law: Lab BFe(s)

Fe(s)

NO3-1

e-1Anode?Cathode?

Lose Mass?Gain Mass?

Fe+3

3e-1

Fe+3

NO3-1 NO3

-1

F = 96500 C/mole-

Amp x second = C

Faraday’s Law Calculation One

3.0 amp x

Au+3+3e-1Au(s)

60 min1 hour

1.5 Hour x 60 Sec1 min

=16000C

16000C x1mole e-1

96500C= 0.17 mol e-1

0.17 mol e-1x1 m Au(s)3mol e-1 =0.056mol Au(s)

0.056mol Au(s)x197.0gAu1mol Au = 11g Au

1. Balanced Equation2. Calculate Coulombs.

3. Calculate moles e-1.

4. Calculate moles of substance.

5. Calculate grams.

How many grams of Gold will be plated, using a current of 3.0 amps for 1.5 hours?

x

Faraday’s Law Calculation Two

2.0 amp x

Ag+1+1e-1Ag(s)

45 Hour x 60 Sec1 min

=5400C

5400C x1mole e-1

96500C= 0.056 mol e-1

0.056 mol e-1x1 m Ag(s)1mol e-1 =0.056mol Au(s)

0.05596mol Au(s)x107.9gAg1mol Ag = 6.0 g Ag

1. Balanced Equation2. Calculate Coulombs.

3. Calculate moles e-1.

4. Calculate moles of substance.

5. Calculate grams.

How many grams of Silver will be plated, using a current of 2.0 amps for 45 minutes?

Salt Bridge

Cu+2

Cu+2

Cu+2

Cu+2

Cu+2

Cu+2

SO4-2

SO4-2

SO4-2

SO4-2

SO4-2

SO4-2

Cr+3

SO4-2

Cr+3

SO4-2

SO4-2

Cathode

Anode Cu+2 + 2e-1 Cu

Cr Cr+3+3e-1

Overall rxn:2Cr+

x3

x2

36

3

26

2

3Cu+22Cr+3+3Cu

+0.34

+0.74

emf= 1.08volts

1.08

Cr+3

Cr+3

3e-1

3e-1

SO4-2

Na+1Na+1

Cu(s) Cr(s)Salt Bridge

e-1 ?rxn:

rxn:

Ion-Electron Method for Balancing

Ion-Electron Method for Balancing

UO2+2 + I2 U+4 + IO3

-1 (Acid)

UO2+2

I2 U+4 + H2O

#2. Bal Non-O Elem.#3. + H2O.#4. + H+1 .#5. + e-1 to bal +/-.

2+ H+14 + e-12

#1. Separate Half-rxn.

IO3-1 2+ H2O6 + H+112

111 + e-110

5X1X

UO2+2

I2

U+4 + H2O10+ H+120 + e-110

IO3-1 2+ H2O6 + H+112

55

1 + e-110

8 4

UO2+25 + U+45 + H2O4+ H+18 + I2 1 IO3

-1 2

Ion-Electron Method for Balancing

IO3-1 + Ti+3 I2 + TiO2

+1 (Acid)

IO3-1

Ti+3 I2 + H2O

#2. Bal Non-O Elem.#3. + H2O.#4. + H+1 .#5. + e-1 to bal +/-.

6+ H+112 + e-110

#1. Separate Half-rxn.

TiO2+11+ H2O2 + H+14

121 + e-12

1X5X

IO3-1

Ti+3

I2+ H2O6+ H+112 + e-110

TiO2+1 5+ H2O10 + H+120

12

5 + e-11084

IO3-12 + I21 + H+18+ H2O4 + Ti+3 5 TiO2

+1 5