chapter 20 electrochemistry -...

Electrochemistry

Mr. Matthew Totaro Legacy High School

AP Chemistry © 2012 Pearson Education, Inc.

Copyright © 2011 Pearson Education, Inc.

Electricity from Chemistry • Many chemical reactions involve the

transfer of electrons between atoms or ions – electron transfer reactions

• all single displacement, and combustion reactions

• some synthesis and decomposition reactions • The flow of electrons is associated

with electricity • Basic research into the nature of this

relationship may, in the near future, lead to cheaper, more efficient ways of generating electricity – fuel cell 2

Electrochemistry

Electrochemical Reactions In electrochemical reactions, electrons are

transferred from one species to another.

© 2012 Pearson Education, Inc.

Electrochemistry

Oxidation Numbers

In order to keep track of what loses

electrons and what gains them, we assign

oxidation numbers.


Electrochemistry

Oxidation and Reduction

• A species is oxidized when it loses electrons. – Here, zinc loses two electrons to go from neutral zinc

metal to the Zn2+ ion.


Electrochemistry


• A species is reduced when it gains electrons. – Here, each of the H+ gains an electron, and they

combine to form H2.


Electrochemistry


• What is reduced is the oxidizing agent. – H+ oxidizes Zn by taking electrons from it.

• What is oxidized is the reducing agent. – Zn reduces H+ by giving it electrons.


Electrochemistry

Oxidation–Reduction • Reactions where electrons are transferred

from one atom to another are called oxidation–reduction reactions – redox reactions for short

• Atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na → Na+ + 1 e– oxidation

Cl2 + 2 e– → 2 Cl– reduction

Oil Rig

8

Electrochemistry

Rules for Assigning Oxidation States • Rules are in order of priority 1. free elements have an oxidation state = 0

– Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g) 2. monatomic ions have an oxidation state equal

to their charge – Na = +1 and Cl = −1 in NaCl

3. (a) the sum of the oxidation states of all the atoms in a compound is 0

– Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0

9

Electrochemistry

Rules for Assigning Oxidation States 3. (b) the sum of the oxidation states of all the atoms

in a polyatomic ion equals the charge on the ion – N = +5 and O = −2 in NO3

–, (+5) + 3(−2) = −1

4. (a) Group I metals have an oxidation state of +1 in all their compounds

– Na = +1 in NaCl 4. (b) Group II metals have an oxidation state of +2 in

all their compounds – Mg = +2 in MgCl2

10

Electrochemistry

Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation

states according to the table below – nonmetals higher on the table take priority

Nonmetal Oxidation State Example F −1 CF4 H +1 CH4 O −2 CO2

Group 7A −1 CCl4 Group 6A −2 CS2 Group 5A −3 NH3

11

Electrochemistry

Example: Determine the oxidation states of all the atoms in a propanoate ion, C3H5O2

– • There are no free elements or free ions in

propanoate, so the first rule that applies is Rule 3b (C3) + (H5) + (O2) = −1

• Because all the atoms are nonmetals, the next rule we use is Rule 5, following the elements in order – H = +1 – O = −2

(C3) + 5(+1) + 2(−2) = −1 (C3) = −2 C = −⅔

Note: unlike charges, oxidation states can be fractions!

12

Electrochemistry

Practice – Assign an oxidation state to each element in the following

• Br2

• K+

• LiF

• CO2

• SO42−

• Na2O2

Br = 0 (Rule 1)

K = +1 (Rule 2)

Li = +1 (Rule 4a) & F = −1 (Rule 5)

O = −2 (Rule 5) & C = +4 (Rule 3a)

O = −2 (Rule 5) & S = +6 (Rule 3b)

Na = +1 (Rule 4a) & O = −1 (Rule 3a) 13

Electrochemistry

Example: Assign oxidation states, determine the element oxidized and reduced, and determine the

oxidizing agent and reducing agent in the following reaction

Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O

0 −2 +7 +1 +3 −2 +4 +1 −2

Oxidation Reduction

Oxidizing Agent

Reducing Agent

14

Electrochemistry

Practice – Assign oxidation states, determine the element oxidized and reduced, and determine

the oxidizing agent and reducing agent in the following reactions

Sn4+ + Ca → Sn2+ + Ca2+

F2 + S → SF4

+4 0 +2 +2 Ca is oxidized, Sn4+ is reduced Ca is the reducing agent, Sn4+ is the oxidizing agent

0 0 +4−1 S is oxidized, F is reduced S is the reducing agent, F2 is the oxidizing agent

15

Electrochemistry

Balancing Oxidation-Reduction Equations

Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method.


Electrochemistry

Half-Reactions • We generally split the redox reaction into two separate

half-reactions – a reaction just involving oxidation or reduction – the oxidation half-reaction has electrons as products – the reduction half-reaction has electrons as reactants

3 Cl2 + I− + 3H2O → 6 Cl− + IO3− + 6 H+

0 −1 +1 −2 −1 +5 −2 +1

oxidation: I− → IO3− + 6 e−

reduction: Cl2 + 2 e− → 2 Cl− 17

Electrochemistry

The Half-Reaction Method

1. Assign oxidation numbers to determine what is oxidized and what is reduced.

2. Write the oxidation and reduction half-reactions.


Electrochemistry


3. Balance each half-reaction.

a. Balance elements other than H and O.

b. Balance O by adding H2O. c. Balance H by adding H+. d. Balance charge by adding

electrons.


Electrochemistry

The Half-Reaction Method 4. Multiply the half-reactions by integers

so that the electrons gained and lost are the same.

5. Add the half-reactions, subtracting things that appear on both sides.

6. Make sure the equation is balanced according to mass.

7. Make sure the equation is balanced according to charge.


Electrochemistry


Consider the reaction between MnO4− and C2O4

2−: MnO4

−(aq) + C2O42−(aq) → Mn2+(aq) + CO2(aq)


Electrochemistry


First, we assign oxidation numbers:

MnO4− + C2O4

2− → Mn2+ + CO2

+7 +3 +4 +2

Since the manganese goes from +7 to +2, it is reduced.

Since the carbon goes from +3 to +4, it is oxidized.


Electrochemistry

Oxidation Half-Reaction

C2O42− → CO2

To balance the carbon, we add a

coefficient of 2:

C2O42− → 2CO2


Electrochemistry

Oxidation Half-Reaction

C2O42− → 2CO2

The oxygen is now balanced as well.

To balance the charge, we must add 2 electrons to the right side:

C2O4

2− → 2CO2 + 2e−


Electrochemistry

Reduction Half-Reaction

MnO4− → Mn2+

The manganese is balanced; to balance

the oxygen, we must add 4 waters to the right side:

MnO4

− → Mn2+ + 4H2O


Electrochemistry


MnO4− → Mn2+ + 4H2O

To balance the hydrogen, we add

8H+ to the left side:

8H+ + MnO4− → Mn2+ + 4H2O


Electrochemistry


8H+ + MnO4− → Mn2+ + 4H2O

To balance the charge, we add 5e− to

the left side:

5e− + 8H+ + MnO4− → Mn2+ + 4H2O


Electrochemistry

Combining the Half-Reactions Now we evaluate the two half-reactions

together:

C2O42− → 2CO2 + 2e−

5e− + 8H+ + MnO4− → Mn2+ + 4H2O

To attain the same number of electrons

on each side, we will multiply the first reaction by 5 and the second by 2:


Electrochemistry

Combining the Half-Reactions

5C2O42− → 10CO2 + 10e−

10e− + 16H+ + 2MnO4− → 2Mn2+ + 8H2O

When we add these together, we get: 10e− + 16H+ + 2MnO4

− + 5C2O42− →

2Mn2+ + 8H2O + 10CO2 +10e−


Electrochemistry

Combining the Half-Reactions 10e− + 16H+ + 2MnO4

− + 5C2O42− →

2Mn2+ + 8H2O + 10CO2 +10e−

The only thing that appears on both sides are the

electrons. Subtracting them, we are left with: 16H+ + 2MnO4

− + 5C2O42− →

2Mn2+ + 8H2O + 10CO2


Electrochemistry

1. assign oxidation states and determine element oxidized and element reduced

2. separate into oxidation & reduction half-reactions

Fe2+ + MnO4– → Fe3+ + Mn2+

+2 +7 −2 +3 +2 oxidation

ox: Fe2+ → Fe3+

red: MnO4– → Mn2+

Example: Balancing redox reactions in acidic solution

reduction

31

Electrochemistry

3. balance half-reactions by mass

a) first balance atoms other than O and H

b) balance O by adding H2O to side that lacks O

c) balance H by adding H+ to side that lacks H

Fe2+ → Fe3+

MnO4– → Mn2+

MnO4– → Mn2+ + 4H2O

MnO4– + 8H+ → Mn2+ + 4H2O


32

Electrochemistry

4. balance each half-reaction with respect to charge by adjusting the numbers of electrons

a) electrons on product side for oxidation

b) electrons on reactant side for reduction

Fe2+ → Fe3+ + 1 e−

MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O

MnO4– + 8H+ → Mn2+ + 4H2O

+7 +2


33

Electrochemistry

5. balance electrons between half-reactions

6. add half-reactions, canceling electrons and common species

7. Check that numbers of atoms and total charge are equal

Fe2+ → Fe3+ + 1 e−

MnO4– + 8H+ + 5 e− → Mn2+ + 4H2O

} x 5

5 Fe2+ → 5 Fe3+ + 5 e− MnO4

– + 8H+ + 5 e− → Mn2+ + 4H2O

5 Fe2+ + MnO4– + 8H+ → Mn2+ + 4H2O + 5 Fe3+


reactant side Element product side 5 Fe 5 1 Mn 1 4 O 4 8 H 8

+17 charge +17 34

Electrochemistry

Practice – Balance the following equation in acidic solution

I– + Cr2O72− → Cr3+ + I2

35

Electrochemistry

Practice – Balancing redox reactions 1. assign oxidation

states and determine element oxidized and element reduced


I− + Cr2O72– → I2 + Cr3+

−1 +6 −2 0 +3

oxidation

reduction

ox: I− → I2

red: Cr2O72– → Cr3+

36

Electrochemistry

Practice – Balancing redox reactions 3. balance half-

reactions by mass




ox: I− → I2 ox: 2 I− → I2

red: Cr2O72– → Cr3+

red: Cr2O72– → 2 Cr3+

red: Cr2O72– → 2Cr3+ +7H2O

Cr2O72– +14H+ → 2Cr3+ +7H2O

37

Electrochemistry

Practice – Balancing redox reactions 4. balance each

half-reaction with respect to charge by adjusting the numbers of electrons a) electrons on

product side for oxidation


2 I− → I2 2 I− → I2 + 2e−

Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O Cr2O72– +14H+ → 2Cr3+ +7H2O

38

Electrochemistry

Cr2O72– +14H+ + 6e−→ 2Cr3+ +7H2O

Practice – Balancing redox reactions 5. balance

electrons between half-reactions


7. check

} x 3

Cr2O72– +14H+ + 6 I−→ 2Cr3+ +7H2O + 3 I2

2 I− → I2 + 2e− 6 I− → 3 I2 + 6e−

reactant side Element product side 2 Cr 2 6 I 6 7 O 7

14 H 14 +6 charge +6

39

Electrochemistry

Balancing in Basic Solution

• If a reaction occurs in a basic solution, one can balance it as if it occurred in acid.

• Once the equation is balanced, add OH− to each side to “neutralize” the H+ in the equation and create water in its place.

• If this produces water on both sides, you might have to subtract water from each side.


Electrochemistry

assign oxidation states

separate into half-reactions

ox: I−(aq) → I2(aq)

red: MnO4−(aq) → MnO2(s)

Example: Balance the equation: I−

(aq) + MnO4−

(aq) → I2(aq) + MnO2(s) in basic solution

assign oxidation states

I−(aq) + MnO4

−(aq) → I2(aq) + MnO2(s)

separate into half-reactions

ox: red:

−1 +7 −2 0 +4 −2

Oxidation Reduction

41

Electrochemistry

balance half-reactions by mass

ox: I−(aq) → I2(aq)


balance half-reactions by mass first, elements other than H and O

ox: 2 I−(aq) → I2(aq)


balance half-reactions by mass then O by adding H2O

ox: 2 I−(aq) → I2(aq)

red: MnO4−(aq) → MnO2(s) + 2 H2O(l)

balance half-reactions by mass then H by adding H+

ox: 2 I−(aq) → I2(aq)

red: 4 H+(aq) + MnO4

−(aq) → MnO2(s) + 2 H2O(l)

balance half-reactions by mass in base, neutralize the H+ with OH−

ox: 2 I−(aq) → I2(aq)


−(aq) → MnO2(s) + 2 H2O(l)

4 H+

(aq) + 4 OH−(aq) + MnO4

−(aq) → MnO2(s) + 2 H2O(l) + 4 OH−

(aq)


ox: 2 I−(aq) → I2(aq)


−(aq) → MnO2(s) + 2 H2O(l)

4 H+

(aq) + 4 OH−(aq) + MnO4

−(aq) → MnO2(s) + 2 H2O(l) + 4 OH−

(aq)

4 H2O(aq) + MnO4−(aq) → MnO2(s) + 2 H2O(l) + 4 OH−

(aq)


ox: 2 I−(aq) → I2(aq)


−(aq) → MnO2(s) + 2 H2O(l)

4 H+

(aq) + 4 OH−(aq) + MnO4

−(aq) → MnO2(s) + 2 H2O(l) + 4 OH−

(aq)

4 H2O(aq) + MnO4−(aq) → MnO2(s) + 2 H2O(l) + 4 OH−

(aq)

2 H2O(l) + MnO4−(aq) → MnO2(s) + 4 OH−

(aq)


(aq) + MnO4−


42

Electrochemistry

balance Half-reactions by charge

ox: 2 I−(aq) → I2(aq) + 2 e−

red: MnO4−(aq) + 2 H2O(l) + 3 e−→ MnO2(s) + 4 OH−

(aq)

Balance electrons between half-reactions

ox: 2 I−(aq) → I2(aq) + 2 e− } x3

red: MnO4−(aq) + 2 H2O(l) + 3 e− → MnO2(s) + 4 OH−

(aq) }x2 ox: 6 I−

(aq) → 3 I2(aq) + 6 e−

red: 2 MnO4−(aq) + 4 H2O(l) + 6 e−→ 2 MnO2(s) + 8 OH−

(aq)


ox: 2 I−(aq) → I2(aq) + 2 e−

red: MnO4−(aq) + 2 H2O(l) + 3 e−→ MnO2(s) + 4 OH−

(aq)



ox: 2 I−(aq) → I2(aq)

red: MnO4−(aq) + 2 H2O(l) → MnO2(s) + 4 OH−

(aq)



(aq) + MnO4−


43

Electrochemistry

add the half-rxns

ox: 6 I−(aq) → 3 I2(aq) + 6 e−

red: 2 MnO4−(aq) + 4 H2O(l) + 6 e− → 2 MnO2(s) + 8 OH−

(aq) tot: 6 I−

(aq)+ 2 MnO4−(aq) + 4 H2O(l) → 3 I2(aq)+ 2 MnO2(s) + 8 OH−

(aq)

check

add the half-rxns

ox: 6 I−(aq) → 3 I2(aq) + 6 e−

red: 2 MnO4−(aq) + 4 H2O(l) + 6 e− → 2 MnO2(s) + 8 OH−

(aq)

check


(aq) + MnO4−


Reactant Count Element

Product Count

6 I 6 2 Mn 2

12 O 12 8 H 8 8− charge 8−

44

Electrochemistry

Practice – Balance the following equation in basic solution

I– + CrO42− → Cr3+ + I2

45

Electrochemistry

Practice – Balancing redox reactions 1. assign oxidation

states and determine element oxidized and element reduced


I− + CrO42– → I2 + Cr3+

−1 +6 −2 0 +3

oxidation

reduction

ox: I− → I2

red: CrO42– → Cr3+

46

Electrochemistry

Practice – Balancing redox reactions 3. balance half-

reactions by mass




d) if in basic solution, neutralize the H+ with OH−

ox: I− → I2 ox: 2 I− → I2

red: CrO42– → Cr3+

red: CrO42– → Cr3+ +4H2O

CrO42– +8H+ → Cr3+ +4H2O

CrO42– +8H+ +8OH− → Cr3+ +4H2O +8OH− CrO42– +8H2O → Cr3+ +4H2O +8OH−

CrO42– +4H2O → Cr3+ +8OH−

47

Electrochemistry

Practice – Balancing redox reactions 4. balance each

half-reaction with respect to charge by adjusting the numbers of electrons

a) electrons on product side for oxidation


2 I− → I2 2 I− → I2 + 2e−

CrO42– +4H2O → Cr3+ +8OH− CrO42– +4H2O+ 3e− → Cr3+ +8OH−

48

Electrochemistry

Practice – Balancing redox reactions

2CrO42– +8H2O+ 6e− → 2Cr3+ +16OH− CrO4

2– +4H2O+ 3e− → Cr3+ +8OH−

5. balance electrons between half-reactions


7. check

} x 3

2CrO42– +8H2O + 6 I− → 2Cr3+ +16OH− + 3 I2

2 I− → I2 + 2e− 6 I− → 3 I2 + 6e−

reactant side Element product side 2 Cr 2 6 I 6

16 O 16 16 H 16

−10 charge −10

} x 2

49

Electrochemistry

Voltaic Cells

In spontaneous oxidation-reduction

(redox) reactions, electrons are

transferred and energy is released.


Electrochemistry

Voltaic Cells

• We can use that energy to do work if we make the electrons flow through an external device.

• We call such a setup a voltaic cell or a galvanic cell.


Electrochemistry

Voltaic Cells • A typical cell looks

like this. • The oxidation occurs

at the anode. • The reduction occurs

at the cathode. • Think Red Cat


Electrochemistry

Voltaic Cell the salt bridge is

required to complete the

circuit and maintain charge

balance

53

Electrochemistry

Electrochemical Cells

54

• Oxidation and reduction half-reactions are kept separate in half-cells

• Electron flow through a wire along with ion flow through a solution constitutes an electric circuit

• Requires a conductive solid electrode to allow the transfer of electrons through external circuit metal or graphite

• requires ion exchange between the two half-cells of the system electrolyte

Electrochemistry

Voltaic Cells Once even one electron flows from

the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons

would stop.


Electrochemistry

Voltaic Cells • Therefore, we use a

salt bridge, usually a U-shaped tube that contains a salt solution, to keep the charges balanced. – Cations move toward

the cathode. – Anions move toward

the anode.


Electrochemistry

Voltaic Cells • In the cell, then,

electrons leave the anode and flow through the wire to the cathode.

• As the electrons leave the anode, the cations formed dissolve into the solution in the anode compartment.


Electrochemistry

Voltaic Cells • As the electrons reach

the cathode, cations in the cathode are attracted to the now negative cathode.

• The electrons are taken by the cation, and the neutral metal is deposited on the cathode.


Electrochemistry

Electromotive Force (emf) • Water only

spontaneously flows one way in a waterfall.

• Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.


Electrochemistry

Electromotive Force (emf)

• The potential difference between the anode and cathode in a cell is called the electromotive force (emf).

• It is also called the cell potential and is designated Ecell.


Electrochemistry

Cell Potential

Cell potential is measured in volts (V).

1 V = 1 J C


Electrochemistry

Standard Reduction Potentials Reduction

potentials for many electrodes have

been measured and tabulated.


Electrochemistry

Standard Hydrogen Electrode

• Their values are referenced to a standard hydrogen electrode (SHE).

• By definition, the reduction potential for hydrogen is 0 V:

2 H+(aq, 1M) + 2e− → H2(g, 1 atm)


Electrochemistry

Standard Cell Potentials

The cell potential at standard conditions can be found through this equation:

Ecell ° = Ered (cathode) − Ered (anode) ° °

Because cell potential is based on the potential energy per unit of charge, it is an intensive property.


Electrochemistry

Cell Potentials • For the oxidation in this cell,

• For the reduction,

Ered = −0.76 V °

Ered = +0.34 V °


Electrochemistry

Cell Potentials

Ecell ° = Ered ° (cathode) − Ered ° (anode) = +0.34 V − (−0.76 V) = +1.10 V


Electrochemistry

Which Way Will Electrons flow?

Zn → Zn2+ + 2 e− E°= +0.76 Cu → Cu2+ + 2 e− E°= −0.34

Under standard conditions, zinc has a stronger tendency to oxidize than copper

67

Therefore the electrons flow from zinc; making zinc the anode

Zn → Zn2+ + 2 e−

Cu2+ + 2 e− → Cu

Pote

ntia

l Ene

rgy

Electron Flow

Electrochemistry

Oxidizing and Reducing Agents • The strongest oxidizers

have the most positive reduction potentials.

• The strongest reducers have the most negative reduction potentials.


Electrochemistry

Oxidizing and Reducing Agents

The greater the difference between the

two, the greater the voltage of the cell.


Electrochemistry 70

Electrochemistry 71

Electrochemistry

Example: Calculate E°cell for the reaction at 25 °C Al(s) + NO3

−(aq) + 4 H+

(aq) → Al3+(aq) + NO(g) + 2 H2O(l)

separate the reaction into the oxidation and reduction half-reactions

ox (anode): Al(s) → Al3+(aq) + 3 e−

red: NO3

−(aq) + 4 H+

(aq) + 3 e− → NO(g) + 2 H2O(l)

find the E° for each half-reaction and sum to get E°cell

E°ox of Al = −E°red of Al3+ = +1.66 V

E°red of NO3− = +0.96 V

E°cell = (+1.66 V) + (+0.96 V) = +2.62 V

72

Electrochemistry

Practice – Calculate E°cell for the reaction at 25 °C IO3

–(aq) + 6 H+(aq) + 5 I−(aq) → 3 I2(s) + 3 H2O(l)

Reduction Half-Reaction E°red, V F2(g) + 2e−→ 2 F−(aq) +2.87 IO3

−(aq) + 6 H++ 5e−→ ½I2(s) + 3H2O(l) +1.20 Ag+(aq) + 1e−→ Ag(s) +0.80 I2(s) + 2e−→ 2 I−(aq) +0.54 Cu2+(aq) + 2e−→ Cu(s) +0.34 Cr3+(aq) + 1e−→ Cr2+(aq) −0.50 Mg2+(aq) + 2e−→ Mg(s) −2.37

73

Electrochemistry

Practice – Calculate E°cell for the reaction at 25 °C 2 IO3

– + 12 H+ + 10 I−→ 6 I2 + 6 H2O

separate the reaction into the oxidation and reduction half-reactions

ox (anode): 2 I−(s) → I2(aq) + 2 e− red: IO3

−(aq) + 6 H+

(aq) + 5 e− → ½ I2(s) + 3 H2O(l)

find the E° for each half-reaction and sum to get E°cell

E°ox of I− = −E°red of I = −0.54 V

E°red of IO3− = +1.20 V

E°cell = (−0.54 V) + (+1.20 V) = +0.66 V

2

74

Electrochemistry

Practice – Sketch and label a voltaic cell in which one half-cell has Ag(s) immersed in 1 M AgNO3, and the

other half-cell has a Pt electrode immersed in 1 M Cr(NO3)2 and 1 M Cr(NO3)3 Write the half-reactions and overall reaction, and

determine the cell potential under standard conditions.



75

Electrochemistry

ox: Cr2+(aq) → Cr3+(aq) + 1 e− E = +0.50 V red: Ag+(aq) + 1 e− → Ag(s) E = +0.80 V

tot: Cr2+(aq) + Ag+(aq) → Cr3+(aq) + Ag(s) E = +1.30 V

anode = Pt

Cr2+

Cr3+

e− →

e− →

e− →

e− →

cathode = Ag

Ag+

salt bridge

76

Electrochemistry

Cell Notation • Shorthand description of a voltaic cell • Electrode | electrolyte || electrolyte |

electrode • Oxidation half-cell on the left, reduction half-

cell on the right • Single | = phase barrier

– if multiple electrolytes in same phase, a comma is used rather than |

– often use an inert electrode • Double line || = salt bridge

77

Electrochemistry

Voltaic Cell

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

Anode = Zn(s) the anode is oxidized to Zn2+

Cathode = Cu(s) Cu2+ ions are reduced at the cathode

78

Electrochemistry Zn(s) | Zn2+(1 M) || H+(1 M) | H2(g) | Pt(s) 79

Electrochemistry

Predicting Spontaneity of Redox Reactions

• A spontaneous reaction will take place when a reduction half-reaction is paired with an oxidation half-reaction lower on the table

• If paired the other way, the reverse reaction is spontaneous

Cu2+(aq) + 2 e− → Cu(s) E°red = +0.34 V Zn2+(aq) + 2 e− → Zn(s) E°red = −0.76 V

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) spontaneous Cu(s) + Zn2+(aq) → Cu2+(aq) + Zn(s) nonspontaneous

80

Electrochemistry

Example: Predict if the following reaction is spontaneous under standard conditions

Fe(s) + Mg2+(aq) → Fe2+

(aq) + Mg(s) separate the reaction into the oxidation and reduction half-reactions

ox: Fe(s) → Fe2+(aq) + 2 e−

red: Mg2+

(aq) + 2 e− → Mg(s)

look up the relative positions of the reduction half-reactions

red: Fe2+(aq) + 2 e− → Fe(s)

red: Mg2+(aq) + 2 e− → Mg(s)

because Mg2+ reduction is below Fe2+ reduction, the reaction is NOT spontaneous as written

81

Electrochemistry

the reaction is spontaneous in the reverse direction

Mg(s) + Fe2+(aq) → Mg2+

(aq) + Fe(s) ox: Mg(s) → Mg2+

(aq) + 2 e− red: Fe2+

(aq) + 2 e− → Fe(s)

sketch the cell and label the parts – oxidation occurs at the anode; electrons flow from anode to cathode

82

Electrochemistry

Practice – Decide whether each of the following will be spontaneous as written or in the reverse direction


F2(g) + 2e−→ 2 F−(aq)

IO3−(aq) + 6 H++ 5e−→ I2(s) + 3H2O(l)

Ag+(aq) + 1e−→ Ag(s)

I2(s) + 2e−→ 2 I−(aq)

Cu2+(aq) + 2e−→ Cu(s)

Cr3+(aq) + 1e−→ Cr2+(aq)

Mg2+(aq) + 2e−→ Mg(s)

F2(g) + 2 I−(aq) → I2(s) + 2 F−(aq)

Mg(s) + 2 Ag+(aq) → Mg2+

(aq) + 2 Ag(s)

Cu2+(aq) + 2 I−(aq) → I2(s) + Cu(s)

Cu2+(aq) + 2 Cr2+

(aq) → Cu(s) + 2 Cr3+(aq)

spontaneous as written



spontaneous in reverse

83

Electrochemistry

Free Energy

∆G for a redox reaction can be found by using the equation

∆G = −nFE

where n is the number of moles of

electrons transferred, and F is a constant, the Faraday:

1 F = 96,485 C/mol = 96,485 J/V-mol © 2012 Pearson Education, Inc.

Electrochemistry

red: I2(l) + 2 e− → 2 I−(aq) E° = +0.54 V

ox: 2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 V

tot: I2(l) + 2Br−(aq) → 2I−(aq) + Br2(l) E° = −0.55 V

Example: Calculate ∆G° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)

85

because ∆G° is +, the reaction is not spontaneous in the forward direction under standard conditions

Answer:

Solve:

Conceptual Plan:

Relationships:

I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)

DG°, (J) Given:

Find:

E°ox, E°red E°cell ∆G°

Electrochemistry

Practice – Calculate ∆G° for the reaction at 25°C 2IO3

–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)



86

Electrochemistry

ox : 2 I−(s) → I2(aq) + 2 e− Eº = −0.54 V red: IO3

−(aq) + 6 H+(aq) + 5 e− → ½ I2(s) + 3 H2O(l) Eº = 1.20 V tot: 2IO3

−(aq) + 12H+

(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V

Practice – Calculate ∆G° for the reaction at 25 °C 2IO3

–(aq) + 12H+(aq) + 10 I−(aq) →6 I2(s) + 6 H2O(l)

because ∆G° is −, the reaction is spontaneous in the forward direction under standard conditions

Answer:

Solve:

Conceptual Plan:

Relationships:

2IO3−

(aq)+12H+(aq)+10I−(aq) → 6 I2(s) + 6 H2O(l)

∆G°, (J) Given:

Find:

E°ox, E°red E°cell ∆G°

87

Tro: Chemistry: A Molecular Approach, 2/e

Electrochemistry

• For a spontaneous reaction – one that proceeds in the forward direction with

the chemicals in their standard states -∆G° < 1 (negative) – E° > 1 (positive) – K > 1

• ∆G° = −RTlnK = −nFE°cell – n is the number of electrons – F = Faraday’s Constant = 96,485 C/mol e−

E°cell, ∆G° and K

88

Electrochemistry

tot: Cu(s) + 2H+(aq) → Cu2+

(aq) + H2(g) E° = −0.34 V

Example: Calculate K at 25 °C for the reaction Cu(s) + 2 H+

(aq) → H2(g) + Cu2+(aq)

89

because K < 1, the position of equilibrium lies far to the left under standard conditions

Answer:

Solve:

Conceptual Plan:

Relationships:

Cu(s) + 2 H+(aq) → H2(g) + Cu2+

(aq) K

Given: Find:

E°ox, E°red E°cell K

ox: Cu(s) → Cu2+(aq) + 2 e− E° = −0.34 V

red: 2 H+(aq) + 2 e− → H2(aq) E° = +0.00 V

Tro: Chemistry: A Molecular Approach, 2/e

Electrochemistry

Practice – Calculate K for the reaction at 25°C 2IO3

–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)



90

Electrochemistry

ox : 2 I−(s) → I2(aq) + 2e− Eº = −0.54 V red: IO3

−(aq) + 6 H+(aq) + 5e− → ½ I2(s) + 3H2O(l) Eº = 1.20 V tot: 2IO3

−(aq) + 12H+

(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V

Practice – Calculate K for the reaction at 25 °C 2IO3

–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)

because K >> 1, the position of equilibrium lies far to the right under standard conditions

Answer:

Solve:

Conceptual Plan:

Relationships:

2 IO3−

(aq)+12 H+(aq)+10 I−(aq) → 6 I2(s) + 6 H2O(l)

K Given:

Find:

E°ox, E°red E°cell K

91

Electrochemistry

Cell Potential when Ion Concentrations Are Not 1 M

• We know there is a relationship between the reaction quotient, Q, the equilibrium constant, K, and the free energy change, ∆Gº

• Changing the concentrations of the reactants and products so they are not 1 M will affect the standard free energy change, ∆Gº

• Because ∆Gº determines the cell potential, Ecell, the voltage for the cell will be different when the ion concentrations are not 1 M 92

Electrochemistry

Ecell When Ion Concentrations are Not 1 M

93

Electrochemistry

Ecell When Ion Concentrations are Not 1 M

• The standard cell potential, Eo, corresponds to the standard conditions of Q = 1. As the system approaches equilibrium, the magnitude (i.e., absolute value) of the cell potential decreases, reaching zero at equilibrium (when Q K). Deviations from standard conditions that take the cell further from equilibrium than Q = 1 will increase the magnitude of the cell potential relative to E. Deviations from standard conditions that take the cell closer to equilibrium than Q = 1 will decrease the magnitude of the cell potential relative to E. In concentration cells, the direction of spontaneous electron flow can be determined by considering the direction needed to reach equilibrium.

Electrochemistry

Deriving the Nernst Equation

95

Electrochemistry

Electrochemical Cells • In all electrochemical cells, oxidation occurs at the

anode, reduction occurs at the cathode • In voltaic cells

– anode is the source of electrons and has a (−) charge – cathode draws electrons and has a (+) charge

• In electrolytic cells – electrons are drawn off the anode, so it must have a place

to release the electrons, the + terminal of the battery – electrons are forced toward the anode, so it must have a

source of electrons, the − terminal of the battery

96

Electrochemistry 97

Electrochemistry 98

Electrolysis • Electrolysis is the process of

using electrical energy to break a compound apart

• Electrolysis is done in an electrolytic cell

• Electrolytic cells can be used to separate elements from their compounds

Electrochemistry

Electrolysis • In electrolysis we use electrical energy to overcome

the energy barrier of a non-spontaneous reaction, allowing it to occur

• The reaction that takes place is the opposite of the spontaneous process

2 H2(g) + O2(g) → 2 H2O(l) spontaneous 2 H2O(l) → 2 H2(g) + O2(g) electrolysis • Some applications are (1) metal extraction from

minerals and purification, (2) production of H2 for fuel cells, (3) metal plating

99

Electrochemistry

Electrolytic Cells • The electrical energy is supplied by a Direct

Current power supply – AC alternates the flow of electrons so the

reaction won’t be able to proceed • Some electrolysis reactions require more

voltage than Ecell predicts. This is called the overvoltage

100

Electrochemistry

Electrolytic Cells • The source of energy is a battery or DC power supply • The + terminal of the souce is attached to the anode • The − terminal of the source is attached to the cathode • Electrolyte can be either an aqueous salt solution or a

molten ionic salt • Cations in the electrolyte are attracted to the cathode

and anions are attracted to the anode • Cations pick up electrons from the cathode and are

reduced, anions release electrons to the anode and are oxidized

101

Electrochemistry

Electrolysis of Pure Compounds • The compound must be in molten (liquid)

state • Electrodes normally graphite • Cations are reduced at the cathode to metal

element • Anions oxidized at anode to nonmetal

element

102

Electrochemistry

Electrolysis of Water

103

Electrochemistry

Electroplating In electroplating, the work piece is the cathode

Cations are reduced at cathode and plate to the surface of the work piece The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution.

104

Electrochemistry

Current • Current is the number of electrons that

flow through the system per second – unit = Ampere

• 1 A of current = 1 Coulomb of charge flowing by each second – 1 A = 6.242 x 1018 electrons per second

• Electrode surface area dictates the number of electrons that can flow – larger batteries produce larger currents

105

Electrochemistry

Stoichiometry of Electrolysis • In an electrolytic cell, the amount of product

made is related to the number of electrons transferred – essentially, the electrons are a reactant

• The number of moles of electrons that flow through the electrolytic cell depends on the current and length of time – 1 Amp = 1 Coulomb of charge/second – 1 mole of e− = 96,485 Coulombs of charge

• Faraday’s constant 106

Electrochemistry

Faraday’s Laws – The first law of electrolysis

• We can link this product to the mass of metal produced at the cathode as being directly proportional to the quantity of electricity passed through the cell.

• This relationship is known as Faraday’s first law of electrolysis.

• It may be written as: m

Electrochemistry

Example: Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction

Au3+(aq) + 3 e− → Au(s)

units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e−

Check:

Solve:

Conceptual Plan:

Relationships:

3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min mass Au, g

Given:

Find:

t(s), amp charge (C) mol e− mol Au g Au

108

Electrochemistry

Practice – Calculate the amperage required to plate 2.5 g of gold in 1 hour (3600 sec) Au3+(aq) + 3 e− → Au(s)

109

Electrochemistry

Practice – Calculate the amperage required to plate 2.5 g of Au in 1 hour (3600 sec)

Au3+(aq) + 3 e− → Au(s)

units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e−

Check:

Solve:

Conceptual Plan:

Relationships:

3 mol e− : 1 mol Au, mass = 2.5 g, time = 3600 s current, A

Given:

Find:

g Au mol Au mol e− charge amps

110

Electrochemistry

Applications of Oxidation-Reduction

Reactions


Electrochemistry

Batteries


Electrochemistry

Alkaline Batteries


Electrochemistry

Hydrogen Fuel Cells


Electrochemistry

Corrosion and…


Electrochemistry

…Corrosion Prevention


chapter 20 electrochemistry -...

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