Electrochemistry

Oxidation – Reduction Reactions

Consider the reaction of Copper wire and AgNO3(aq)

AgNO3(aq)

Ag(s)

Cu(s)

Oxidation – Reduction Reactions If you leave the reaction a long time the solution

goes blue!

The blue is due to Cu2+(aq)

Oxidation-Reduction Reactions So when we mix Ag+

(aq) with Cu(s) we get Ag(s) and Cu2+

(aq)

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

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

The electrons gained by Ag+ must come from the Cu2+

Can’t have reduction without oxidation (redox) Each Cu can reduce 2 Ag+

2Ag+(aq) + 2e- 2Ag(s)

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

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

(aq) + 2e-

lose electrons = oxidation

gain electrons = reduction

Redox Cu/Ag

Cu

Ag+ Ag+

E

electron flow

Redox Cu/Ag

Cu2+

Ag Ag

E

ΔE = e.Ecell

• e = charge on an electron• Ecell = Voltage in a electrochemical cell• If we could separate the two reactions we

could use the energy gained by the e to do work

Redox Cu/Ag

Cu2+

Ag Ag

E

• The maximum amount of energy available to do work is a definition of the free energy ΔG

Free energy for redox rxn

cell voltage

# electrons in redox rxn

F = Faraday’s constant= 96485 C/mol

ΔEcell: the cell potential

In a redox reaction electrons are transferred to a more stable state

Most of the free energy of the reaction is due to these electrons

Can we access this energy?

Yes, by conducting the two half reactions in separate cells2Ag+

(aq) + 2e- 2Ag(s)

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

This is called the electrochemical cell

The cell potential can be measured in such a cell with a voltmeter

Electrochemical/Voltaic Cell Cu/Ag

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

(aq) + 2e-

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Electric Current Flowing Directly Between Atoms

11

Electric Current Flowing Indirectly Between Atoms

12

Electrochemistry and Cells

electrochemistry is the study of redox reactions that produce or require an electric current

the conversion between chemical energy and electrical energy is carried out in an electrochemical cell

spontaneous redox reactions take place in a voltaic cell (galvanic cell)

nonspontaneous redox reactions can be made to occur in an electrolytic

cell by the addition of electrical energy

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Electrodes

Anode (donates electrons to the cathode) electrode where oxidation occurs In a Galvanic cell it is the –ve terminal and in an electrolytic

cell it is the +ve terminal anions attracted to it connected to positive end of battery in electrolytic cell loses weight in electrolytic cell

Cathode (attracts electrons from the anode) electrode where reduction occurs In a Galvanic cell it is the +ve terminal and in an electrolytic

cell it is the -ve terminal cations attracted to it connected to negative end of battery in electrolytic cell gains weight in electrolytic cell

electrode where plating takes place in electroplating

14

Current and Voltage

the number of electrons that flow through the system per second is the current unit = Ampere 1 A of current = 1 Coulomb of charge flowing by each second 1 A = 6.242 x 1018 electrons/second Electrode surface area dictates the number of electrons that

can flow

the difference in potential energy between the reactants and products is the potential difference (the potential for an electric field to cause an electrical current) unit = Volt 1 V of force = 1 J of energy/Coulomb of charge the voltage needed to drive electrons through the external

circuit amount of force pushing the electrons through the wire is

called the electromotive force, emf

15

Cell Potential

the difference in potential energy between the anode the cathode in a voltaic cell is called the cell potential

the cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode

the cell potential under standard conditions is called the standard emf, E°cell

25°C, 1 atm for gases, 1 M concentration of solution sum of the cell potentials for the half-reactions

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Cell Notation

shorthand description of Voltaic cell

electrode | electrolyte || electrolyte | electrode

oxidation half-cell on left (anode), reduction half-cell on the right (cathode)

single | = phase barrier if multiple electrolytes in same phase, a comma is

used rather than | often use an inert electrode

double line || = salt bridge

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Fe(s) | Fe2+(aq) || MnO4-(aq), Mn2+(aq), H+(aq) | Pt(s)

Keeping track of electrons

Redox reactions involve the transfer of electrons from the donor to the acceptor

The donor loses electrons and is oxidized

The acceptor acquires electrons and is reduced

To know if a reaction is a redox reaction we need a way to keep track of how many valence electrons each element has

We define the oxidation state of an element to be +n (n integer) if it has n less electrons than it does as the free atom, and –p if it has p more electrons than it does in the free atom 18

21

Assigning Oxidation Numbers

1. The sum of the oxidation numbers Q of all the atoms in a compound/ion at up to the charge on the compound/ion. This means ….

a. free elements have an oxidation state = 0 QNa = 0 and QCl = 0 in 2 Na(s) + Cl2(g)

b. monatomic ions have an oxidation state equal to their charge QNa = +1 and QCl = -1 in NaCl

c. The sum of the oxidation numbers of all the atoms in a neutral compound is 0

d. the sum of the oxidation numbers of all the atoms in a polyatomic ion equals the charge on the ion

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

Na = +1 in NaCl(b) Group II metals have an oxidation state of +2 in all their compounds

Mg = +2 in MgCl23. F has an oxidation number QF = -1

4. H has an oxidation number QH = +1

5. O has an oxidation number QO = -2

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Oxidation and Reduction

oxidation occurs when an atom’s oxidation state increases during a reaction

reduction occurs when an atom’s oxidation state decreases during a reaction

CH4 + 2 O2 → CO2 + 2 H2O-4 +1 0 +4 –2 +1 -2

oxidationreduction

rule 1rule 4 rule 4rule 5

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Oxidation–Reduction oxidation and reduction must occur simultaneously

if an atom loses electrons another atom must take them

the reactant that reduces an element in another reactant is called the reducing agent the reducing agent contains the element that is oxidized

the reactant that oxidizes an element in another reactant is called the oxidizing agent the oxidizing agent contains the element that is reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)

Na is oxidized, Cl is reducedNa is the reducing agent, Cl2 is the oxidizing agent

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Identify the Oxidizing and Reducing Agents in Each of the Following

3 H2S + 2 NO3– + 2 H+ 3 S + 2 NO + 4 H2O

MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O

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Identify the Oxidizing and Reducing Agents in Each of the Following

3 H2S + 2 NO3– + 2 H+ 3 S + 2 NO + 4 H2O

MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O

+1 -2 +5 -2 +1 0 +2 -2 +1 -2

ox agred ag

+4 -2 +1 -1 +2 -1 0 +1 -2

oxidationreduction

oxidation

reduction

red agox ag

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Common Oxidizing Agents

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Common Reducing Agents

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Balancing Redox Reactions

1. assign oxidation numbers• determine element oxidized and element reduced

2. write ox. & red. half-reactions, including electrons• ox. electrons on right, red. electrons on left of arrow

3. balance half-reactions by massa) first balance elements other than H and Ob) add H2O where need Oc) add H+1 where need Hd) neutralize H+ with OH- in base

4. balance half-reactions by charge• balance charge by adjusting electrons

5. balance electrons between half-reactions

6. add half-reactions

7. check

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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:

Assign Oxidation States

Separate into half-reactions

ox: I-(aq) I2(aq)

red: MnO4-(aq) MnO2(s)

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Ex 18.3 – Balance the equation:I-(aq) + MnO4

-(aq) I2(aq) + MnO2(s) in basic

solution

Balance half-reactions by mass

ox: I-(aq) I2(aq)

red: MnO4-(aq) MnO2(s)

Balance half-reactions by mass

ox: 2 I-(aq) I2(aq)

red: MnO4-(aq) MnO2(s)

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)

red: 4 H+(aq) + MnO4

-(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)

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

(aq)

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Ex 18.3 – Balance the equation:I-(aq) + MnO4

-(aq) I2(aq) + MnO2(s) in basic

solution

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- } x 3

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

(aq) } x 2ox: 6 I-

(aq) 3 I2(aq) + 6 e-

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

(aq)

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Balance the equation:I-(aq) + MnO4

-(aq) I2(aq) + MnO2(s) in basic

solutionAdd the Half-reactions

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 Reactant

Count ElementProductCount

6 I 6

2 Mn 2

12 O 12

8 H 8

8- charge 8-

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Practice - Balance the Equation H2O2 + KI + H2SO4 K2SO4 + I2 +

H2O

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Practice - Balance the Equation H2O2 + KI + H2SO4 K2SO4 + I2 +

H2O+1 -1 +1 -1 +1 +6 -2 +1 +6 -2 0 +1 -2

oxidationreduction

ox: 2 I-1 I2 + 2e-1

red: H2O2 + 2e-1 + 2 H+ 2 H2Otot 2 I-1 + H2O2 + 2 H+ I2 + 2 H2O

1 H2O2 + 2 KI + H2SO4 K2SO4 + 1 I2 + 2 H2O

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Practice - Balance the EquationClO3

-1 + Cl-1 Cl2 (in acid)

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Practice - Balance the EquationClO3

-1 + Cl-1 Cl2 (in acid)

+5 -2 -1 0

oxidationreduction

ox: 2 Cl-1 Cl2 + 2 e-1 } x 5red: 2 ClO3

-1 + 10 e-1 + 12 H+ Cl2 + 6 H2O} x 1tot 10 Cl-1 + 2 ClO3

-1 + 12 H+ 6 Cl2 + 6 H2O

1 ClO3-1 + 5 Cl-1 + 6 H+1 3 Cl2

+ 3 H2O

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Standard Reduction Potential

a half-reaction with a strong tendency to occur has a large half-cell potential

when two half-cells are connected, the electrons will flow so that the half-reaction with the stronger tendency will occur

we cannot measure the absolute tendency of a half-reaction, we can only measure it relative to another half-reaction

we select as a standard half-reaction, the reduction of H+ to H2 under standard conditions, which we assign a potential difference = 0 V

standard hydrogen electrode, SHE

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Half-Cell Potentials

SHE reduction potential is defined to be exactly 0 V

half-reactions with a stronger tendency toward reduction than the SHE have a + value for E°red

half-reactions with a stronger tendency toward oxidation than the SHE have a - value for E°red

E°cell = E°oxidation + E°reduction

E°oxidation = -E°reduction

when adding E° values for the half-cells, do not multiply the half-cell E° values, even if you need to multiply the half-reactions to balance the equation

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42

Calculate E°cell for the reaction at 25°CAl(s) + NO3

−(aq) + 4 H+

(aq) Al3+(aq) + NO(g) + 2 H2O(l)

Separate the reaction into the oxidation and reduction half-reactions

ox: Al(s) Al3+(aq) + 3 e−

red: NO3−

(aq) + 4 H+(aq) + 3 e− NO(g) + 2 H2O(l)

find the Eo for each half-reaction and sum to get Eo

cell

Eoox = −Eo

red = +1.66 vEo

red = +0.96 vEo

cell = (+1.66 v) + (+0.96 v) = +2.62 v

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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: Mg2+(aq) + 2 e− Mg(s) -2.37V

red: Fe2+(aq) + 2 e− Fe(s) -0.44V

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

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the reaction is spontaneous in the reverse direction tot: 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

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Practice - Sketch and Label the Voltaic CellFe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s)

Write the Half-Reactions and Overall Reaction, and Determine the Cell Potential under Standard Conditions.

46

ox: Fe(s) Fe2+(aq) + 2 e− Eox = +0.45 V

red: Pb2+(aq) + 2 e− Pb(s) Ered = −0.13 V

tot: Pb2+(aq) + Fe(s) Fe2+(aq) + Pb(s) Ecell = +0.32 V

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Predicting Whether a Metal Will Dissolve in an Acid

acids dissolve in metals if the reduction of the metal ion is easier than the reduction of H+

(aq)

metals whose ion reduction reaction lies below H+ reduction on the table will dissolve in acid

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Electrochemical Cell Summary

A device for either harnessing the electrical power of a redox reaction (battery), or a device for using electricity to induce non-spontaneous redox reactions (electrolytic cell)

The cell consists of two half cells, one where oxidation occurs, and one where reduction occursEach half cell consists of an electrode and an electrolyteThe electrode which loses electrons we call the anode, and the electrode that gains electrons we call the cathode

The anode and cathode are connect by a wire through which electrons can flow, and the electrolytes in each half cell are connected by a salt bridge through which cations and anions can flow

salt bridgee-

anode

Zn (s)--> Zn2+ (aq)+ 2e-

cathode

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

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Electrochemical Cell Summary

salt bridgee-

anode

Zn (s)--> Zn2+ (aq)+ 2e-

cathode

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

The differing stability of reactants, (Zn(s), Cu2+(aq)), and products (Zn2+, and Cu(s)), creates a potential energy gradient through which the charges migrate (from high energy to low).

This manifests as a potential difference Ecell, across the electrodes. Where -qEcell is the change in potential energy when an amount of negative charge (-q) passes from the anode to the cathode

The cell potential is related to the free energy of the reaction according to the relation Gcell = -nFEcell

The cell potential can be calculated knowing the standard reduction potentials. These can be used to find Eo

red for the reaction at the cathode, and

Eoox (= - Eo

red). Then Eocell = Eo

ox+ Eored

Zn2+(aq) + 2e- --> Zn(s) -0.76V

Cu2+(aq) + 2e- --> Cu(s) Ered=0.34V

Zn(s) --> Zn2+(aq) + 2e- Eox= 0.76V

Ecell=1.1 VEcell = 0.76V+0.34V = 1.1V

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E°cell, ΔG° and K

for a spontaneous reaction one proceeds in the forward direction with the chemicals in their standard

states ΔG° < 1 (negative) E° > 1 (positive) K > 1

ΔG° = −RTlnK = −nFE°cell

n is the number of electrons F = Faraday’s Constant = 96,485 C/mol e−

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The Relationship between ΔGo and Eocell

Remember ΔGo is the maximum work the system can do Eo

cell is the cell potential energy (standard emf) per unit of charge

Since the potential energy is the maximum amount of work that can be done on the surrounding we can write

The total charge q = nF, n = number moles of electrons in the balanced equation and F is Faraday’s constant. Then we can write

52

Example 18.6- Calculate ΔG° for the reaction

I2(s) + 2 Br−(aq) → Br2(l) + 2 I−

(aq)

since ΔG° is +, the reaction is not spontaneous in the forward direction under standard conditions

Answer:

Solve:

Concept Plan:

Relationships:

I2(s) + 2 Br−(aq) → Br2(l) + 2 I−

(aq)

ΔG°, (J)

Given:Find:

E°ox, E°red E°cell ΔG°

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

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

tot: I2(l) + 2Br−(aq) → 2I−

(aq) + Br2(l) E° = −0.55 V

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Example 18.7- Calculate K at 25°C for the reaction

Cu(s) + 2 H+(aq) → H2(g) + Cu2+

(aq)

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

Answer:

Solve:

Concept 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

tot: Cu(s) + 2H+(aq) → Cu2+

(aq) + H2(g) E° = −0.34 V

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Nonstandard Conditions - the Nernst Equation

ΔG = ΔG° + RT ln Q

E = E° - (0.0592/n) log Q at 25°C

when Q = K, E = 0

use to calculate E when concentrations not 1 M

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E° at Nonstandard Conditions

56

Example 18.8- Calculate Ecell at 25°C for the reaction3 Cu(s) + 2 MnO4

−(aq) + 8 H+

(aq) → 2 MnO2(s) + 3Cu2+(aq) + 4

H2O(l)

units are correct, Ecell > E°cell as expected because [MnO4−] > 1 M

and [Cu2+] < 1 MCheck:

Solve:

Concept Plan:

Relationships:

3 Cu(s) + 2 MnO4−

(aq) + 8 H+(aq) → 2 MnO2(s) + 3Cu2+

(aq) + 4 H2O(l)

[Cu2+] = 0.010 M, [MnO4−] = 2.0 M, [H+] = 1.0 M

Ecell

Given:

Find:

E°ox, E°red E°cell Ecell

ox: Cu(s) → Cu2+(aq) + 2 e− }x3 E° = −0.34 V

red: MnO4−

(aq) + 4 H+(aq) + 3 e− → MnO2(s) + 2 H2O(l) }x2 E° = +1.68 V

tot: 3 Cu(s) + 2 MnO4−

(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+

(aq) + 4 H2O(l)) E° = +1.34 V

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Concentration Cells

it is possible to get a spontaneous reaction when the oxidation and reduction reactions are the same, as long as the electrolyte concentrations are different

the difference in energy is due to the entropic difference in the solutions the more concentrated solution has lower entropy than

the less concentrated

electrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution till the concentrations are equal

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when the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow

Concentration Cell

when the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode)

Cu(s)| Cu2+(aq) (0.010 M) || Cu2+

(aq) (2.0 M)| Cu(s)

59

LeClanche’ Acidic Dry Cell

electrolyte in paste form ZnCl2 + NH4Cl

or MgBr2

anode = Zn (or Mg)Zn(s) Zn2+(aq) + 2 e-

cathode = graphite rod

MnO2 is reduced2 MnO2(s) + 2 NH4

+(aq) + 2 H2O(l) + 2 e-

2 NH4OH(aq) + 2 Mn(O)OH(s) cell voltage = 1.5 V

expensive, nonrechargeable, heavy, easily corroded

60

Alkaline Dry Cell same basic cell as acidic dry cell, except

electrolyte is alkaline KOH paste

anode = Zn (or Mg)Zn(s) + 2OH- Zn(OH)2(s) + 2 e-

cathode = graphite

MnO2 is reduced

2 MnO2(s) + 2 H2O(l) + 2 e-

2 Mn(O)OH(s) + 2 OH-(aq) Overall reaction

Zn(s)+2MnO2(s)+2H2O(l) Zn(OH)2(s) + 2MnO(OH)(s)

cell voltage = 1.54 V longer shelf life than acidic dry cells and

rechargeable, little corrosion of zinc

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Lead Storage Battery

6 cells in series

electrolyte = 30% H2SO4

anode = PbPb(s) + SO4

2-(aq) PbSO4(s) + 2 e-

cathode = Pb coated with PbO2

PbO2 is reduced

PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e-

PbSO4(s) + 2 H2O(l)

cell voltage = 2.09 V rechargeable, heavy

Tro, Chemistry: A Molecular Approach

62

NiCad Battery

electrolyte is concentrated KOH solution

anode = CdCd(s) + 2 OH-1(aq) ® Cd(OH)2(s) + 2 e-1 E0 = 0.81 V

cathode = Ni coated with NiO2

NiO2 is reducedNiO2(s) + 2 H2O(l) + 2 e-1 ® Ni(OH)2(s) + 2OH-1 E0

= 0.49 V

cell voltage = 1.30 V

rechargeable, long life, light – however recharging incorrectly can lead to battery breakdown

Tro, Chemistry: A Molecular Approach

63

Ni-MH Battery

electrolyte is concentrated KOH solution

anode = metal alloy with dissolved hydrogen oxidation of H from H0 to H+1

M∙H(s) + OH-1(aq) ® M(s) + H2O(l) + e-1 E° = 0.89 V

cathode = Ni coated with NiO2

NiO2 is reducedNiO2(s) + 2 H2O(l) + 2 e-1 ® Ni(OH)2(s) + 2OH-1 E0 = 0.49

V cell voltage = 1.30 V

rechargeable, long life, light, more environmentally friendly than NiCad, greater energy density than NiCad

64

Lithium Ion Battery

electrolyte is concentrated KOH solution

anode = graphite impregnated with Li ions

cathode = Li - transition metal oxide

work on Li ion migration from anode to cathode causing a corresponding migration of electrons from anode to cathode

rechargeable, long life, very light, more environmentally friendly, greater energy density

65

66

Fuel Cells

like batteries in which reactants are constantly being added so it never runs down!

Anode and Cathode both Pt coated metal

Electrolyte is OH– solution

Anode Reaction: 2 H2 + 4 OH–

→ 4 H2O(l) + 4 e-

Cathode Reaction: O2 + 4 H2O + 4 e-

→ 4 OH–

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Electrolytic Cell

uses electrical energy to overcome the energy barrier and cause a non-spontaneous reaction must be DC source

the + terminal of the battery = anode

the - terminal of the battery = cathode

cations attracted to the cathode, anions to the anode

cations pick up electrons from the cathode and are reduced, anions release electrons to the anode and are oxidized

some electrolysis reactions require more voltage than Etot, called the overvoltage

68

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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

70

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

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Electrolysis

electrolysis is the process of using electricity to break a compound apart

electrolysis is done in an electrolytic cell

electrolytic cells can be used to separate elements from their compounds generate H2 from water for fuel cells recover metals from their ores

72

Electrolysis of Water

73

Electrolysis of Pure Compounds

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

74

Electrolysis of NaCl(l)

75

Mixtures of Ions

when more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode least negative or most positive E°red

when more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode least negative or most positive E°ox

76

Electrolysis of Aqueous Solutions

Complicated by more than one possible oxidation and reduction possible cathode reactions

reduction of cation to metal reduction of water to H2

2 H2O + 2 e-1 ® H2 + 2 OH-1 E° = -0.83 V @ stand. cond.

E° = -0.41 V @ pH 7 possible anode reactions

oxidation of anion to element oxidation of H2O to O2

2 H2O ® O2 + 4e-1 + 4H+1 E° = -1.23 V @ stand. cond.

E° = -0.82 V @ pH 7 oxidation of electrode

particularly Cu graphite doesn’t oxidize

half-reactions that lead to least negative Etot will occur unless overvoltage changes the conditions

77

Electrolysis of NaI(aq) with Inert Electrodes

possible oxidations2 I-1 I2 + 2 e-1 E° = −0.54 v2 H2O O2 + 4e-1 + 4H+1 E° = −0.82 v

possible reductionsNa+1 + 1e-1 Na0 E° = −2.71 v2 H2O + 2 e-1 H2 + 2 OH-1 E° = −0.41 v

possible oxidations2 I-1 I2 + 2 e-1 E° = −0.54 v2 H2O O2 + 4e-1 + 4H+1 E° = −0.82 v

possible reductionsNa+1 + 1e-1 Na0 E° = −2.71 v2 H2O + 2 e-1 H2 + 2 OH-1 E° = −0.41 v

overall reaction2 I−

(aq) + 2 H2O(l) I2(aq) + H2(g) + 2 OH-1(aq)

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Faraday’s Law

the amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current, and the length of time the cell runs charge that flows through the cell =

current x time

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Example 18.10- 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 since 10 A running for 1 hr ~ 1/3 mol e−

Check:

Solve:

Concept 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

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Corrosion

corrosion is the spontaneous oxidation of a metal by chemicals in the environment

since many materials we use are active metals, corrosion can be a very big problem

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Rusting

rust is hydrated iron(III) oxide moisture must be present

water is a reactant required for flow between cathode

and anode electrolytes promote rusting

enhances current flow acids promote rusting

lower pH = lower E°red

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Preventing Corrosion

one way to reduce or slow corrosion is to coat the metal surface to keep it from contacting corrosive chemicals in the environment paint some metals, like Al, form an oxide that

strongly attaches to the metal surface, preventing the rest from corroding

another method to protect one metal is to attach it to a more reactive metal that is cheap sacrificial electrode

galvanized nails

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Sacrificial Anode

Zn Zn2+ + 2e- 0.76V

Fe Fe2+ + 2e- 0.45V

The Final

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Long Questions on the Following:

• Kinetics • Method of Initial Rates to find rate law• Arrhenius plot to find Activation energy Ea and the

collision factor A

• Acid-Base Titration of a Weak Acid• Calculate the initial pH of the acid given Ka

• Calculate the pH in the buffer region using Henderson-Hasselbalch

• Calculate the pH at the end point using Kb

• Galvanic and Electrolytic Cell

• An extra credit question on kinetics mechanisms (worth 15% of the final)

Multiple Choice Redo for Electrochemistry

• A short (optional multiple choice on electrochemistry for those who want a second chance)