ELECTROCHEMISTRY

CHARGE (Q) – A property of matter which causes it to experience the electromagnetic force

COULOMB (C) – The quantity of charge equal to 6.241 × 1018

electrons

ELECTROMOTIVE FORCE or POTENTIAL or VOLTAGE (Ɛ) – The potential difference between 2 substances, causing electrons to

flow from one to the other

VOLT (V) – One joule of potential energy per coulomb

ELECTRIC CURRENT or AMPERAGE (I) – The rate of flow of electric charge

AMPERE (A) – Flow rate of one coulomb of electric charge per second

3D-1 (of 16)

Spontaneous oxidation-reduction reaction:

Fe (s) + Cu2+

(aq) → Fe2+

(aq) + Cu (s)

Redox reactions can be written as the sum of 2 half-reactions

Oxidation:

Reduction:

Fe (s) → Fe2+

(aq)

Cu2+

(aq) → Cu (s)

If the Fe (s) and Cu2+

(aq) are separated, the electron transfer can happen through a wire

3D-2 (of 16)

Iron is more reactive then copper, iron atoms will release their valence

electrons to the copper (II) ions

+ 2e-

2e- +

Fe (s) + 2e- + Cu

2+ (aq) → Fe

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

-

Fe (s) + Cu2+

(aq) → Fe2+

(aq) + Cu (s)

ANODE – The electrode where oxidation occurs

CATHODE – The electrode where reduction occurs

0.78 V is called the cell potential, the cell voltage, or the cell emf

3D-3 (of 16)

Cu

GALVANIC CELL – An electrochemical cell that produces electric current from a chemical reaction

Shorthand notation:

Anode | Anode Solution || Cathode Solution | Cathode

Fe | Fe2+

(1 M) || Cu2+

(1 M) | Cu

3D-4 (of 16)

The ΔG of a reaction occurring in a Galvanic cell is related to Ɛ

ΔG = -nFƐ

n = number of moles of electrons transferred in the redox reaction

F = the Faraday constant

the charge of 1 mole of electrons, equal to 96,485 C

For Galvanic cells with 1 M concentrations

ΔGº = -nFƐº

J = (mol) (C/mol) (V) (J/C)

3D-5 (of 16)

Calculate ΔGº for the reaction

Fe (s) + Cu2+

(aq) → Fe2+

(aq) + Cu (s) Ɛº = 0.78 V

ΔGº = -nFƐº

= -(2 mol)(96,485 C/mol)(0.78 V)

= -(2 mol)(96,485 C/mol)(0.78 J/C) = -150,000 J

ΔG < 0 and Ɛ > 0 means a spontaneous process

The more negative ΔG, the more spontaneous the process

The more positive Ɛ, the more spontaneous the process

3D-6 (of 16)

REDUCTION AND OXIDATION POTENTIALS

The sum of a reduction potential and an oxidation potential must equal the potential for the overall redox reaction

3D-7 (of 16)

REDUCTION POTENTIAL (Ɛred) – The electric potential for a reduction half-reaction

OXIDATION POTENTIAL (Ɛox) – The electric potential for an oxidation half-reaction

Ɛºred and Ɛºox are for a standard state reactions

The more positive the Ɛred or Ɛox, the more spontaneous the reaction

The potential of an overall redox reaction in a Galvanic cell can be measured with a voltmeter

Unfortunately, the potential of a half-reaction cannot be measured,

so we make one up!

This standard hydrogen half-reaction is assigned a potential of 0.00 V

3D-8 (of 16)

H2 (g, 1 atm) → 2H+

(aq, 1 M) + 2e-

All other standard reduction potentials are measured relative to this one

Fe (s) → Fe

2+ (aq) + 2e

-

3e- + Cr

3+ (aq) → Cr (s)

USES FOR STANDARD REDUCTION POTENTIALS

1) Predicting the spontaneity of a reaction

Determine if the following standard state reaction is spontaneous

3Fe (s) + 2Cr

3+ (aq) → 3Fe

2+ (aq) + 2Cr

(s)

Find the 2 reduction potentials that can be used to make the reaction

2e- + Fe

2+ (aq) → Fe (s)

3e- + Cr

3+ (aq) → Cr (s)

Ɛºred = -0.44 V

Ɛºred = -0.73 V

Add a reduction and oxidation half-reaction to make the desired reaction

3D-9 (of 16)

Ɛºox = 0.44 V

Ɛºred = -0.73 V

Ɛº = -0.29 V

Ɛº is negative, not spontaneous

Fe (s) → Fe

2+ (aq) + 2e

-

3e- + Cr

3+ (aq) → Cr (s)

USES FOR STANDARD REDUCTION POTENTIALS

1) Predicting the spontaneity of a reaction

( ) x 2

( ) x 3

Determine if the following standard state reaction is spontaneous

3Fe (s) + 2Cr

3+ (aq) → 3Fe

2+ (aq) + 2Cr

(s)

3D-10 (of 16)

Ɛºox = 0.44 V

Ɛºred = -0.73 V

Ɛº = -0.29 V

Ɛº is negative, not spontaneous

3Fe (s) + 2Cr3+

(aq) → 3Fe2+

(aq) + 2Cr (s)

USES FOR STANDARD REDUCTION POTENTIALS

2) Predicting strong oxidizing and reducing agents

e- + Ag

+ (aq) → Ag (s)

2e- + Cu

2+ (aq) → Cu (s)

2e- + Ni

2+ (aq) → Ni (s)

3e- + Al

3+ (aq) → Al (s)

Reduction Half-Reactions Stand. Reduction Potentials

Ɛºred = 0.80 V

Ɛºred = 0.34 V

Ɛºred = -0.23 V

Ɛºred = -1.71 V

A large, positive reduction potential means the forward reaction is spontaneous (the REACTANT has a strong tendency to be

REDUCED)

Best oxidizing agent from the list? Ag+

(aq)

Good oxidizing agents? Halogens (X2) → X-

O2 → H2O

3D-11 (of 16)

USES FOR STANDARD REDUCTION POTENTIALS

2) Predicting strong oxidizing and reducing agents

e- + Ag

+ (aq) → Ag (s)

2e- + Cu

2+ (aq) → Cu (s)

2e- + Ni

2+ (aq) → Ni (s)

3e- + Al

3+ (aq) → Al (s)

Reduction Half-Reactions Stand. Reduction Potentials

Ɛºred = 0.80 V

Ɛºred = 0.34 V

Ɛºred = -0.23 V

Ɛºred = -1.71 V

A large, negative reduction potential means the reverse reaction is spontaneous (the PRODUCT has a strong tendency to be

OXIDIZED)

3D-12 (of 16)

Best reducing agent from the list? Al (s)

Good reducing agents? (Alkali Metals) M → M+

C → CO2

USES FOR STANDARD REDUCTION POTENTIALS

3) Predicting the potential and spontaneous reaction in a Galvanic cell

3D-13 (of 16)

For a Galvanic cell with silver and nickel electrodes in 1 M solutions of Ag+

and Ni2+

respectively, determine the (a) standard

cell potential, (b) spontaneous reaction, and (c) anode and cathode

(a) Find the 2 reduction potentials to produce the Galvanic cell

e- + Ag

+ (aq) → Ag (s)

2e- + Ni

2+ (aq) → Ni (s)

Ɛºred = 0.80 V

Ɛºred = -0.23 V

The largest positive potential is the spontaneous process, and will be the reduction

the other must be reversed, and will be the oxidation

e- + Ag

+ (aq) → Ag (s)

Ni (s) → Ni

2+ (aq) + 2e

-

Ɛºred = 0.80 V

Ɛºox = 0.23 V

Add the reduction and oxidation potentials to get the cell potential

0.80 V + 0.23 V = 1.03 V

3D-14 (of 16)

For a Galvanic cell with silver and nickel electrodes in 1 M solutions of Ag+

and Ni2+

respectively, determine the (a) standard

cell potential, (b) spontaneous reaction, and (c) anode and cathode

(b) Equalize e-s and add the reduction and oxidation half-reactions together

e- + Ag

+ (aq) → Ag (s)

Ni (s) → Ni

2+ (aq) + 2e

-

( ) x 2

2e- + 2Ag

+ (aq) → 2Ag (s)

Ni (s) → Ni

2+ (aq) + 2e

-

2Ag+

(aq) + Ni (s) → 2Ag

(s) + Ni

2+ (aq)

3D-15 (of 16)

For a Galvanic cell with silver and nickel electrodes in 1 M solutions of Ag+

and Ni2+

respectively, determine the (a) standard

cell potential, (b) spontaneous reaction, and (c) anode and cathode

2Ag+

(aq) + Ni (s) → 2Ag

(s) + Ni

2+ (aq)(c)

Ag+

is reduced ∴ Ag is the cathode

Ni is oxidized ∴ Ni is the anode

3D-16 (of 16)

For a Galvanic cell with iron and chromium electrodes in 1 M solutions of Fe2+

and Cr3+

respectively, determine the (a)

standard cell potential, (b) spontaneous reaction, and (c) anode and cathode

2e- + Fe

2+ (aq) → Fe (s)

3e- + Cr

3+ (aq) → Cr (s)

Ɛºred = -0.44 V

Ɛºred = -0.52 V

2e- + Fe

2+ (aq) → Fe (s)

Cr (s) → Cr

3+ (aq) + 3e

-

Ɛºred = -0.44 V

Ɛºox = 0.52 V

-0.44 V + 0.52 V = 0.08 V

3E-1 (of 13)

For a Galvanic cell with iron and chromium electrodes in 1 M solutions of Fe2+

and Cr3+

respectively, determine the (a)

standard cell potential, (b) spontaneous reaction, and (c) anode and cathode

2e- + Fe

2+ (aq) → Fe (s)

Cr (s) → Cr

3+ (aq) + 3e

-

( ) x 3

6e- + 3Fe

2+ (aq) → 3Fe (s)

2Cr (s) → 2Cr

3+ (aq) + 6e

-

3Fe2+

(aq) + 3Cr (s) → 3Fe

(s) + 2Cr

3+ (aq)

( ) x 2

3E-2 (of 13)

Fe2+

is reduced ∴ Fe is the cathode

Cr is oxidized ∴ Cr is the anode

For nonstandard cells

ΔG = ΔGº + RT ln Q

-nFƐ = -nFƐº + RT ln Q

Ɛ = Ɛº – RT ln Q

____

nF

THE NERNST EQUATION

3E-3 (of 13)

NONSTANDARD STATE CELLS

Calculate the potential for the following cell at 25ºC

Zn (s) | Zn2+

(0.200 M) || Ag+

(0.100 M) | Ag (s)

e- + Ag

+ (aq) → Ag (s)

2e- + Zn

2+ (aq) → Zn (s)

Ɛºred = 0.80 V

Ɛºred = -0.76 V

0.80 V + 0.76 V = 1.56 V

e- + Ag

+ (aq) → Ag (s)

Zn (s) → Zn

2+ (aq) + 2e

-

Ɛºred = 0.80 V

Ɛºox = 0.76 V

3E-4 (of 13)

e- + Ag

+ (aq) → Ag (s)

Zn (s) → Zn

2+ (aq) + 2e

-

( ) x 2

Calculate the potential for the following cell at 25ºC

Zn (s) | Zn2+

(0.200 M) || Ag+

(0.100 M) | Ag (s)

2Ag+

(aq) + Zn (s) → 2Ag

(s) + Zn

2+ (aq)

Ɛ = Ɛº – RT ln Q

____

nF

= 1.56 V – (8.314 CV/K)(298.2 K) ln 0.200

____________________________ ________

(2 mol)(96,485 C/mol) 0.1002

= 1.56 V – 0.0385 V = 1.52 V

3E-5 (of 13)

Q = [Zn2+

]

_________

[Ag+

]2

For a nickel-cadmium cell with solutions of 0.00100 M nickel (II) sulfate and 0.10 M cadmium sulfate, determine the (a) standard

cell potential,

(b) spontaneous reaction, (c) anode and cathode (d) cell potential at 25ºC

2e- + Ni

2+ (aq) → Ni (s)

2e- + Cd

2+ (aq) → Cd (s)

Ɛºred = -0.23 V

Ɛºred = -0.40 V

-0.23 V + 0.40 V = 0.17 V

2e- + Ni

2+ (aq) → Ni (s)

Cd (s) → Cd

2+ (aq) + 2e

-

Ɛºred = -0.23 V

Ɛºox = 0.40 V

3E-6 (of 13)

For a nickel-cadmium cell with solutions of 0.00100 M nickel (II) sulfate and 0.10 M cadmium sulfate, determine the (a) standard

cell potential,

(b) spontaneous reaction, (c) anode and cathode (d) cell potential at 25ºC

2e- + Ni

2+ (aq) → Ni(s)

Cd (s) → Cd

2+ (aq) + 2e

-

Ni2+

(aq) + Cd (s) → Ni

(s) + Cd

2+ (aq)

Ni – cathode

Cd – anode

Ɛ = Ɛº – RT ln Q

____

nF

= 0.17 V – (8.314 CV/K)(298.2 K) ln 0.10

____________________________ __________

(2 mol)(96,485 C/mol) 0.00100

= 0.17 V – 0.059 V = 0.11 V

3E-7 (of 13)

Q = [Cd2+

]

_________

[Ni2+

]

Ɛ = Ɛº – RT ln Q

____

nF

For a reaction at equilibrium

ΔG = 0 ∴ Ɛ = 0

0 = Ɛº – RT ln Keq

____

nF

RT ln Keq = Ɛº

____

nF

Keq = eƐºnF/RT

3E-8 (of 13)

Find the equilibrium constant at 25ºC for

Fe (s) + I2 (s) → Fe

2+ (aq) + 2I

- (aq)

2e- + I2

(s) → 2I

- (aq)

2e- + Fe

2+ (aq) → Fe (s)

Ɛºred = 0.54 V

Ɛºred = -0.41 V

0.54 V + 0.41 V = 0.95 V

2e- + I2

(s) → 2I

- (aq)

Fe (s) → Fe

2+ (aq) + 2e

-

Ɛºred = 0.54 V

Ɛºox = 0.41 V

3E-9 (of 13)

I2 (s) + Fe (s) → 2I

- (aq) + Fe

2+ (aq)

Find the equilibrium constant at 25ºC for

Fe (s) + I2 (s) → Fe

2+ (aq) + 2I

- (aq)

Keq = eƐºnF/RT

= e[(0.95 V)(2 mol)(96,485 C/mol)]

/

[(8.314 CV/K)(298.2 K)]

= 1.3 x 1032

3E-10 (of 13)

Find the solubility product constant at 25ºC for

Hg2Cl2 (s) ⇄ Hg22+

(aq) + 2Cl-

(aq)

2e- + Hg2

2+ (aq) → 2Hg (l)

2e- + Cl2

(g) → 2Cl

- (aq)

Ɛºred = 0.80 V

Ɛºred = 1.36 V

2Hg (l) → Hg2

2+ (aq) + 2e

-

2e- + Cl2

(g) → 2Cl

- (aq)

Ɛºox = -0.80 V

Ɛºred = 1.36 V

2Hg (l) + Cl2 (g) → Hg2

2+ (aq) + 2Cl

- (aq)

3E-11 (of 13)

Find the solubility product constant at 25ºC for

Hg2Cl2 (s) ⇄ Hg22+

(aq) + 2Cl-

(aq)

2e- + Hg2Cl2

(s) → 2Hg (l) + 2Cl

- (aq)

2e- + Hg2

2+ (aq) → 2Hg (l)

Ɛºred = 0.27 V

Ɛºred = 0.80 V

2e- + Hg2Cl2

(s) → 2Hg (l) + 2Cl

- (aq)

2Hg (l) → Hg2

2+ (aq) + 2e

-

Ɛºred = 0.27 V

Ɛºox = -0.80 V

Hg2Cl2 (s) ⇄ Hg22+

(aq) + 2Cl-

(aq)

0.27 V – 0.80 V = -0.53 V

3E-12 (of 13)

Ksp = eƐºnF/RT

= e[(-0.53 V)(2 mol)(96,485 C/mol)]

/

[(8.314 CV/K)(298.2 K)]

= 1.2 x 10-18

Find the solubility product constant at 25ºC for

Hg2Cl2 (s) ⇄ Hg22+

(aq) + 2Cl-

(aq)

3E-13 (of 13)

BATTERIES

BATTERY – One or more electrochemical cells that produce electricity from a chemical reaction

3F-1 (of 16)

Dry Cell

Graphite rod (cathode)

Paste of MnO2, NH4Cl, and H2O

Zinc casing (anode)

Anode:

Cathode:

Zn (s) → Zn2+

(aq) + 2e-

2e- + 2MnO2

(s) + 8NH4

+ (aq) → 2Mn

3+ (aq) + 4H2O (l) + 8NH3 (aq)

Potential or Voltage: 1.5 V Current or Amperage: Depends on battery size

3F-2 (of 16)

Dry Cell

The acidic content tends to corrode the Zn

Fast usage : NH3 insulates the cathode, reducing the voltage

With rest : Zn2+

migrates to center, forming Zn(NH3)42+

to bind the NH3

Graphite rod (cathode)

Paste of MnO2, NH4Cl, and H2O

Zinc casing (anode)

3F-3 (of 16)

Alkaline Battery

Graphite rod (cathode)

Paste of MnO2, KOH, and H2O

Powdered zinc (anode)

Anode:

Cathode:

Zn (s) + OH- (aq) → ZnO (s) + H2O (l) + 2e

-

2e- + 2MnO2

(s) + H2O (l) → 2Mn2O3 (s) + 2OH

- (aq)

Zn resists corrosion in a basic solution

3F-4 (of 16)

Lithium-Ion Battery

Lithium cobalt oxide (anode)

Graphite (cathode)

Anode:

Cathode:

LiCoO2 (s) → CoO2 (s) + Li+

(org) + e-

e- + Li

+ (org) + 6C (s) → LiC6 (s)

Because both products stick to the electrodes, by applying an external source of electricity the reverse reaction will occur,

reforming the reactants

This is called RECHARGING

Separator

3F-5 (of 16)

Lead Storage Battery

Lead (anode)

Lead + lead (IV) oxide (cathode)

4 M sulfuric acid

Anode:

Cathode:

Pb (s) + SO42-

(aq) → PbSO4 (s) + 2e-

2e- + 2PbO2

(s) + 4H

+ (aq) + SO4

2- (aq) → PbSO4 (s) + 2H2O (l)

3F-6 (of 16)

Lead Storage Battery

Potential or Voltage: 2.1 V x 6 cells = 12.6 V

_______

cell

Because products stick to the electrodes, this battery is rechargeable

Lead (anode)

Lead + lead (IV) oxide (cathode)

4 M sulfuric acid

3F-7 (of 16)

Hydrogen Fuel Cell

Platinum Catalyst

Polymer Electrolyte Membrane

Anode: Pt catalyst splits hydrogen atoms into hydrogen ions and electrons

Electrolyte: PEM allows hydrogen ions to pass through to the cathode

Cathode: Oxygen and electrons combine with hydrogen ions to make water

3F-8 (of 16)

ELECTROLYTIC CELL – An electrochemical cell that uses electricity to

produce a chemical reaction

H2O

Anode: Oxidation

H2O (l) → O2 (g)

+1 -2

Cathode: Reduction

H2O (l) → H2 (g)

2 + 4e-

+ 4H+

(aq)

2e- + + OH

- (aq)

3F-9 (of 16)

2 2

Electricity through Anode Cathode

KF (l)

NaCl (l)

NaCl (aq)

KF (aq)

CuBr2 (aq)

HCl (aq)

HNO3 (aq)

H2SO4 (aq)

Na2SO4 (aq)

AgNO3 (aq)

F2 (g)

Cl2 (g)

Cl2 (g)

F2 (g)

Br2 (l)

Cl2 (g)

O2 (g)

O2 (g)

O2 (g)

O2 (g)

K (s)

Na (s)

Na (s)

H2 (g)

Cu (s)

H2 (g)

H2 (g)

H2 (g)

H2 (g)

Ag (s)

H2 (g)

N in HNO3 cannot be oxidized

3F-10 (of 16)

, so O in H2O will be oxidized

Na+

(aq) + e- → Na

(s)

2H2O (l) + 2e- → H2

(g) + 2OH

- (aq)

Ɛºred = -2.71 V

Ɛºred = -0.83 V

Ag

Ag+

Electrolytic cells are used for

(1) producing elements (Na, Cl2, etc.)

(2) purification of metals from ore

(3) electroplating metals (Au, Ag, Pt, etc.)

Anode: Oxidation

Ag (s) → Ag+

(aq)

Cathode: Reduction

Ag+

(aq) → Ag (s)

+ e-

e- +

3F-11 (of 16)

Ag+

FARADAY’S LAWS OF ELECTROLYSIS

(1) Passing the same quantity of electricity through a cell always leads to the same amount of chemical change

(2) It takes 96,485 C of electricity to deposit or liberate 1 mole of a substance that gains or loses 1 e- during the cell reaction

96,485 C = 1 mole e-

= 1 Faraday

3F-12 (of 16)

Calculate the mass of copper deposited by a current of 7.89 amperes flowing for 1.20 x 103 seconds, if the cathode reaction is

Cu2+

(aq) + 2e- ⇄ Cu

(s)

7.89 A x 1 C

______

1 As

x 1.20 x 103 s

x 1 F

___________

96,485 C

x 1 mol e-

__________

1 F

x 1 mol Cu

___________

2 mol e-

x 63.55 g Cu

______________

1 mol Cu

= 3.12 g Cu

3F-13 (of 16)

Calculate the mass of aluminum deposited by a current of 5.00 amperes flowing for 10.0 minutes through an aluminum nitrate

solution.

Al3+

(aq) + 3e- ⇄ Al

(s)

5.00 A x 1 C

______

1 As

x 10.0 min

x 1 F

___________

96,485 C

x 1 mol e-

__________

1 F

x 1 mol Al

___________

3 mol e-

x 26.98 g Al

_____________

1 mol Al

= 0.280 g Al

x 60 s

________

1 min

3F-14 (of 16)

Calculate the current needed to plate 0.150 grams of zinc onto an electrode in 60.0 seconds from a zinc acetate solution.

Zn2+

(aq) + 2e- ⇄ Zn

(s)

0.150 g Zn x 2 mol e-

___________

1 mol Zn

x 1 F

__________

1 mol e-

x 96,485 C

___________

1 F

x 1 As

______

1 C

x 1

________

60.0 s

= 7.38 A

x mol Zn

_____________

65.38 g Zn

3F-15 (of 16)

Calculate the time, in minutes, needed to deposit 0.400 grams of chromium from a chromium (III) nitrate solution with a current of

10.0 amperes.

Cr3+

(aq) + 3e- ⇄ Cr

(s)

0.400 g Cr x 3 mol e-

___________

1 mol Cr

x 1 F

__________

1 mol e-

x 96,485 C

___________

1 F

x 1 As

______

1 C

x 1

________

10.0 A

= 3.71 min

x mol Cr

_____________

52.00 g Cr

x 1 min

_______

60 s

3F-16 (of 16)