Page 1
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)
Page 2
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)
Page 3
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
Page 4
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)
Page 5
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)
Page 6
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)
Page 7
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
Page 8
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
Page 9
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
Page 10
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)
Page 11
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)
Page 12
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
Page 13
USES FOR STANDARD REDUCTION POTENTIALS
3) Predicting the potential and spontaneous reaction in a Galvanic cell
3D-13 (of 16)
Page 14
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)
Page 15
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)
Page 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)
Page 18
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)
Page 19
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
Page 20
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
Page 21
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)
Page 22
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
Page 23
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)
Page 24
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+
]
Page 25
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)
Page 26
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)
Page 27
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)
Page 28
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)
Page 29
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)
Page 30
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)
Page 32
BATTERIES
BATTERY – One or more electrochemical cells that produce electricity from a chemical reaction
3F-1 (of 16)
Page 33
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)
Page 34
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)
Page 35
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)
Page 36
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)
Page 37
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)
Page 38
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)
Page 39
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)
Page 40
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
Page 41
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
Page 42
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+
Page 43
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)
Page 44
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)
Page 45
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)
Page 46
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)
Page 47
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)