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
ELECTRIC CURRENT or AMPERAGE (I) – The rate of flow of electric charge
AMPERE (A) – Flow rate of one coulomb of electric charge per second
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
3D-1 (of 14)
Spontaneous oxidation-reduction reaction:
Fe (s) + Cu2+ (aq) → Fe2+ (aq) + Cu (s)
Iron is more reactive then copper,  iron
atoms will release their valence electrons
to the copper (II) ions
Redox reactions can be written as the sum of 2 half-reactions
Fe (s) → Fe2+ (aq) + 2e-
Oxidation:
2e- + Cu2+ (aq) → Cu (s)
Reduction:
Fe (s) + 2e- + Cu2+ (aq) → Fe2+ (aq) + Cu (s) + 2e-
Fe (s) + Cu2+ (aq) → Fe2+ (aq) + Cu (s)
If the Fe (s) and Cu2+ (aq) are separated, the electron transfer can happen
through a wire
3D-2 (of 14)
Cu
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 14)
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 14)
The ΔG of a reaction occurring in a Galvanic cell is related to Ɛ
ΔG = -nFƐ
n
F
= number of moles of electrons transferred in the redox reaction
= 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 14)
Calculate ΔGº for the reaction
Fe (s) + Cu2+ (aq) → Fe2+ (aq) + Cu (s)
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 14)
Ɛº = 0.78 V
REDUCTION AND OXIDATION POTENTIALS
REDUCTION POTENTIAL (Ɛred) – The electric potential for a reduction halfreaction
OXIDATION POTENTIAL (Ɛox) – The electric potential for an oxidation halfreaction
The more positive the Ɛred or Ɛox, the more spontaneous the half-reaction
Ɛºred and Ɛºox are for a standard state half-reactions
The sum of a reduction potential and an oxidation potential must equal the
potential for the overall redox reaction
3D-7 (of 14)
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!
H2 (g, 1 atm) → 2H+ (aq, 1 M) + 2eThis standard hydrogen half-reaction is assigned a potential of 0.00 V
All other standard reduction potentials are measured relative to this one
3D-8 (of 14)
USES FOR STANDARD REDUCTION POTENTIALS
1) Predicting the spontaneity of a reaction
Determine if the following standard state reaction is spontaneous
3Fe (s) + 2Cr3+ (aq) → 3Fe2+ (aq) + 2Cr (s)
Find the 2 reduction potentials that can be used to make the reaction
2e- + Fe2+ (aq) → Fe (s)
3e- + Cr3+ (aq) → Cr (s)
Ɛºred = -0.44 V
Ɛºred = -0.73 V
Add a reduction and oxidation half-reaction to make the desired reaction
Fe (s) → Fe2+ (aq) + 2e3e- + Cr3+ (aq) → Cr (s)
Ɛºox = 0.44 V
Ɛºred = -0.73 V
Ɛº
= -0.29 V
Ɛº is negative,  not spontaneous
3D-9 (of 14)
USES FOR STANDARD REDUCTION POTENTIALS
2) Predicting strong oxidizing and reducing agents
Reduction Half-Reactions
e2e2e3e-
+
+
+
+
Ag+ (aq)
Cu2+ (aq)
Ni2+ (aq)
Al3+ (aq)
→
→
→
→
Ag (s)
Cu (s)
Ni (s)
Al (s)
Stand. Reduction Potentials
Ɛºred
Ɛºred
Ɛºred
Ɛºred
=
=
=
=
0.80
0.34
-0.23
-1.71
V
V
V
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) → XO2
→ H2O
3D-10 (of 14)
USES FOR STANDARD REDUCTION POTENTIALS
2) Predicting strong oxidizing and reducing agents
Reduction Half-Reactions
e2e2e3e-
+
+
+
+
Ag+ (aq)
Cu2+ (aq)
Ni2+ (aq)
Al3+ (aq)
→
→
→
→
Stand. Reduction Potentials
Ag (s)
Cu (s)
Ni (s)
Al (s)
Ɛºred
Ɛºred
Ɛºred
Ɛºred
=
=
=
=
0.80
0.34
-0.23
-1.71
V
V
V
V
A large, negative reduction potential means the reverse reaction is
spontaneous (the PRODUCT has a strong tendency to be OXIDIZED)
Best reducing agent from the list?
Al (s)
Good reducing agents?
(Alkali Metals) M → M+
C
→ CO2
3D-11 (of 14)
USES FOR STANDARD REDUCTION POTENTIALS
3) Predicting the potential and spontaneous reaction in a Galvanic cell
3D-12 (of 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- + Ni2+ (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) → Ni2+ (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-13 (of 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
(b) Equalize e-s and add the reduction and oxidation half-reactions together
( e- + Ag+ (aq) → Ag (s) ) x 2
Ni (s) → Ni2+ (aq) + 2e2e- + 2Ag+ (aq) → 2Ag (s)
Ni (s) → Ni2+ (aq) + 2e2Ag+ (aq) + Ni (s) → 2Ag (s) + Ni2+ (aq)
(c)
Nickel half-reaction is the oxidation, ∴ Ni is the anode
Silver half-reaction is the reduction, ∴ Ag is the cathode
3D-14 (of 14)
NONSTANDARD STATE CELLS
For nonstandard cells
ΔG = ΔGº + RT ln Q
-nFƐ = -nFƐº + RT ln Q
THE NERNST EQUATION
Ɛ = Ɛº – RT ln Q
____
nF
3E-1 (of 11)
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- + Zn2+ (aq) → Zn (s)
e- + Ag+ (aq) → Ag (s)
Zn (s) → Zn2+ (aq) + 2e0.80 V + 0.76 V = 1.56 V
3E-2 (of 11)
Ɛºred = 0.80 V
Ɛºred = -0.76 V
Ɛºred = 0.80 V
Ɛºox = 0.76 V
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) ) x 2
Zn (s) → Zn2+ (aq) + 2e2Ag+ (aq) + Zn (s) → 2Ag (s) + Zn2+ (aq)
Q = [Zn2+]
_________
[Ag+]2
Ɛ = Ɛº – RT ln Q = 1.56 V – (8.314 CV/K)(298.2 K) ln
0.200
____
____________________________
________
nF
(2 mol)(96,485 C/mol)
0.1002
= 1.56 V – 0.0385 V = 1.52 V
3E-3 (of 11)
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- + Ni2+ (aq) → Ni (s)
2e- + Cd2+ (aq) → Cd (s)
Ɛºred = -0.23 V
Ɛºred = -0.40 V
2e- + Ni2+ (aq) → Ni (s)
Cd (s) → Cd2+ (aq) + 2e-
Ɛºred = -0.23 V
Ɛºox = 0.40 V
-0.23 V + 0.40 V = 0.17 V
3E-4 (of 11)
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- + Ni2+ (aq) → Ni(s)
Cd (s) → Cd2+ (aq) + 2e-
Ni – cathode
Cd – anode
Ni2+ (aq) + Cd (s) → Ni (s) + Cd2+ (aq)
Q = [Cd2+]
_________
[Ni2+]
Ɛ = Ɛº – RT ln Q = 0.17 V – (8.314 CV/K)(298.2 K) ln
0.10
____
____________________________
__________
nF
(2 mol)(96,485 C/mol)
0.00100
= 0.17 V – 0.059 V = 0.11 V
3E-5 (of 11)
For a nickel-cadmium cell with solutions of 0.00100 M nickel (II) sulfate
and 0.10 M cadmium sulfate, determine the (e) the cell potential at 25ºC
once 90.0% of the nickel (II) ions have reacted away
Ni2+ (aq) + Cd (s) → Ni (s) + Cd2+ (aq)
Initial M’s
Change in M’s
Final M’s
0.00100
- 0.00090
0.00010
0.10
+ 0.00090
0.10090
Q = [Cd2+]
_________
[Ni2+]
Ɛ = Ɛº – RT ln Q = 0.17 V – (8.314 CV/K)(298.2 K) ln
0.10090
____
____________________________
__________
nF
(2 mol)(96,485 C/mol)
0.00010
= 0.17 V – 0.089 V = 0.08 V
3E-6 (of 11)
For a reaction at equilibrium
ΔG = 0 ∴ Ɛ = 0
Ɛ = Ɛº – RT ln Q
____
nF
0 = Ɛº – RT ln Keq
____
nF
RT ln Keq = Ɛº
____
nF
Keq = eƐºnF/RT
3E-7 (of 11)
Find the equilibrium constant at 25ºC for
Fe (s) + I2 (s) → Fe2+ (aq) + 2I- (aq)
2e- + I2 (s) → 2I- (aq)
2e- + Fe2+ (aq) → Fe (s)
2e- + I2 (s) → 2I- (aq)
Fe (s) → Fe2+ (aq) + 2eI2 (s) + Fe (s) → 2I- (aq) + Fe2+ (aq)
0.54 V + 0.44 V = 0.98 V
3E-8 (of 11)
Ɛºred = 0.54 V
Ɛºred = -0.44 V
Ɛºred = 0.54 V
Ɛºox = 0.44 V
Find the equilibrium constant at 25ºC for
Fe (s) + I2 (s) → Fe2+ (aq) + 2I- (aq)
Keq = eƐºnF/RT
= e[(0.98 V)(2 mol)(96,485 C/mol)] /[(8.314 CV/K)(298.2 K)]
= 1.3 x 1033
3E-9 (of 11)
Find the solubility product constant for mercury (I) chloride at 25ºC
Hg2Cl2 (s) ⇄ Hg22+ (aq) + 2Cl- (aq)
2e- + Hg2Cl2 (s) → 2Hg (l) + 2Cl- (aq)
2e- + Hg22+ (aq) → 2Hg (l)
Ɛºred =
Ɛºred =
2e- + Hg2Cl2 (s) → 2Hg (l) + 2Cl- (aq)
2Hg (l) → Hg22+ (aq) + 2e-
Ɛºred = 0.33 V
Ɛºox = -0.80 V
Hg2Cl2 (s) ⇄ Hg22+ (aq) + 2Cl- (aq)
0.33 V – 0.80 V = -0.47 V
3E-10 (of 11)
0.33 V
0.80 V
Find the solubility product constant for mercury (I) chloride at 25ºC for
Hg2Cl2 (s) ⇄ Hg22+ (aq) + 2Cl- (aq)
Ksp = eƐºnF/RT
= e[(-0.47 V)(2 mol)(96,485 C/mol)] /[(8.314 CV/K)(298.2 K)]
= 1.3 x 10-16
3E-11 (of 11)
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)
Zn (s) → Zn2+ (aq) + 2e-
Anode:
Cathode:
2e- + 2MnO2 (s) + 8NH4+ (aq) → 2Mn3+ (aq) + 4H2O (l) + 8NH3 (aq)
Potential or Voltage: 1.5 V
3F-2 (of 16)
Current or Amperage: Depends on battery size
Dry Cell
Graphite rod (cathode)
Paste of MnO2, NH4Cl, and H2O
Zinc casing (anode)
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
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) + 2e2e- + 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)
Separator
Graphite (cathode)
Anode:
Cathode:
LiCoO2 (s) → CoO2 (s) + Li+ (org) + ee- + 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
3F-5 (of 16)
Lead Storage Battery
Lead (anode)
Lead + lead (IV) oxide (cathode)
4 M sulfuric acid
Anode:
Cathode:
3F-6 (of 16)
Pb (s) + SO42- (aq) → PbSO4 (s) + 2e2e- + 2PbO2 (s) + 4H+ (aq) + SO42- (aq) → PbSO4 (s) + 2H2O (l)
Lead Storage Battery
Lead (anode)
Lead + lead (IV) oxide (cathode)
4 M sulfuric acid
Potential or Voltage: 2.1 V x 6 cells = 12.6 V
_______
cell
Because products stick to the electrodes, this battery is rechargeable
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
Anode: Oxidation
2 H2O (l) → O2 (g) + 4H+ (aq) + 4eCathode: Reduction
2e- + 2 H2O (l) → H2 (g) +
3F-9 (of 16)
2 OH- (aq)
H2O
+1 -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)
H2 (g)
Cu (s)
H2 (g)
H2 (g)
H2 (g)
H2 (g)
Ag (s)
Na+ (aq) + e- → Na (s)
2H2O (l) + 2e- → H2 (g) + 2OH- (aq)
Ɛºred = -2.71 V
Ɛºred = -0.83 V
N in HNO3 cannot be oxidized , so O in H2O will be oxidized
3F-10 (of 16)
Electrolytic cells are used for
(1) producing elements (Na, Cl2, etc.)
(2) purification of metals from ore
(3) electroplating metals (Au, Ag, Pt, etc.)
Ag
Anode: Oxidation
Ag (s) → Ag+ (aq)
Ag+
+ eAg+
Cathode: Reduction
e- +
3F-11 (of 16)
Ag+ (aq) → Ag (s)
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 mole of e-s 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 1.20 x 103 s x
1C x
______
1 As
x 63.55 g Cu
______________
1 mol Cu
3F-13 (of 16)
= 3.12 g Cu
1F
___________
96,485 C
x 1 mol e- x 1 mol Cu
__________
___________
1F
2 mol e-
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 10.0 min x
x 1 mol Al
___________
3 mol e-
3F-14 (of 16)
60 s
x
1C x
________
______
1 min
1 As
x 26.98 g Al
_____________
1 mol Al
1F
___________
96,485 C
= 0.280 g Al
x 1 mol e__________
1F
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
x 1 As
mol Zn
2 mol e- x
1F
_____________
___________
__________
65.38 g Zn
1 mol Zn
1 mol e-
x
1
______
________
1C
60.0 s
3F-15 (of 16)
x
= 7.38 A
x 96,485 C
___________
1F
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
x 1 As
mol Cr
x
3 mol e- x
_____________
___________
__________
52.00 g Cr
1 mol Cr
1 mol e-
x
1
x 1 min
______
________
_______
1C
10.0 A
60 s
3F-16 (of 16)
1F
= 3.71 min
x 96,485 C
___________
1F