Contributed by:
The highlights are:
1. Electrolytic cells
2. Electrolytic cell reactions
3. Electrolysis of sodium chloride
4. Electroplating
5. Electrolysis of melts
6. Quantitative relationships
7. Electrical units
8. Faraday's Laws
1.
Chemistry 1011
Electrochemistry
TEXT REFERENCE
Masterton and Hurley Chapter 18
Chemistry 1011 Slot 5 1
2.
18.5 Electrolytic Cells
YOU ARE EXPECTED TO BE ABLE TO:
• Construct a labelled diagram to show the structure of an
electrolytic cell and describe its operation
• Identify half reactions that take place at the anode and
cathode of an electrolytic cell
• Predict the most likely products of electrolysis of aqueous
solutions and melts of ionic compounds
• Calculate the minimum voltage necessary for the
electrolysis of a solution under standard conditions
• Carry out calculations related to the current passing through
an electrolytic cell and the quantities of product formed
Chemistry 1011 Slot 5 2
3.
Electrolytic Cells
• In an electrolytic cell, a non-spontaneous
redox reaction is made to occur by pumping
electrical energy into the system
- DC Source +
Could be Voltaic
cell
CATHODE + ANODE
Reduction occurs Oxidation occurs
at the cathode at the anode
M+ + e M X X + e
Chemistry 1011 Slot 5 3
4.
Electrolytic Cells
• The simple electrolytic cell consists of
– two electrodes, dipping into
– a solution containing positive and negative ions
• The external battery or DC source acts as a pump
– It pushes electrons to the cathode and removes electrons from
the anode
• Metal ions in solution travel to the cathode and gain
electrons (reduction)
• Non-metal ions travel to the anode and give up electrons
(oxidation)
• The process is called electrolysis
Chemistry 1011 Slot 5 4
5.
Electrolytic Cell Reactions
• In aqueous solution, a number of anode and cathode
half reactions are possible
• At the cathode:
1. Reduction of a metal cation to the metal:
Cu2+(aq) + 2e Cu(s) Eored = +0.339V
Common when transition metal cations present; does not
occur if cation is difficult to reduce (Eored < -0.828V)
2. Reduction of a water molecule to hydrogen gas:
2H2O + 2e H2(g) + 2OH(aq) Eored = -0.828V
Occurs if cation is difficult to reduce
Chemistry 1011 Slot 5 5
6.
Electrolytic Cell Reactions
• At the anode:
1. Oxidation of an anion to the non-metal:
2I(aq) I2(s) + 2e Eoox = -0.534V
Does not occur if anion is difficult to oxidize
(Eored < -1.229)
2. Oxidation of a water molecule to oxygen gas:
2H2O O2(g) + 4H(aq) + 4e Eored = -1.229V
Occurs if anion is difficult to reduce
3. Oxidation of the anode material
Cu(s) Cu2+(aq) + 2e Eored = -0.339V
Chemistry 1011 Slot 5 6
7.
Electrolysis of Sodium Chloride
• Possible anode half reactions:
2Cl(aq) Cl2(g) + 2e Eoox = -1.360V
2H2O O2(g) + 4H(aq) + 4e Eored = -1.229V
• At high concentrations, the chlorine oxidation half reaction is
favoured
• Possible cathode half reactions:
2H2O + 2e H2(g) + 2OH(aq) Eored = -0.828V
Na+(aq) + e Na(s) Eored = -2.714V
• The water reduction half reaction has a much lower Eored value and
is favoured
• Overall cell reaction:
2Cl(aq) + 2H2O Cl2(g) + H2(g) + 2OH(aq) Eocell = -2.188V
Chemistry 1011 Slot 5 7
8.
Electrolysis of Sodium Chloride
• Chlorine gas is produced at the anode
• Hydrogen gas is produced at the cathode
• Sodium ions remain in solution
• Hydroxide ions are formed in the solution
• Commercially, this process is used to
manufacture
– Chlorine
– Hydrogen
– Sodium hydroxide
Chemistry 1011 Slot 5 8
9.
Electrolysis of Sodium Chloride
Chemistry 1011 Slot 5 9
10.
Electroplating
• In many electrolytic cells, metal is
deposited at the cathode
• An object can be electroplated by making it
the cathode
• Copper, chromium and silver are often
plated onto brass, nickel, or other metals by
this method
• The process is also used to purify copper, as
the last step in the manufacture of copper
from copper ore
Chemistry 1011 Slot 5 10
11.
Electrolysis of melts
• Salts will conduct electricity when melted, as well as when in
solution, e.g.: 2NaCl(l) 2Na (l) + Cl2(g)
• Aluminum is produced by electrolysis of alumina, Al2O3
• The melting point of Al2O3 is too high for direct electrolysis, so it is
dissolved in molten Na3AlF6 (cryolite) at 980oC
– The cathode is iron
– The anode is carbon
• Cathode reaction: Al3+ + 3e Al(l)
• Anode reaction: 2O2 O2(g) + 4e
• (The oxygen reacts with the carbon anode to form carbon dioxide gas)
Chemistry 1011 Slot 5 11
12.
Quantitative Relationships
• The amount of a substance produced or used in an
electrochemical cell will be determined by the
amount of electricity flowing in the cell
• Cathode reaction:
Ag+(aq) + e Ag(s)
– One mole of electrons will produce one mole of
silver
Cu2+(aq) + 2e Cu(s)
– Two moles of electrons will produce one mole of
copper
Chemistry 1011 Slot 5 12
13.
Electrical Units
Quantity Unit
Potential Volt (V)
Current Ampere (A) Rate of flow of electrons
Charge Coulomb (C) 1 Amp for 1 second
Energy Joules (J) #Volts #Coulombs
1 mole of electrons = 96500 Coulombs = 1 Faraday
Chemistry 1011 Slot 5 13
14.
Faraday’s Laws
1. In an electrolysis, the amount of product
produced or reactant consumed is proportional
to the length of time that a constant current is
passed in the circuit
2. To produce one mole of product, or to consume
one mole of reactant, requires n moles of
electrons, (where n is the number of electrons
gained or lost by one atom or ion). This is
96,500n Coulombs
Chemistry 1011 Slot 5 14
15.
Faraday’s Laws
• Faraday’s Laws provide a relationship
between
– The current flowing in an electrolytic cell
(Amperes)
– The time that the current flows (seconds)
– The amount of material deposited or consumed
(moles)
Chemistry 1011 Slot 5 15
16.
Applying Faraday’s Laws
1 mole e
+
AgNO3 CuSO4
solution solution
cathode cathode
1 mole Ag (108g) ½ mole Cu (32g)
deposited on deposited on
cathode cathode
Chemistry 1011 Slot 5 16
17.
Applying Faraday’s Laws
• What current will deposit 0.155g of silver from a solution
of silver ions in 11.0 minutes ( 660 seconds)?
Ag+(aq) + e Ag(s)
# moles silver produced = 0.155g 108 g/mol = 1.44 10-3
mol
# moles of electrons = 1.44 10-3 mol
Total charge = 1.44 10-3 mol 96500 Coulombs/mol = 139C
• Current = Charge (Coulombs) time (seconds)
• Current = 139C 660 seconds = 0.211 Amps
Chemistry 1011 Slot 5 17
18.
Applying Faraday’s Laws
How much copper will be deposited if a current of 0.150A is
passed through a solution of copper sulfate for 20.0 minutes?
Cu2+(aq) + 2e Cu(s)
Two moles of electrons will be required to deposit each mole of copper
• Total charge (C) = current (A) time (s)
= 0.150A 1200s = 180 C
• # moles (Faradays) = 180 C 96500 C/mol = 1.87 x 103 mol
• #moles of copper deposited
= ½ 1.87 103mol = 9.33 x 104mol
• Mass of copper = 9.33 x 104mol 63.5 g/mol
= 5.92 102g
Chemistry 1011 Slot 5 18
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