Download Chapter 20 Worksheet 4

Chapter 19 worksheet 3
Electrolysis
Electrolysis: Forcing a non-spontaneous redox reaction to occur by applying an external voltage that
opposes the spontaneous flow of electrons.
Electrolysis of water (demonstration)
At room temperature, the following reaction is spontaneous (remember the Hindenburg!) and has a large
equilibrium constant.
2 H2 (g) + O2 (g) → 2 H2O (l)
Considering the definition of G (G = H - TS), why is this reaction spontaneous at room temperature?
The reverse reaction would be spontaneous if the temperature was sufficiently __________.
Forward (spontaneous) reaction: _________ is oxidized and __________ is reduced.
Reverse reaction (electrolysis): _________ is oxidized and __________ is reduced.
Reactions that occur at the cathode and anode during the electrolysis of water
Balance the following half-reactions (use H+ and OH-) and look up their Eo values in appendix E:
Reduction (cathode):
H2O (l)
→ H2 (g)
Oxidation (andode):
H2O (l)
→ O2 (g)
Choose the likely half-reactions and calculate the minimum voltage required for electrolysis to occur:
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Stoichiometry of electrolysis
A balanced reduction half-reaction tells you how many moles of electrons must be pumped into the
cathode (or out of the anode) of an electrolytic cell to produce a mole of product. For example:
Al3+(aq) + 3 e- → Al (s) says that it takes 3 moles of electrons to produce a mole of aluminum.
Electrical current
The amount of charge passing through a circuit (current) is measured in amperes (A):
1 A = 1 C/s
The total charge that passes through an electrolytic cell is given by:
charge = (current) x (time)
charge is in coulombs
current is in amps (C/s)
time is in seconds
Electrical work
Electrical power (rate of energy expenditure) is measured in watts (W):
Power = current x potential
1 W = (1 C/s)(1 J/C) = (1 A)(1 V) = 1 J/s
Electrical work = w = (power) x (time).
It is commonly expressed in units of kilowatt-hours (kWh):
1 kWh = 1000 w-h = (1000 J/s)(3600 s) = 3.6 x 106 J
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A voltaic cell utilizes a spontaneous redox reaction to do electrical work. In contrast, electrolysis requires
that work be done on the reaction to make the (normally non-spontaneous) reaction happen.
G is the maximum amount of (non-PV) work that can be done by a reaction. Thus, the maximum
electrical work that can be done by a voltaic cell is:
wmax = G = -nFEcell
(recall that F is the charge on a mole of electrons = 96500 C/mol)
n is the total number of moles of electrons that pass through the cell
If the cell potential is positive, wmax is negative meaning that a spontaneous reaction can do work on
the surroundings.
If the cell potential is negative, wmax is positive meaning that work must be done on a nonspontaneous reaction to make it happen. This is electrolysis.
During electrolysis, the electrical work done on the system is:
w = nFEext
n is the total number of moles of electrons pumped through the cell
Eext is the applied voltage
These ideas let us determine how much energy and time is required to produce a desired amount of
an electrolysis product.
You should be able to do the following:
A. Given the current through an electrolytic cell, calculate the rate of product formation.
B. Given the external potential applied to an electrolytic cell, calculate the amount of electrical
energy required to produce a given amount of reduced product.
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1. Aluminum can be produced by the electrolysis of molten AlCl2 (Why not aqueous AlCl2?)
a. Draw and label a diagram of this process. Calculate the minimum voltage that must be
applied in order to produce Al.
b. Calculate the mass of aluminum produced in 1.00 hr if the electrical current is 10.0 A.
c. Calculate the minimum number of kilowatt-hours of electricity required to produce 1.0 x
103 kg of aluminum if the applied emf is 4.50 V.
d. Why is the answer to part c the minimum energy required?
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2. During the electrolysis of molten MgCl2:
a. Calculate the minimum time required to produce 30.0 g of magnesium if the current is
60.0 A.
b. Calculate the minimum number of kilowatt-hours of electricity required to produce 1.00 kg
of magnesium if the applied emf is 5.00 V.
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