Faraday’s Laws 2010
Faraday’s Laws:
The first law of electrolysis
When electroplating is to be used there are a number of questions that you may
need to ask
How can I determine how much metal is being plated?
How long should I leave the object being plated in the electroplating
cell?
 What size electric current should be used?
To answer these questions you need to understand how the electroplating cell
would work:


We can use this cell to study the relationship between the mass of metal
formed at the cathode and the quantity of charge that passes through the cell.
While the mass of metal formed is easily measured by weighing the cathode,
electric charge is determined indirectly.
Electric charge, symbol
Q, is measured using the unit coulomb. The electric
charge passing through a cell may be calculated from measurements of the
current,
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I, through the cell and the time, t, for which the current flows.
Faraday’s Laws 2010
Current (A) Time (s
Current (A)
Time (s)
Charge (C)
I
t
Q=Ixt
2.0
200
400
2.0
400
800
2.0
600
1200
2.0
800
1600
2.0
1000
2000
 Experimental data from a Silver plating cell
Mass of Silver
formed (g)
0.45
0.90
1.34
1.79
2.24
The results of the experiment above show clearly that the more charge that
passes through the cell, the more metal is formed at the cathode. These
mass
of metal produced at the cathode is directly proportional
to the quantity of electricity passed through the cell.
This relationship is known as Faraday’s first law of electrolysis
experimental results may be summarised by saying that the
the quantity of elements
separated by passing an electrical current through a
molten or dissolved salt is proportional to the quantity of
electric charge passed through the circuit.
In 1832, Michael Faraday reported that
This became the basis of the first law of electrolysis:
charge (coulombs) = current (amps) × time (seconds)
Q = I × t
The charge of 1 electron is 1.602 x 10-19C
1 mole of electrons, therefore , has a charge of
(6.023 x 1023 x 1.602 x 10-19) = 9.649 x 104 C
This quantity of charge carried by 1 mole of electrons as known as the
faraday (F), or Faraday Constant. Generally with a value of
96,500 C mol-1
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Faraday’s Laws 2010
The second law of electrolysis
Many experiments can be repeated as with the Silver model using various
metals. When the results are plotted the graphs appear similar to that of the
silver, however if we graph the amount of each metal, measured in moles,
against charge the results are more interesting. All data lie along just three
lines.
One mole of silver is produced by about 96 500 coulomb of charge,
whereas one mole of copper, tin or lead is produced by 193 000 C.
What chemical reason can you give for this observation?
The number of faraday’s (moles of electrons) needed to liberate 1 mole of an
element during electrolysis is found from the equation for the electrode
reaction.
The electrode reactions when copper, tin or lead are electroplated are:
Cu2+(aq) + 2e– →
Sn2+(aq) + 2e– →
Pb2+(aq) + 2e– →
Cu(s)
Sn(s)
Pb(s)
For one mole of each of these metals to be deposited, two moles of electrons
are required. Two faradays of charge (193 000 C) must therefore be passed
through the cell.
To deposit one mole of chromium, three faradays of charge (289 500 C) are
required, since the equation for the electrode reaction is:
Cr3+(aq) + 3e– → Cr(s)
in order to
produce one mole of a metal, one, two, three, or
another whole number of moles of electrons must be
These observations may be summarised by saying that
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Faraday’s Laws 2010
consumed according to the relevant half-reactions. This is a
modernised statement of Faraday’s second law of electrolysis.
This can be described by the equation:
n(electrons) = Q/F
or
Q = I x t
Questions
Q1.
A silver-plating cell operates with a steady current of 30.0 A for 20.0
minutes. What mass of silver is plated on the object at the cathode?
Quantity of charge passed through the cell:
Q =I×t
= 30.0 A × (20.00 × 60) s = 36 000 C
(Note that time must be expressed in seconds.)
Amount, in mol, of electrons passed through the cell:
n(e–) = Q/F
= 36 000 C / 96 500 C mol–1
= 0.3731 mol
Since the cathode reaction is: Ag+(aq) + e– –—Ü Ag(s), 1 mol of electrons will
deposit 1 mol of silver metal, so
n(Ag) = n(e–) = 0.3731 mol
Mass of silver plated at the cathode:
m(Ag) = n × M
= 0.3731 mol × 107.9 g mol–1
= 40.25 g
The mass of silver plated on the cathode is 40.3 g.
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Faraday’s Laws 2010
Q2.
How long would it take to deposit 50.0 g of copper at the cathode of a
copper-plating cell operating at a current of 8.00 A?
Amount of copper to be deposited:
n(Cu) = m/M
= 50.0 g / 63.5 g mol–1
= 0.787 mol
Since the cathode reaction is Cu2+(aq) + 2e– → Cu(s),
2 mol of electrons will deposit 1 mol of copper metal, so:
n(e–) = 2 × n(Cu)
= 2 × 0.787
= 1.57 mol
Quantity of charge passed through the cell:
Q = n(e–) × F
= 1.57 mol × 96 500 C mol–1
= 152 000 C
Time required:
t = Q/I
= 152 000 C / 8.00 A
= 19 000 s
= 5.27 hours
Questions refer to Study On Chemistry 2
Page 415
Questions 4 – 9
During electrolysis:
• The highest reaction in the electrochemical series that can occur in the
forward direction is likely to occur at the cathode, that is,
the strongest
oxidant will usually react at the cathode.
• The lowest reaction in the series that can occur in reverse is likely to occur at
the strongest reductant will usually react
at the anode.
the anode, that is,
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Faraday’s Laws 2010
Q.
Predict the products of electrolysis of 1 M nickel sulphate
solution using copper electrodes.
Solution
The chemicals present at the electrodes are Ni2+(aq), SO42–(aq), and H2O(l). The
electrodes are made of copper. By referring to the electrochemical series we
can see that a number of reactions need to be considered as the cell contains
two oxidants, Ni2+(aq), and H2O(l)., and two reductants, Cu and H2O.
O2 + 4H+ + 4e– → 2H2O
Cu2+ + 2e– → Cu
Ni2+ + 2e– → Ni
2H2O + 2e– → H2 + 2OH–
1.23 V
0.34 V
–0.23 V
–0.83 V
At the cathode (–) we see that the reduction reactions that could occur are:
Ni2+(aq) + 2e– → Ni(s)
2H2O(l) + 2e– → H2(g) + 2OH–(aq)
We predict that nickel metal will plate onto the electrode, since the highest
reduction reaction in the electrochemical series is favoured; that is, the
strongest oxidant, Ni2+(aq), will be reduced.
At the anode (+), oxidation reactions that could occur are:
2H2O (l) → O2(g)+ 4H+(aq) + 4e–
Cu(s) → Cu2+ (aq) + 2e–
The copper anode is therefore likely to dissolve as the oxidation reaction lowest
in the series is favoured; that is, the strongest reductant, Cu, will be oxidised.
The overall equation is:
Ni2+(aq) +
2e– →
Cu(s) →
Ni(s)
Cu2+(aq) + 2e–
Ni2+(aq) + Cu(s) → Ni(s) + Cu2+(aq)

Note that experiments are the only sure way to find out what products
form in a particular instance. In practice, reactions are affected by
factors such as electrolyte concentration, gas pressures, current,
voltage, and even the nature of the electrodes. Some electrodes permit
particular reactions to occur more readily than others.
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