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The
faraday and electrochemical calculations
Francine
Taylor-Campbell, Contributor
Main points
- The
Faraday (F) is the quantity of electricity
needed to remove one mole of electrons
from the cathode during electrolysis
or to deposit one mole of electrons
on the anode during electrolysis.
- The
Faraday Constant is the amount of
electric charge carried by one mole
of electrons, that is, 96,500C.
During
electrolysis, electrons are taken
from the cathode by positive ions
called cations.
Eg.
2H+,(aq),+ 2e
= H2(g) Cu2+
(aq)+ 2e = Cu(s)
The
ions are said to be discharged.
Electrons
are deposited on the anode (+ve) by
negative ions called anions.
Eg.
2Cl-(aq)= Cl2(g)
+ 2e 4OH- = 2H20 + 02
+ 4e
Electrons
may also leave from the anode if the
anode dissolves.
Eg.
Cu = Cu2+ + 2e-
One
Faraday = 96,500 Coulombs i.e. 1F
= 96,500 C
The
coulomb is the unit of electrical
charge and is one ampere flowing for
one second.
i.e.
coulombs = amps x sec, quantity of
electricity = current x time (Q =
I x t)
Eg.
When two amps flow for one minute,
the quantity of electricity flowing
(Q)
Q
= 2X60 = 120C
Note
The
Faraday may also be regarded as the
charge on one mole of electrons.
Thus
F = Le L = Avogadro's number e = the
charge on one electron
Further
examples
1.
What mass of copper would be deposited
during electrolysis by 0.5F?
Cu2+
+ 2e = Cu
2F
64g
thus
0.5F = 16g Cu
2.
What mass of lead would be produced
by a current of 5A, passed through
molten lead bromide for one hour?
C
= A x s
C = 5 x 60 x 60 = 18,000C
Now
Pb2+(l) + 2e = Pb(l)
2F
207g
193,000C
= 207g Pb
Thus,
18,000C = (207/ 193,000) x 18,000
= 19.2g Pb
3.
What volumes of (a) H2 (b) O2 would
be liberated at RTP when 0.1 F is
passed through dilute sulphuric acid?
4H+
+ 4e- = 2H24OH- = 2H2O
+ O2 + 4e-
Calculate
the volume of H2, for example 2H+
+ 2e = H2
2F
1mole = 24 dm3 at RTP
Thus
0.1F = (24/2) x 0.1 dm3
= 1.2 dm3 H2 at R.T.P
And
volume of O2 = 0.6dm3
at RTP (from above equations)
Examples
of examination questions
Q1.
A current of three amperes was passed
for one minute and 36 seconds through
an electrolyte containing aluminium
ions. Calculate the number of moles
of aluminium deposited at the cathode.
Answers
1.
First of all the amount of electricity
flowing through the electrolyte needs
to be determined. Q = current x time(in
sec) Q = 3 x (60 + 36) = 3 x 96 =
288C
Aluminium
ions are discharged according to the
equation Al3+ + 3e == Al
This
means that to deposit one mole of
aluminium, that is 27g requires 3F
or 289500C.
1
mol Al ===== 289500C
X
mol Al ===== 288C, X = 288/289500
== 0.00099mol = 0.001 mol aluminium.
Q2.
A solution of copper (II) nitrate
was used to electroplate a silver
coin with copper. (a). Write an equation
to represent the reaction taking place
at the anode.
(b)
Originally, you were given a silver
coin that weighed 1.00g. Calculate
the total mass of the coin after electrolysis,
when a current of 5A was passed through
the copper (II) nitrate solution for
five minutes.
(c)
What would you expect to observe if
the electroplated coin were placed
in diluted sulphuric acid?
Answers
2.
To electroplate the silver coin, the
anode is made of copper, which dissolves
in a solution of copper ions to be
later deposited on to the silver coin.
At
the anode: Cu (s) === Cu2+(aq)
+ 2e
(b)
Quantity of electricity == 5 x (5x60)
= 1500C
1
mol Cu or 63.5g is deposited by 2F
or 193000C
X
mol Cu is deposited by 1500C
X
== 0.00777 mol Cu
Mass
of Cu deposited = 0.00777 x 63.5 =
0.494g
The
total mass of the silver coin would
be 1.00 + 0.494 = 1.494g
(c)
The electroplated coin would not react
with sulphuric acid since the copper
coat does not react with dilute acids.
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(From
left) Bridgeport High School's
4x100m Class three boys' relay
team of Romario Campbell, Jazeel
Murphy, Kemar Philpotts and
Kamal Atkins pose for the camera
on the third and penultimate
day of the Inter-secondary Schools'
Sports Association/GraceKennedy
Boys' and Girls' Athletic Championships
at the National Stadium on March
14.
- Anthony Minott/Freelance
Photographer
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Francine
Taylor-Campbell is an independent
contributor.
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