Half-reactions indicate the mole ratio of electrons to ions involved in redox reactions. The document discusses how current, time, and charge are related based on 1 mole of electrons equating to 96,500 Coulombs of charge. It provides examples calculating the mass of copper produced from electrolysis and the volume of chlorine gas produced from an industrial electrolysis process based on given values of current and time.

1. ELECTROLYSIS
CALCULATIONS

2. A LOOK AT HALF-EQUATIONS
 A half equation really just tells you the number of electrons
removed or added to an ion:
 The following half-equation tells us that 1 mole of electrons is lost
from sodium metal to form1 mole of sodium ions:
Na – 1e- → Na+

3. THIS MEANS THAT
Mg – 2e- → Mg2+
Al – 3e- → Al3+
Cl + 1e- →Cl-
1 mole of Magnesium needs
to lose 2 moles of electrons
to form 1 mole of
magnesium ion
1 mole of aluminium needs
to lose 3 moles of electrons
to form 1 mole of
aluminium ion
1 mole of chlorine atoms
need to gain 1 mole of
electrons to form 1 mole of
chloride ion

4. Let’s assume that for
sodium metal, only
half the number of
moles of electrons
are supplied.
BUT SUPPOSE YOU DON’T HAVE THE
SUFFICIENT AMOUNT OF MOLES?
 Since the ratio between moles of
electrons to moles of sodium metal is 1:1,
 then 0.5 moles of electrons should
supply 0.5 moles of Na+ ions
In essence, the number of
moles of electrons is directly
proportional to the number
of ions in electrolysis and
vice-versa

5. HOW DO YOU KNOW THE AMOUNT
OF ELECTRONS IN THE CELL?
It has been determined that 1 mole of
electrons has a charge of 96,500 Coloumbs
(C).
This number of Coulombic charge is
equal to 1 Faraday (F)
Charge of 1 mole of electrons = 96,500 C = 1F

6. WHAT IS CURRENT?
The word, current, refers to the rate of
flow of electricity
One ampere (amp) is equal to a rate of
flow of charge of 1 Coulomb every second
Therefore,
amps = Coulombs/second
and
Coulombs = amps * seconds

7. WHAT IS CURRENT?
This can be denoted by the following equation:
Q = I x T
Where
Q = quantity of charge in coloumbs
I = current in Amps
T = time in seconds

8. RELATIONSHIPS BETWEEN
Number of
moles of
electrons
Amount of
electric
current
Time
current
passed
through the
system
Mass of
product at
electrodes

9. WORKED EXAMPLES

10. EXAMPLE 1
Calculate the mass of copper produced
in 1.5 hours by the electrolysis of
molten CuCl2 if the electrical current is
12.3 Amps

11. STEP 1: DETERMINE CHARGE
Since
the charge in Couloumbs = amps * seconds or
Q = I x TQ = 12.3 amps * 1.5 hours * 60 min/hour * 60 sec/min
Q = 6.64 x 104 C

12. STEP 2:DETERMINE NUMBER
OF ELECTRONS
Therefore, 6.64 x 104 Coulombs total pass into the reduction cell
(cathode) during the reaction.
Since 96,500 C is the charge for 1 mole of e-,
96,500 C = 1 mole of e-
1 C = (1/ 96,500) mole of e-
6.64 x 104 C = 6.64 x 104 C * (1 mole e- / 96,500C)
= 0.688 moles e-

13. STEP 3:DETERMINE NUMBER OF
MOLES OF ION/METAL PRODUCED
How many moles of Cu can be reduced with 0.688 moles of
electrons?
The half-equation for this reaction is
Cu2+ + 2e- → Cu
This means that it takes
2 moles of e- to produce 1 mole of Cu or
1 mole of e- to produce 1/2 mole of Cu
Therefore
0.688 moles e- = 0.688 * (1/2) = 0.344 moles Cu

14. STEP 4:DETERMINE MASS OF
ION/METAL PRODUCED
Finally, how many grams of Copper are there per mole?
The relative atomic mass of copper, Ar = 63.5 g/mole
This means that
1 mole of Cu has a mass of 63.5 g
Therefore
0.344 moles of Cu = 0.344 * 63.5 = 21.8 g

15. EXAMPLE 2
In the industrial production of chlorine gas, a
current of 50,000 A was passed through a salt
solution for 1 hour. Calculate the volume of gas
which would be produced.
Take the molar volume to be 24 dm3 / mol.

16. STEP 1: DETERMINE CHARGE
Since
the charge in Couloumbs = amps * seconds or
Q = I x T
I = 50000 A
t = 3600 s
Q = I × t
= 50000 × 3600
= 180000000
= 1.8 × 108 C

17. STEP 2:DETERMINE NUMBER
OF ELECTRONS
Therefore, 1.8 × 108 Coulombs total pass into the cell (anode) during
the reaction.
Since 96,500 C is the charge for 1 mole of e-,
96,500 C = 1 mole of e-
1 C = (1/ 96,500) mole of e-
1.8 × 108 C = 1.8 × 108 C * (1 mole e- / 96,500C)
= 1,865.28 moles e-

18. STEP 3:DETERMINE NUMBER OF
MOLES OF ION/METAL PRODUCED
How many moles of Cl2 can be produced from 1,865.28 moles of
electrons?
The half-equation for this reaction is
2Cl-(aq)→Cl2(g) + 2e-
This means that it takes
2 moles of e- for every 1 mole of Cl2(g) or
1 mole of e- for every 1/2 mole of Cl2(g)
Therefore
1,865.28 moles e- = 1,865.28 * (1/2) = 932.64 moles Cl2(g)

19. STEP 4:DETERMINE MASS/ VOLUME
OF ION/METAL/ GAS PRODUCED
Now, assuming room temperature and pressure (RTP), (20◦C and 1
atmosphere presssure)
Molar volume of chlorine gas , Cl2(g)= 24 dm3
This means that
1 mole of Cl2(g) occupies a volume of 24 dm3
Therefore
932.64 moles Cl2(g) = 932.64 x 24 = 22, 383.36 dm3