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Electrolysis

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kin kin

on 10 December 2013

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Transcript of Electrolysis


Electrolysis
+ -
oxidation reduction
Electrolytic cell
D.C. power supply
Electrolyte: molten or
aqueous solution

decomposed

Red Cat An Ox
Factor affecting preferential discharge of ions:

1. Position of ions in E.C.S.
Cations lower in E.C.S. --> stronger oxidizing agents

Anions higher in E.C.S. --> stronger reducing agents
2. Concentration of ions in solution
When two ions not too far apart in E.C.S. move towards an electrode, the one at a much higher concentration would be discharged.
Relative concentration of hyroxide ions << that of chloride ions

Result: chloride ions discharge into chlorine gas

Chloride is preferentially discharged !

2 Cl --------> Cl + 2 e

- -
(aq) 2(g)
3. Material of electrode
Inert electrodes (e.g. graphite, platinum) do not take part in any reaction in electrolysis.

Some electrodes (e.g. copper) may involve in electrolysis, it affects the preferential discharge of ions and hence the products formed.

3a. Electrolysis of molten electrolyte:
Pb
2+
Pb
2e
-
Pb + 2e --------> Pb
Reduction
2+ -
(l) (l)
- cathode
Lead(II) ions are attracted to the cathode.

Lead(II) ion gains 2 electrons and discharges into molten lead metal
2 Br
-
2e
-
2Br --------> Br + 2e
Oxidation
- -
(l) 2(g)
Bromide ions are attracted to the anode.

Bromide ion loses 2 electrons and discharges into bromine gas.
Br - Br
+ anode
Pb + --------> Pb
2+
(l) (l)
2Br + Br

-
(l) 2(g)
electrons gained = electrons lost
Redox
b. i. Electrolysis of dilute sulphuric acid
Hofmann voltameter
Cathode (-):

Formation of hydrogen gas


Anode (+):

Formation of oxygen gas

Volume ratio:
hydrogen : oxygen
= 2 : 1
2H
+
Pb
2e
-
2H + 2e --------> H
Reduction
+ -
(aq) 2(g)
- cathode
Hydrogen ions are attracted to the cathode.

Hydrogen ion gains 2 electrons and discharges into hydrogen gas
H - H
4 OH
-
4e
-
4OH --------> O + 2H O + 4e
Oxidation
- -
(aq) 2(g) 2 (l)
Hydroxide ions are attracted to the anode. They are preferentially discharged (oxidized)

Hydroxide ion is a stronger reducing agent than sulphate (hydroxide ion is higher in E.C.S.)

Hydroxide ion loses electrons and discharges into oxygen gas.
O = O
+ anode
+ 2 H O
2
2H O --------> 2 H + O
2 (l) 2(g) 2(g)
Water is decomposed !

=> As electrolysis goes on,
concentration of sulphuric acid
increases !
At the same conditions,

mole ratio of gases = their volume ratio
Electrolysis of very dilute NaCl solution
+ -
Oxygen gas
Hydrogen gas
Very dilute sodium chloride soluton
Graphite
Graphite
+
2e
-
2H + 2e --------> H
Reduction
+ -
(aq) 2(g)
Hydrogen ions are attracted to the cathode.

Hydrogen ion is a stronger oxidizing agent than sodium ion, hydrogen ion gains electrons and discharges into hydrogen gas
H - H
2H
4 OH
-
4e
-
4OH --------> O + 2H O + 4e
Oxidation
Hydroxide ions are attracted to the anode. They are preferentially discharged (oxidized)

Hydroxide ion is a stronger reducing agent than chloride (hydroxide ion is higher in E.C.S.)

Hydroxide ion loses electrons and discharges into oxygen gas.
O = O
+ 2 H O
2
- -
(aq) 2(g) 2 (l)
- cathode
+ anode
electrolysis
2H O --------> 2 H + O
2 (l) 2(g) 2(g)
Water is decomposed !

=> As electrolysis goes on,
concentration of dil. NaCl solution
increases !
At the same conditions,

mole ratio of gases = their volume ratio
electrolysis
Volume ratio = 2 : 1
Electrolysis of concentrated NaCl solution
concentrated sodium chloride soluton (brine)
+
2e
-
2H + 2e --------> H
Reduction
+ -
(aq) 2(g)
Hydrogen ions are attracted to the cathode.

Hydrogen ion is a stronger oxidizing agent than sodium ion, hydrogen ion gains electrons and discharges into hydrogen gas
H - H
2H
- cathode
2H + 2Cl --------> H + Cl
+ -
(aq) (aq) 2(g) 2(g)
NaOH is formed !

=> As electrolysis goes on,
hydrogen ions and chloride ions
are consumed !
electrolysis
Volume ratio = 1 : 1
2 Cl
-
2e
-
2 Cl --------> Cl + 2e
Oxidation
We would expect hydroxide ions to be preferentially discharged because hydroxide ion is a stronger reducing agent than chloride ion.

However, the concentration of chloride ion is much higher than that of hydroxide while the position of them are not too far away.

Due to the concentration effect, the discharge of chloride become more favourable.
Cl - Cl
- -
(aq) 2(g)
+ anode
Don't use platinum electrodes! Chlorine produced may attack platinum.
H OH
Na Cl
+ -
+ -
Electrolysis of concentrated NaCl solution with mercury electrode
Cathode (-)

Na + e ---------> Na (reduction)


Na + Hg ---------> Na/Hg

In the presence of mercury, sodium ion is preferentially discharges to form sodium metal



which dissolve in mercury cathode to form sodium amalgam
+ -
(aq) (s)




(s) (l) (l)
Na
+
Na
e
-
Na/Hg
Hg electrode
2 Cl
-
2e
-
2 Cl --------> Cl + 2e
Oxidation
We would expect hydroxide ions to be preferentially discharged because hydroxide ion is a stronger reducing agent than chloride ion.

However, the concentration of chloride ion is much higher than that of hydroxide while the position of them are not too far away.

Due to the concentration effect, the discharge of chloride become more favourable.
Cl - Cl
- -
(aq) 2(g)
+ anode
2Hg + 2Na + 2Cl --------> 2Na/Hg + Cl
+ -
(l) (aq) (aq) (l) 2(g)
The concentrated NaCl solution gradually changes to dilute NaCl.

=> As electrolysis goes on,
sodium ions and chloride ions
are consumed !
electrolysis
H OH
Na Cl
+ -
+ -
* Na can be regenerated by vaporizing the
mercury into vapour.
* Passing Na/Hg into water, Na can be
converted into NaOH.
Electrolysis of copper(II) sulphate with Cu electrodes
2e
-
Hydroxide ions and sulphate ions are attracted to the anode. But neither of them is discharged because Cu is a stronger reducing agent than hydroxide and sulphate ion

(Cu is higher than hydroxide and sulphate in E.C.S.)

Cu loses electrons and discharges into copper(II) ion.
2+ -
(s) (aq)
+ anode
Cu + 2e --------> Cu
Reduction
2+ -
(aq) (s)
Copper(II) ion is lower than hydrogen ion in the E.C.S.

Copper(II) ion is a stronger oxidizing agent than hydrogen ion.

Therefore copper(II) ions are preferentially discharged at cathode.
2+
2e
-
Cu
Cu
- cathode
Cu
2+
Cu ----------> Cu + 2e
(Cu metal)
OH
-
SO
2-
4
Oxidation
The copper cathode gets bigger (thicker) and heavier as copper deposited on it. On the other hand, the copper anode becomes smaller (thinner) and lighter , as it dissolves to form hydrated copper(II) ion.
As electrolysis goes on, the amount of copper(II) ions discharged is equal to that of copper metal formed
=> concentration and colour intensity of blue colour of solution also remains unchanged.
Application: purification of copper
Impure copper is connected to the
side.
positive (i.e. anode)
Electrolysis of copper(II) sulphate with graphite electrodes
Cu + 2e -----> Cu (Reduction)

Copper(II) ion is lower than hydrogen ion in the E.C.S.
Copper(II) ion is a stronger oxidizing agent than hydrogen ion.
Copper(II) ions are preferentially discharged at cathode.
2+ -
(aq) (s)
4OH -----> O + 2H O + 4e (oxidation)

Hydroxide ion is discharged because it is a stronger reducing agent than sulphate.

- -
(aq) 2(g) 2 (l)
overall: 2Cu + 4OH -----> 2Cu + O + 2H O
2+ -
(aq) (aq) (s) 2(g) 2 (l)
Change in electrodes: The graphite cathode gets bigger and heavier as
copper deposited on it. on the other hand,
colourless gas bubbles (oxygen) are formed on
the graphite anode.
Changes in the solution: As electrolysis goes on, copper(II) ion and
hydroxide ion are consumed. The copper(II)
sulphate solution gradually changes to dilute
sulphuric acid solution. The colour intensity of
blue colour of solution decreases.
Cu SO
H OH
2+ 2-

+ -
Comparing copper containing electrolytic cell
Check point:
1. State the products produced at
both electrodes.
2. State the change in electrolyte.
3. Write equations involved.
1. Electrolysis of dilute copper(II)
chloride solution with graphite
electrodes.

2. Electrolysis of dilute sodium
nitrate solution with graphite
electrodes.
1. Cathode: brown solid (copper) was produced



Anode: Colourless gas (Oxygen) was
produced











Cu + 2e --------> Cu
Reduction
2+ -
(aq) (s)
4OH --------> O + 2H O + 4e
Oxidation

- -
(aq) 2(g) 2 (l)
overall: 2Cu + 4OH -----> 2Cu + O + 2H O
2+ -
(aq) (aq) (s) 2(g) 2 (l)
Change in electrodes: The cathode gets bigger and heavier as
copper deposited on it. on the other hand,
colourless gas bubbles (oxygen) are formed on
the anode.
Changes in the solution: As electrolysis goes on, copper(II) ion and
hydroxide ion are consumed. The copper(II)
chloride solution gradually changes to dilute
shydrochloric acid solution. The colour intensity
of bluish green colour of solution decreases.
Almost no change in colour intensity
(Concentration of copper(II) ion decreases a little)
Check point:
Q. How to plate a thin
layer of metal on
glass or plastics
objects?
A. These objects are first
spray with a layer of
powdered graphite or
metal (e.g. Al powder)
Electroplating factories produce waste solution containing acids, alkalis, metal salts, toxic chemicals(e.g. Cyanide ions CN ), organic solvents and plating sludge.

Q. How to removed them in the
electroplating effluents ?
-
A.
Dilute Na CO solution is added in order to neutralize the acids.
2 3
Past Paper
A.
Dilute H SO solution is added in order to neutralize the alkalis.
2 4
Dilute Na CO or NaOH is added in order to change soluble heavy metal ions into insoluble precipitate of metal hydroxides. The heavy metal ions in precipitate can be removed by filtration.
2 3
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