1. Chapter 10
Electrochemistry

2. Electrochemistry Reduction REDOX REACTION Oxidation
Is the study of the relationship between electricity
and chemical reaction
Chemical reactions involved in electrochemistry are :
Reduction
REDOX REACTION
Oxidation
One type of reaction cannot occur without the other.

3. REDOX Reaction REDUCTION OXIDATION Remember… RED CAT Mg  Mg2+ + 2e-
gain of electron
loss of electron
Oxidation no. decrease
Oxidation no. increase
Reaction at cathode
Reaction at anode
Example:
Remember…
RED CAT
= REDuction at CAThode
Mg  Mg2+ + 2e-
Example:
Oxidation no. 
Cu e-  Cu
Oxidation no. 

4. Example Electrochemical reaction consists of reduction and oxidation.
These two reactions are called ‘half-cell reactions’
The combination of 2 half reactions are called ‘cell reaction’
Example
Reduction :
Cu2+(aq) + 2e-  Cu(s)
Half-cell
reaction
Oxidation :
Zn(s)  Zn2+(aq) + 2e-
Overall cell
reaction :
Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq)

5. Cells Electrochemical Electrolytic Cells Cells
There are 2 type of cells
Electrochemical
Cells
Electrolytic
Cells
Uses electricity to produce chemical reaction
where chemical reaction produces electricity
Also called;
Galvanic cell or Voltaic cell
Chemical
Energy
Electrical
Electrical
Energy
Chemical

6. Component and Operation of Galvanic cell
Consists of :
1) Zn metal in an aqueous solution of Zn2+
2) Cu metal in an aqueous solution of Cu2+
The 2 metals are connected by a wire
The 2 containers are connected by a salt bridge.
A voltmeter is used to detect voltage generated.

7. Galvanic cell Voltmeter Cu electrode Zn electrode Salt bridge Zn2+
CuSO4(aq)
solution
ZnSO4(aq)
solution

8. What happens at the zinc electrode ?
Zinc is more electropositive than copper.
Tendency to release electrons: Zn > Cu.
Zn (s)  Zn2+ (aq) + 2e-
Zinc dissolves.
Oxidation occurs at the Zn electrode.
Zn2+ ions enter ZnSO4 solution.
Zn is the –ve electrode since it is a source of
electrons  anode.

9. What happens at the copper electrode ?
The electron from the Zn metal moves out through
the wire enter the Cu metal
Cu2+ ions from the solution accept electrons.
Cu2+ (aq) + 2e-  Cu (s)
Copper is deposited.
Reduction occurs at the Cu electrode.
Cu is the +ve electrode  cathode

10. Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)
Reactions Involved:
Anode :
Zn (s)  Zn2+ (aq) + 2e-
Cu2+ (aq) + 2e-  Cu (s)
Cathode :
Overall cell
reaction :
Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)

11. Salt bridge
An inverted U tube containing a gel permeated with solution of an inert electrolyte such as KCl, Na2SO4, NH4NO3.
Functions
Salt bridge helps to maintain electrical neutrality
Completes the circuit by allowing ions carrying charge to move from one half-cell to the other.

12. This excess charge build-up can be reduced by adding a salt bridge
What happened if there is no salt bridge? V Cu Zn e e e e
ZnSO4(aq)
Zn2+ e e
Cu2+
CuSO4(aq)
As the zinc rod dissolves, the concentration of Zn2+
in the left beaker increase.
The reaction stops because the nett increase in
positive charge is not neutralized.
This excess charge build-up can be reduced by adding a salt bridge

13. Cu Zn Zn (s)  Zn2+ (aq) + 2e- Cu2+ (aq) + 2e-  Cu (s) CATHODE (+)
ANODE (-)
E = V Cu Zn - + e e
ZnSO4(aq)
Zn2+
CuSO4(aq)
Cu2+
Salt bridge (KCl)

14. How does the cell maintains its electrical neutrality?
Left Cell
Right Cell
Zn (s)  Zn2+ (aq) + 2e-
Cu2+ (aq) + 2e-  Cu (s)
Zn2+ ions enter the solution.
Causing an overall excess of tve charge.
Cu 2+ ions leave the solution.
Causing an overall excess of
-ve charge.
Cl- ions from salt bridge move into Zn half cell
K+ ions from salt bridge move into Cu half cell
Electrical neutrality is maintained

15. Electrochemical Cells
anode
oxidation
cathode
reduction
spontaneous
redox reaction
half
19.2

16. Cell notation Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
Also can be represented as:
Salt bridge
Phase boundary
Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s)
anode
cathode
Cell notation

17. Zn (s) | Zn2+(aq) || Cr3+ (aq) | Cr (s)
Exercise
For the cell below, write the reaction at anode and cathode and also the overall cell reaction.
Cell notation
Zn (s) | Zn2+(aq) || Cr3+ (aq) | Cr (s)
X 3
Anode : 3
Zn(s) → Zn2+(aq) + 2e- 3
6e-
Cathode : 2
Cr3+(aq) + 3e- → Cr(s) 6e 2
X 2
Overall cell
reaction:
3Zn(s) + 2Cr3+(aq) + → 3Zn2+ (aq) +2Cr(s)

19. The difference in electrical potential between the anode and cathode is called:
cell voltage
electromotive force (emf)
cell potential
measured by a voltmeter
Acts as ‘electrical pressure’ that pushes electron through the wire.

20. Electrode Potential Cu2+(aq) + 2e → Cu(s) Eored = +0.34 V
A measure of the ability of a half-cell to attract electrons towards it.
Cu2+(aq) + 2e → Cu(s) Eored = V
Zn2+(aq) + 2e → Zn(s) Eored = V
Standard reduction potential
The more positive the half-cell’s electrode potential, the stronger the attraction for electrons.
Tendency for reduction ↑ (cathode)
Standard reduction potential of copper half-cell is more positive compared to zinc.
Zinc half-cell becomes anode.

21. Cu2+(aq) + 2e → Cu(s) Eored = +0.34 V
Zn2+(aq) + 2e → Zn(s) Eored = V
Cell Potential (Eocell)= Eocatode — Eoanode
= – (-0.76)
= +1.1 V

22. or Zn2+ (aq) + 2e-  Zn (s) Cu2+ (aq) + 2e-  Cu (s)
E0 = -0.76V
Cu2+ (aq) + 2e-  Cu (s)
E0 = +0.34V
E0cell = E0cathode - E0anode
= – (-0.76)
= V or
E0cell = E0red + E0ox
Change the sign
= (+0.76)
= V
Half-cell equation at:
Anode :
Zn (s)  Zn2+ (aq) + 2e-
Cu2+ (aq) + 2e-  Cu (s)
Cathode :

23. Standard Electrode Potentials (Eo)
A measure of the ability of half-cell to attract electrons towards it
at 25oC, the pressure is 1 atm (for gases), and the concentration of electrolyte is 1M.
The sign of E0 changes when the reaction is reversed
Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

24. For example:
Cl2(g) + 2e-  2Cl-(aq)
E0 = V
½Cl2(g) + e-  Cl-(aq)
E0 = V
Cl-(aq)  ½Cl2(g) + e-
E0 = V

25. Standard Hydrogen Electrode The standard reduction of SHE is 0 V
Made up of a platinum electrode, immersed in an aqueous solution of H+ (1 M) and bubbled with hydrogen gas at 1 atm pressure, and temperature at 25oC
H2 gas
at 1 atm Pt
electrode
H+ (aq)
1 M
The standard reduction of SHE is 0 V

26. Standard reduction potential of Zinc half cell is measured by setting up the electrochemical
cell as below.
H2 (g), 25oC,1 atm. Zn e
Zn2+
ZnSO4(aq)
1 M
H+(aq),1 M V Pt
E0 = 0
E0 = +0.76 - +

27. Standard Electrode Potentials
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
Anode (oxidation):
Zn (s) Zn2+ (1 M) + 2e-
Cathode (reduction):
2e- + 2H+ (aq,1 M) H2 (g,1 atm)
Cell reaction
Zn (s) + 2H+ (aq,1 M) Zn2+(aq) + H2 (g,1 atm)
E0 = EH /H - EZn /Zn
cell + 2+ 2
0.76 V = 0 - EZn /Zn 2+
EZn /Zn = V 2+
Zn2+ + 2e- Zn E0 = V

28. Standard reduction potential of Copper half cell is measured by setting up the electrochemical
cell as below. V
E0 = 0 + - -
H2 (g) 25oC 1 atm. + Cu
H+(aq)1 M
Cu2
CuSO4(aq) 1M Pt

29. Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
Anode (oxidation):
H2 (1 atm) H+ + 2e-
Cathode (reduction):
2e- + Cu2+ (1 M) Cu (s)
H2 + Cu Cu (s) + 2H+
E0 = Ecathode - Eanode
cell
Ecell = ECu /Cu – EH /H 2+ + 2
0.34 = ECu /Cu - 0 2+
ECu /Cu = 0.34 V 2+

30. In either case, E0 of SHE remains 0
The direction of half-reaction of SHE depends on the other
half-cell connected on it.
The cell notation for SHE is either:
H+(aq) | H2(g) | Pt(s) when it is cathode
Pt(s) | H2(g) | H+ (aq) when it is anode
In either case, E0 of SHE remains 0

31. or Zn2+ (aq) + 2e-  Zn (s) Cu2+ (aq) + 2e-  Cu (s)
E0 = -0.76V
Cu2+ (aq) + 2e-  Cu (s)
E0 = +0.34V
E0cell = E0cathode - E0anode
= – (-0.76)
= V or
E0cell = E0red + E0ox
Change the sign
= (+0.76)
= V
Half-cell equation at:
Anode :
Zn (s)  Zn2+ (aq) + 2e-
Cu2+ (aq) + 2e-  Cu (s)
Cathode :

32. E0cell = E0cathode - E0anode
At standard-state condition
E0cell = E0cathode E0anode or
E0cell = E0red E0ox

33. Exercise Answer E0cell = E0cathode - E0anode
Calculate the standard cell potential of the following
electrochemical cell.
Co(s) | Co2+(aq) || Ag+(aq) | Ag(s)
Ag+(aq) + e-  Ag(aq)
E0 = +0.80V
Answer
Co2+(aq) + 2e-  Co(s)
E0 = -0.28V
Cathode (Red) :
Anode (Ox) :
Co(s)  Co2+(aq) + 2e-
Ag+(aq) + e-  Ag(aq)
E0 = +0.80V
E0ox = +0.28V
E0cell = E0cathode - E0anode
= – (-0.28)
= V

34. Refer to the list of Standard Reduction Potential:
Oxidation agent → left of the half cell equation
Reduction agent → right of the half cell equation
Example :
Increase strength as
reducing agent
Ni2+ (aq) + 2e- → Ni (s)
E0 = V
Cu2+ (aq) + 2e- → Cu (s)
E0 = V
Ag+ (aq) + e- → Ag (s)
E0 = V
Reducing agent
Oxidation agent
The more -ve the value of E0 → the stronger the reducing agent
The more +ve the value of E0 → the stronger the oxidizing agent

35. Exercise Answer : L < X < Y
Arrange the 3 elements in order of increasing strength of reducing agents
X e-  X
E0 = V
Y e-  Y
E0 = V
L e-  L
E0 = V
Answer :
L < X < Y

36. E0cell = Ecathode - Eanode
Example
Calculate the E0 cell for the reaction ::
Mg(p) | Mg2+(ak) || Sn4+(ak),Sn2+(ak) | Pt(p)
Given :
Mg2+(ak) e→ Mg(p) Eθ = V
Sn4+(ak) e → Sn2+(ak) Eθ= V
Oxidation : Mg(p) → Mg2+(ak) e Eoox = V
Reduction : Sn4+(ak) e → Sn2+(ak) Eo = V
Mg(p) + Sn4+(ak) → Mg2+(ak) + Sn2+(ak) Ecell = V
E0cell = Ecathode - Eanode
= (-2.38)
=+2.53V
Eθcell = E o red E o ox =
= V

37. Exercise A cell is set up between a chlorine electrode and a
hydrogen electrode
Pt | H2(g, 1 atm) | H+(aq, 1M) || Cl2(g, 1atm) | Cl-(aq, 1M) | Pt
E0cell = V
Draw a diagram to show the apparatus and chemicals
used.
Discuss the chemical reactions occurring in the
electrochemical cell.

38. Answer V E0cell =1.36V H2 (g), 1 atm. Cl2 (g), 1 atm. Pt Pt H+(aq), 1M- + - + Pt Pt
H+(aq), 1M
Cl-(aq), 1M

39. 1. Show the process occur at anode and cathode
- Half-cell reaction
2. Overall reaction

40. Answer Reduction (cathode) Cl2 (g) + 2e-  2Cl- (aq) Oxidation(anode)
H2 (g)  2H+ (aq) + 2e-
E0 = 0
Eocell =+1.36 V
Eocell = Eocathode - E0anode
+1.36 = Eocathode – 0
E0cathode = V
So the standard reduction potential for Cl2 is:
Eo = V

41. Eθ cell = 0 The reaction is at equilibrium
Spontaneous & Non-Spontaneous reactions
- Redox reaction is spontaneous when
Ecell is +ve.
- Non spontaneous is when Ecell is –ve.
Eθ cell = The reaction is at equilibrium

42. Eo Zn + Sn4+ → Zn2+ + Sn2+ Eocell = Eo → Eo
Predict whether the following reactions occur spontaneously or non-spontaneously under standard condition. Zn + Sn4+ → Sn2+ + Zn2+ The two half-cells involved are:- Anod : Zn → Zn2+ + 2e Eoox = V Cathode: Sn4+ + 2e → Sn2+ Eo = V
Eo =+0.15V
Sn4+/Sn2+ Eo
= V.
Zn/Zn2+
Zn + Sn →
Zn Sn2+
Eocell = Eo
→ Eo
Eocell= Eored + Eoox
Sn4+/Sn2+
Zn/Zn2+
= (+0.15) + (0.76)
= – (-0.76 ) Or
= V
= V
spontaneous

43. Predict : Spontaneous or non-spontaneous?
Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g)

44. Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g)
Reduction
Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g)
Oxidation
cathode: Pb2+(aq) + 2e → Pb(s)
Eo = V
anode: 2Cl-(aq) → Cl2(g) + 2e
Eoox = -1.36
Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g)
Eocell= Eored + Eoox
= (-1.36) + (-0.13)
= V
Non-spontaneous
No Reaction

45. standard reduction potential
Example :
Predict whether the following reactions occur spontaneously :
2Ag(s) + Br2(aq) Ag+(aq) Br-(aq)
EAg /Ag = +0.8 V + 2
EBr /Br = V -
standard reduction potential
Answer :
2Ag(s) Ag+(aq) e Eθox = V
Br2(aq) e Br -(aq) Eθ = V
2Ag(s) + Br2(aq) Ag+(aq) + 2Br-(aq) Esel = V
The reaction is spontaneous

46. Exercise E0cell = E0cathode - E0anode
A cell consists of silver and tin in a solution of 1 M
silver ions and tin (II) ions. Determine the spontaneity
of the reaction and calculate the cell voltage of this
reaction.
Ag+ (aq) + e- → Ag (s)
Sn2+ (aq) + 2e- → Sn (s)
E0 = V
E0 = V
(cathode)
(anode)
E0cell = E0cathode - E0anode
= – (-0.14)
= V
E0cell = +ve ( reaction is spontaneous)

47. Nernst equation
Nernst equation can be used to calculate the E cell for any chosen concentration :
Ecell = Eocell – RT ln [ product ]x
nF [ reactant]y
At 298 K and R = J K-1 mol-1 , 1 F = C
Ecell = Eocell – [ product ]x
n [ reactant]y
2.303 log

48. = Ecell = Eocell – 0.0592 [ product ]x n [ reactant]y log [ product ]xQ =
Ecell = Eocell – n
log Q
n = no of e- that are involved Q = reaction quotient

49. Example 1 Calculate the Ecell for the following cell
Zn(s) / Zn2+ (aq, 0.02M) // Cu2+(aq, 0.40 M) / Cu(s)
Answer
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Eocell = Eored + Eoox @
= V V
= V
Eocell = Eocathode – Eoanode
= V - ( V)
= V

50. E = Eo – log [ Zn2+]
n [ Cu2+]
E = V – log (0.02 )
( 0.40)
= V – ( )
= V

51. At equilibrium:
~ No net reaction occur (Q=K)
~ Ecell = 0
Ecell = Eocell – n
log K
0 = Eocell – n
log K
Eθcell = n
log K

52. Example 2
Calculate the equilibrium constant (K) for the following reaction.
Cu(s) Ag+(ak) Cu2+(ak) Ag(s)
Answer
At equilibrium, E cell = 0
Eocell = Eo cathode - Eo anode
= – ( +0.34)
= V

53. Ecell = Eocell – log K 2
0 = 0.46 – log K 2
log K = 0.46 2
log K = 15.54
K = x 1015

54. Electrolysis Electrolytic Cell
Electrolysis is a chemical process that uses electricity
for a non-spontaneous redox reaction to occur.
Such reactions take place in electrolytic cells.
Electrolytic Cell
It is made up of 2 electrodes immersed in an
electrolyte.
A direct current is passed through the electrolyte
from an external source.
Molten salt and aqueous ionic solution are commonly
used as electrolytes.

55. Electrolytic Cell + - Oxidation Reduction Electrolyte (M+X-) Anion
X-,OH- M+,H+
Anion
Cation

56. Electrons flow from anode to cathode
Positive electrode
The electrode which is connected to the
positive terminal of the battery
Anode
Oxidation takes place
Electrons flow from anode to cathode
Negative electrode
The electrode which is connected to the
negative terminal of the battery
Cathode
Reduction takes place

57. Electrode
as circuit connectors
as sites for the precipitation of insoluble
products
example: Platinum , Graphite (inert electrode)
Electrolyte
a liquid that conducts electricity due to
the presence of +ve and –ve ions
must be in molten state or in aqueous
solution so that the ions can move freely
example: KCl(l), HCl(aq), CH3COOH(aq)

58. Comparison between an electrochemical cell and an electrolytic cell+ - e- - + e-
Anode
Anode
Cathode
Cathode

59. Electrolytic Cell Electrochemical Cell Cathode = negative
Cathode = positive
Anode = positive
Anode = negative
Non-spontaneous redox
reaction requires energy
to drive it
Spontaneous redox
reaction releases energy
Similarities:
Oxidation occurs at anode, reduction occurs
at cathode
Anions move towards anode, cations move towards cathode.
Electrons flow from anode to cathode in an external circuit.

60. Electrolysis of molten salt
Electrolysis of molten salt requires high temp.
Electrolysis of molten NaCl
Cation : Na+
Anion : Cl-
Anode :
Cl- (l) → Cl2(g) + 2e-
Cathode :
Na+ (l) + e- → Na (s)
2Na+ (l) + 2e- → 2Na (s)
Overall :
2Na+ (l) + 2Cl-(l) → Cl2(g) + 2Na(s)

61. Electrolysis of molten NaCl gives sodium metal
deposited at cathode and chlorine gas evolved at
anode.
Electrolysis of molten NaCl is industrially important.
The industrial cell is called ‘Downs Cell’

62. Electrolysis of Aqueous Salt
Electrolysis of aqueous salt is more complex
because the presence of water.
Aqueous salt solutions contains anion, cation and water.
Water is an electro-active substance that may be
oxidised or reduced in the process depending on
the condition of electrolysis.
Reduction :
2H2O (l) + 2e H2 (g) + 2OH- (aq)
E0 = V
Oxidation :
2H2O (l) H+ (aq) + O2 (g) + 4e-
E0 = V

63. Factors influencing the products :
Predicting the products of electrolysis
Factors influencing the products :
1. Reduction/oxidation potential of the species in electrolyte
Concentrations of ions
Types of electrodes used – active or inert

64. Electrolysis of Aqueous NaCl
NaCl aqueous solution contains Na+ cation, Cl- anion
and water molecules
On electrolysis,
the cathode attracts Na+ ion and H2O molecules
the anode attracts Cl- ion and H2O molecules
The electrolysis of aqueous NaCl depends on the
concentration of electrolyte.

65. Electrolysis of diluted NaCl solution
Cathode
Na+ (aq) + e Na (s)
E0 = V
2H2O (l) + 2e H2 (g) + 2OH- (aq)
E0 = V
E0 for water molecules is more positive.
H2O easier to reduce.
Anode
Cl2 (g) + 2e Cl- (aq)
E0 = V
O2 (g) + 4H+ (aq) + 4e H2O (l)
E0 = V
In dilute solution, water will be selected for oxidation because of its lower Eo.

66. Reactions involved 2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V
Cathode:
4H2O (l) + 4e H2 (g) + 4OH- (aq)
Anode:
2H2O (l) O2 (g) + 4H+ (aq) + 4e-
E0 = V
Cell
reaction:
6H2O(l) O2(g) + 2H2(g) + 4OH-(aq) + 4H+(aq)
4 H2O
2H2O(l) O2(g) + 2H2(g)
E0cell = V

67. Electrolysis of Concentrated NaCl solution
Cathode
Na+ (aq) + e Na (s)
E0 = V
2H2O (l) + 2e H2 (g) + 2OH- (aq)
E0 = V
E0 for water molecules is more positive
H2O easier to be reduce
Anode
Cl2 (g) + 2e Cl- (aq)
E0 = V
O2 (g) + 4H+ (aq) + 4e H2O (l)
E0 = V
In concentrated solution, chloride ions will be oxidised because of its high concentration.

68. Reactions involved 2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V
Cathode:
2H2O (l) + 2e H2 (g) + 2OH- (aq)
E0 = V
Anode:
2Cl- (aq) Cl2 (g) + 2e-
E0 = V
Cell reaction:
2H2O(l) + 2Cl Cl2(g) + H2(g) + 2OH-(aq)
E0cell = V

69. Exercise Solution Predict the electrolysis reaction when
Na2SO4 solution is electrolysed using platinum electrodes.
Solution
Na2SO4 aqueous solution contains Na+ ion, SO42- ion
and water molecules
On electrolysis,
the cathode attracts Na+ ion and H2O molecules
the anode attracts SO42- ion and H2O molecules

70. Cathode Anode Na+ (aq) + e- Na (s) E0 = -2.71 V
2H2O (l) + 2e H2 (g) + 2OH- (aq)
E0 = V
E0 for water molecules is more positive
H2O easier to reduce
Anode
S2O82- (aq) + 2e SO42- (aq)
E0 = V
O2 (g) + 4H+ (aq) + 4e H2O (l)
E0 = V
E0 for water molecules is less positive
H2O easier to oxidise

71. Equation 2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V
Cathode:
2H2O (l) + 2e H2 (g) + 2OH- (aq)
E0 = V
Anode:
2H2O (l) O2 (g) + 4H+ (aq) + 4e-
E0 = V
Cell Reaction:
2H2O(l) O2(g) + 2H2(g)
E0cell = V
Cathode = H2 gas is produced and solution become
basic at cathode because OH- ions are formed
Anode = O2 gas is produced and solution become
acidic at anode because H+ ions are formed

72. Faraday’s Law of Electrolysis
Describes the relationship between the amount of
electricity passed through an electrolytic cell and
the amount of substances produced at electrode.
Faraday’s First Law
States that the quantity of substance formed at an
electrode is directly proportional to the quantity of
electric charge supplied.

73. m α Q Faraday’s 1st Law Q = electric charge in coulombs (C)
m = mass of substance discharged

74. The End

75. is the charge on 1 mole of electron
Q = It
Q = electric charge in coulombs (C)
I = current in amperes (A)
t = time in second (s)
Faraday constant (F)
is the charge on 1 mole of electron
1 F = C

76. Example Answer An aqueous solution of CuSO4 is electrolysed using a
current of A for 5 hours. Calculate the mass
of copper deposited at the cathode.
Answer
Electric charge, Q = Current (I) x time (t)
Q = (0.150 A) x ( 5 x 60 x 60 )s
Q = 2700 C
1 mole of electron Ξ 1 F Ξ C
2700
96 500
No. of e- passed through =
= mol

77. Cu2+ (aq) + 2e-  Cu (s)
From equation:
2 mol electrons  1 mol Cu
0.028 mol electrons  mol Cu
Mr for Cu = 63.5
Mass of Copper deposited = x 63.5
= g