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Chapter 10ElectrochemistryElectrochemistry
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Electrochemistry
Is the study of the relationship between electricityand chemical reaction
Chemical reactions involved in electrochemistry are :Chemical reactions involved in electrochemistry are :
Oxidation
ReductionREDOX REACTION
One type of reaction cannot occur withoutOne type of reaction cannot occur without
the other.the other.
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REDUCTION
gain of electron
Oxidation no. decrease
Reaction at cathode
RememberRED CATRED CAT
= REDREDuctionatCATCAThode
Example:
Cu2+ + 2e- p Cu
Oxidation no.q
OXIDATIONloss of electron
Oxidation no. increase
Reaction at anodeExample:
Mgp Mg2+ +2e-
Oxidationno.o
REDOX Reaction
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Reduction :
Oxidation :
Cu2+(aq) + 2e- p Cu(s)
Zn(s) p Zn2+(aq) +2e-Half-cellreaction
Cu2+(aq) + Zn(s) p Cu(s) + Zn2+(aq)
Example
Overall cell
reaction :
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
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CellsThere are 2 type of cells
Electrochemical
CellsElectrolytic
Cells
where chemical reaction
produces electricity
Uses electricity to
produce chemicalreaction
ChemicalEnergy
ElectricalEnergy
ElectricalEnergy
ChemicalEnergy
Also called;
Galvanic cell or Voltaic cell
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Component and Operation ofGalvanic cell
Consists of :Consists of :
1) Zn1) Zn metal in an aqueous solution of Zn2+
2) Cu2) 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.
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Znelectrode
Cuelectrode
Salt
bridge
Zn2+ Cu2+
Voltmeter
Galvaniccell
ZnSO4(aq)solution
CuSO4(aq)solution
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What happens at the zinc electrode ?
Zinc is more electropositive than copper. Tendency to release electrons: Zn > Cu.
Zn (s) p 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
electronsp
anode.
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Cu2+(aq) + 2e- p Cu(s)
Copper is deposited. Reduction occurs at the Cu electrode.
Cu is the +ve electrode p cathode
What happens at the copper electrode ?
The electron from the Zn metal moves out through
the wire enter the Cu metal
Cu
2+
ions from the solution accept electrons.
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Reactions Involved:
Anode :Cathode :
Zn (s) p Zn2+(aq) + 2e-Cu2+(aq) + 2e- p Cu(s)
Zn (s) + Cu2+(aq) p Zn2+(aq) + Cu(s)Overall cellreaction :
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Functions
Salt bridge helps to maintain electrical neutrality
Completes the circuit by allowing ions carrying chargeto move from one half-cell to the other.
Salt bridge
An inverted U tube containing a gelAn inverted U tube containing a gelpermeated with solution of anpermeated with solution of an inertelectrolyte such as KCl, Naelectrolyte such as KCl, Na22SOSO44, NH, NH44NONO33..
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What happened if there is no salt bridge?
ZnCu
eZn2+ Cu2+
V
e
e
e
e
e
ZnSO4(aq) 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
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Salt bridge
(KCl)
Zn2+
- +Zn
ZnSO4(aq) CuSO4(aq)
Cu
ee
Zn (s) p Zn2+(aq) + 2e- Cu2+(aq) + 2e- p Cu(s)
ANODE(-) CATHODE(+)
Cu2+
E = +1.10 V
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How does the cell maintains its electrical neutrality?
Left Cell Right Cell
Cl- ions from saltbridge move into Zn
half cell
K+ ions from saltbridge move into Cu
half cell
Electrical neutrality is maintained
Zn (s) p Zn2+(aq) + 2e- Cu2+(aq) + 2e- p Cu(s)
Zn2+ ions enter the solution.
Causing an overall excess oftve charge.
Cu2+ ions leave the solution.
Causing an overall excess of-ve charge.
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Electrochemical Cells
19.2
spontaneousredox reaction
anodeoxidation
cathodereduction
half
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Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s)
anode cathode
Alsocan be represented as:
Cell notation
Phase boundarySalt bridge
Cell notation
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Exercise
For the cell below, write the reaction at anodeand cathode and also the overall cell reaction.
Zn (s) | Zn2+(aq) || Cr3+ (aq) | Cr (s)
Zn(s) Zn2+(aq) + 2e-
Cr3+(aq) + 3e- Cr(s)
3Zn(s) +2Cr3+(aq) + 3Zn2+ (aq) +2Cr(s)
Cathode :
Anode :
Overall cellreaction:
X 2
X 3
Cell notation
3 3
2 26e
6e-
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The difference in electrical potential between the anode
and cathode is called:
cell voltage
electromotive force (emf)
cell potential
Acts as electrical pressure that
pushes electron through the wire.
measured by a voltmeter
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Electrode PotentialElectrode Potential
A measure of the ability of a halfA measure of the ability of a half--cell to attract electrons towardsit.cell to attract electrons towardsit.
Cu2+(aq) +2e Cu(s) Eored= +0.34 V
Zn2+(aq) +2e Zn(s) Eored = -0.76 V
Standard reduction potential of copper half-cell is
more positive compared to zinc.
Standard reduction
potential
The more positive the half-cells electrode potential, the
stronger the attraction for electrons.
Tendency for reduction
(cathode)
Zinc half-cell becomesanode.
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Cell Potential (Eocell)= Eocatode E
oanode
= +0.34 (-0.76)
= +1.1 V
Cu2+(aq) + 2e Cu(s) Eored = +0.34 V
Zn
2+
(aq) + 2e Z
n(s) Eo
red = -0.76 V
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Zn2+(aq) + 2e- p Zn (s)
Cu2+
(aq) + 2e-
p Cu(s)
E0 = -0.76V
E0
= +0.34V
E0cell = E0
cathode - E0
anode
= +0.34 (-0.76)
= +1.10 Vor E0cell = E0red + E0ox
= +0.34+ (+0.76)
= +1.10 V
Change the sign
Half-cell equation at:
Anode :
Cathode :
Zn (s) p Zn2+(aq) + 2e-
Cu
2+
(aq) + 2e- p
Cu(s)
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StandardElectrode Potentials(Eo)
at 25oC, the pressure is1 atm (for
gases), and the concentrationof
electrolyte is1M.
A measure of the ability of half-cell
to attract electrons towardsit
The signofE0 changes when the
reactionis reversed
Changing the stoichiometriccoefficientsof a half-cell reaction
doesnot change the value ofE0
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For example:
Cl2(g) + 2e- p 2Cl-(aq) E0 = +1.36 V
Cl2(g) + e- p Cl-(aq) E0 = +1.36 V
Cl-(aq)p Cl2(g) + e- E0 = -1.36 V
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Standard HydrogenElectrode (SHE)
Made up of a platinum electrode, immersedin an
aqueoussolutionof H+ (1 M) and bubbled with
hydrogengas at 1 atm pressure, and temperature at
25oC
PtelectrodeH+(aq)
1M
H2 gasat1atm
The standard reductionof SHEis 0 V
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Standard reduction potential of Zinc half cell ismeasured by setting up the electrochemical
cell as below.
H2(g), 25oC,1 atm.
Zn
eZn2+
ZnSO4(aq)
1 M
H+(aq),1 M
V
ee
e
Pt
E0 = 0E0 = +0.76- +
-+
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Standard Electrode Potentials
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
2e- + 2H+ (aq,1 M) H2 (g,1 atm)
Zn (s) Zn2+ (1 M) + 2e-Anode (oxidation):
Cathode (reduction):
Zn (s) + 2H+ (aq,1 M) Zn2+(aq) + H2 (g,1 atm)Cell reaction
0.76 V = 0 - EZn /Zn0
2+
EZn /Zn = -0.76 V0
2+
Zn2+ + 2e- Zn E0= -0.76 V
E0 = EH /H
- EZn /Zncell
0 0+ 2+
2
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Standard reduction potential of Copper half cell ismeasured by setting up the electrochemical
cell as below.
H2(g) 25oC1 atm.
Cu
Cu2
CuSO4(aq)
1M
H+(aq)
1 M
V
Pt
E0 = 0+ -
+ -
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Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
2e- + Cu2+ (1 M) Cu (s)
H2 (1 atm) 2H+ + 2e-Anode (oxidation):
Cathode (reduction):
H2 + Cu2+ Cu (s) + 2H+
E0 = Ecathode - Eanodecell0 0
Ecell= ECu /Cu EH /H2+ + 20 0 0
0.34 = ECu /Cu - 00
2+
ECu /Cu = 0.34 V2+0
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The direction of half-reaction of SHE depends on the
otherhalf-cell connected on it.
The cell notation for SHE is either:
Pt(s)| H2(g)| H+ (aq)whenitisanode
H+(aq)| H2(g)| Pt(s) whenitiscathode
In either case, E0 of SHE remains 0
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Zn2+(aq) + 2e- p Zn (s)
Cu2+
(aq) + 2e- p
Cu(s)
E0 = -0.76V
E0 =
+0.34V
E0cell = E0
cathode - E0
anode
= +0.34 (-0.76)
= +1.10 Vor E0cell = E0red + E0ox
= +0.34+ (+0.76)
= +1.10 V
Change the sign
Half-cell equation at:
Anode :
Cathode :
Zn (s) p Zn2+(aq) + 2e-
Cu2+(aq)
+
2e
- p Cu(s)
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At standard-state condition
E0cell = E0red + E
0ox
E0cell = E0cathode - E
0anode
or
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ExerciseCalculate the standard cell potential of the following
electrochemical cell.Co(s) | Co2+(aq) || Ag+(aq) | Ag(s)
AnswerCathode (Red) :
Anode (Ox) : Co(s) p Co2+(aq) + 2e-
Ag+(aq) + e- p Ag(aq) E0 =+0.80V
E0ox =+0.28V
E0cell = E0
cathode - E0
anode
= +0.80 (-0.28)
= +1.08 V
Ag+(aq) + e- p Ag(aq)
Co2+(aq) + 2e-p Co(s) E0 = -0.28V
E0 =+0.80V
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Oxidationagent leftofthe halfcellequation
Reductionagent rightofthe halfcellequationExample :
Oxidationagent
Reducingagent
Refer to the list of Standard Reduction Potential:
Ag+ (aq) + e- Ag (s) E0 =+0.80 V
Cu2+ (aq) +2e- Cu(s) E0 =+0.34 VNi2+ (aq) +2e- Ni (s) E0 = -0.25 V Increase
strength as
reducingagent
The more +ve the value of E0 the strongerthe oxidizing agent
The more -ve the value of E0 the strongerthe reducing agent
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Arrange the 3 elements in order of increasingstrength of reducing agents
X3+ + 3e- p X E0 = -1.66 V
Y2+ + 2e- p Y E0 = -2.87 V
L2+ + 2e- p L E0 = +0.85 V
Answer :L < X < Y
Exercise
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Calculate the E0 cell for the reaction ::
Mg(p) | Mg2+
(ak) || Sn4+
(ak),Sn2+
(ak) | Pt(p)
Given :
Mg2+(ak) + 2e Mg(p) E = -2.38 V
Sn4+(ak) + 2e Sn2+
(ak) E= +0.15 V
Oxidation : Mg(p) Mg2+
(ak) + 2e Eoox = +2.38 V
Reduction : Sn4+(ak) + 2e Sn2+
(ak) Eo = +0.15 V
Mg(p) + Sn4+
(ak) Mg2+
(ak) + Sn2+
(ak) Ecell = +2.53 V
Example
Ecell = Eo red + Eo ox
E0cell = Ecathode - Eanode
=+0.15- (-2.38)
=+2.53V
= +2.38 + 0.15
= +2.53 V
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Exercise
A cell is set up between a chlorine electrode and ahydrogen electrode
(a) Draw a diagram to show the apparatus and chemicals
used.(b) Discuss the chemical reactions occurring in theelectrochemical cell.
Pt | H2(g, 1 atm) | H+(aq, 1M) || Cl2(g, 1atm) | Cl-(aq, 1M) | Pt
E0cell = +1.36 V
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Answer
H2(g),
1 atm.
Pt
H+(aq), 1M
Pt
E0cell =1.36V
- +
V- + Cl2(g),1 atm.
Cl-(aq), 1M
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1. Show the processoccur at anode and
cathode
2. Overall reaction
- Half-cell reaction
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Answer Reduction (cathode)
Cl2(g) + 2e- p
2Cl-
(aq) Oxidation(anode)
H2 (g) p 2H+ (aq) + 2e-
Eocell =+1.36 V
E0 = 0
Eocell = Eocathode - E0anode
+1.36= Eocathode 0
E0cathode =+1.36 VSo the standard reduction potential forCl2 is: Eo =+1.36 V
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Spontaneous & NonSpontaneous & Non--Spontaneous reactionsSpontaneous reactions
-- Redox reaction is spontaneous whenRedox reaction is spontaneous when
EEcellcell is +ve.is +ve.
-- Non spontaneous is when ENon spontaneous is when Ecellcell isis ve.ve.
E cell = 0 The reaction is at equilibrium
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Sn4+/Sn2+
Predict whether the following reactionsoccurPredict whether the following reactionsoccur
spontaneously ornonspontaneously ornon--spontaneously understandardspontaneously understandard
condition.condition.
Zn+ SnZn+ Sn4+ 4+
SnSn2+2+
+Zn+Zn2+2+
The two halfThe two half--cellsinvolved are:cellsinvolved are:--
Anod : ZnAnod : Zn ZnZn2+2+ +2e E+2e Eoooxox= +0.76 V= +0.76 V
Cathode: SnCathode: Sn4+4+ +2e+2e SnSn2+2+ EEoo = +0.15 V= +0.15 V
Zn + Sn4+ Zn2+ + Sn2+
Eocell = Eo
= +0.91 V
spontaneous
Zn/Zn2
+
Eo
= +0.15 (-0.76 )
Eocell= Eo
red +Eoox
= (+0.15) +(0.76)
= +0.91 VOr
Sn4+/Sn2+Eo =+0.15V
EoZn/Zn2+
= - 0.76V.
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PbPb2+(2+(aq) + 2Claq) + 2Cl--(aq) Pb(s) + Cl(aq) Pb(s) + Cl22(g)(g)
Predict : Spontaneous or nonPredict : Spontaneous or non--
spontaneous?spontaneous?
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PbPb2+2+(aq) +2Cl(aq) +2Cl--(aq) Pb(s) + Cl(aq) Pb(s) + Cl22(g)(g)Pb2+(aq) +2Cl-(aq) Pb(s) + Cl2(g)
cathode: Pb2+(aq) +2e Pb(s)
anode: 2Cl-(aq) Cl2(g) + 2e
Pb2+(aq) +2Cl-(aq) Pb(s) + Cl2(g)
Eo= -0.13 V
Eoox = -1.36
Eocell= Eo
red +Eoox
Reduction
Oxidation
= (-1.36) +(-0.13)
= -1.48 V
Non-spontaneous
NoReaction
l
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Example :
2Ag(s) 2Ag+
(aq) + 2e Eox = - 0.80 V
Br2(aq) + 2e 2Br-(aq) E
= +1.07 V
2Ag(s) + Br2(aq) 2Ag+
(aq) + 2Br-(aq) Esel = + 0.27 V
The reaction is spontaneous
Answer :
Predict whether the following reactions occurspontaneously :
2Ag(s) + Br2(aq) 2Ag+
(aq) + 2Br-(aq)
E Ag /Ag= +0.8 V+0
2EBr /Br = +1.07 V
0-
standard reduction
potential
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Exercise
A cell consists of silver and tin in a solution of 1 Msilver ions and tin (II) ions. Determine the spontaneityof the reaction and calculate the cell voltage of thisreaction.
Ag+ (aq) + e- Ag (s)
Sn2+ (aq) +2e- Sn (s)
E0 =+0.80 VE0 = -0.14 V
(cathode)
(anode)
E0
cell = E0
cathode - E0
anode
=+0.80 (-0.14)=+0.94 V
E0cell
= +ve ( reaction is spontaneous)
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Nernst equation
Nernst equation can be used to calculate the E cell
for any chosen concentration :
Ecell = Eo
cell RT ln [ product ]x
nF [ reactant]y
At 298 K and R = 8.314 J K-1 mol-1 , 1 F = 96500 C
Ecell = Eocell 0.0257 [ product ]x
n [ reactant]y2.303 log
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Ecell = Eocell 0.0592 [ product ]
x
n [ reactant]ylog
Ecell = Eo
cell 0.0592
n
log Q
n = no of e- that are involvedQ = reaction quotient
[ product ]x
[ reactant]yQ=
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Example 1
Calculate the Ecell for the following cell
Zn(s) / Zn2+ (aq, 0.02M) // Cu2+(aq, 0.40 M) / Cu(s)
Answer
n(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Eocell = Eored + Eoox @
= +0.34 V + 0.76 V
= +1.10 V
Eocell = Eocathode Eoanode
= +0.34 V - (- 0.76 V)
= +1.10 V
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E = Eo 0.0592 log [ Zn2+]
n [ Cu2+
]
E = +1.10 V 0.0592 log (0.02 )
2 ( 0.40)
= +1.10 V (-0.0385)
= +1.139 V
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At equilibrium:
~ No net reaction occur (Q=K)
~ Ecell = 0
Ecell = Eocell 0.0592
n
log K
0 = Eocell 0.0592
n
log K
Ecell = 0.0592
nlog K
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Example 2
Calculate the equilibrium constant (K) for thefollowing reaction.
Cu(s) + 2Ag+
(ak) Cu2+
(ak) + 2Ag(s)
At equilibrium, E cell = 0
Eocell = Eo cathode - Eo anode
= +0.80 ( +0.34)
= +0.46 V
Answer
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Ecell = Eocell 0.0592 log K
2
0 = 0.46 0.0592 log K
2
0.0592 log K= 0.46
2
log K= 15.54
K= 3.467 x 1015
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Electrolysis
Electrolysis is a chemical process that uses electricityfor a non-spontaneous redox reaction to occur.Such reactions take place in electrolyticcells.
Electrolytic Cell It is made up of 2 electrodes immersed in an
electrolyte.
A direct current is passed through the electrolytefrom an external source.
Molten salt and aqueous ionic solution are commonlyused as electrolytes.
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Anion Cation
Oxidation Reduction
Electrolytic Cell
Electrolyte
(M+
X-
)X-,OH- M+,H+
+ -
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A
n
od
e
C
a
t
ho
d
e
Positive electrode
The electrode which is connected to thepositive terminal of the battery
Oxidation takes place
Negative electrode
The electrode which is connected to thenegative terminal of the battery
Reduction takes place
Electronsflow fromanodetocathode
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Electrode
as circuit connectors as sites for the precipitation of insoluble
products example: Platinum , Graphite (inert electrode)
Electrolyte
a liquid that conducts electricity due tothe presence of +ve and ve ions must be in molten state or in aqueous
solution so that the ions can move freelyexample: KCl(l), HCl(aq), CH3COOH(aq)
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Comparison between an electrochemical celland an electrolytic cell
CathodeAnode
- +e- e-
Anode Cathode
+ -
e- e-
+ -
Electrolytic Cell Electrochemical Cell
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Electrolytic Cell Electrochemical Cell
Cathode
=negative
Anode = positive
Cathode
=positive
Anode = negative
Non-spontaneous redoxreaction requires energy
to drive it
Spontaneous redoxreaction releases energy
Oxidation occurs at anode, reduction occursat cathode
Anions move towards anode, cations movetowards cathode. Electrons flow from anode to cathode in anexternal circuit.
Similarities:
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Electrolysis of molten salt
Electrolysis of molten salt requires high temp. Electrolysis of molten NaCl
Cation : Na+
Anion : Cl-
Anode:
Cathode: Na+ (l) + e- Na (s)
Cl- (l) Cl2(g) + 2e-
Overall: 2Na+ (l) + 2Cl-(l) Cl2(g) + 2Na(s)
2Na+ (l) + 2e- 2Na (s)
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Electrolysis of molten NaCl gives sodium metaldeposited at cathode and chlorine gas evolved at
anode.Electrolysis of molten NaCl is industrially important.The industrial cell is called Downs Cell
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Electrolysis of Aqueous Salt Electrolysis of aqueous salt is more complexbecause 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 :
Oxidation :
2H2O (l) + 2e- H2 (g) + 2OH
- (aq)
2H2O (l) 4H+ (aq) + O2 (g) + 4e
-
E0 = -0.83 V
E0 = -1.23 V
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Predicting the productsof electrolysis
Factors influencing the products :
1. Reduction/oxidation potential of the
speciesin electrolyte2. Concentrationsofions
3. Typesof electrodesused active or
inert
l l f A l
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Electrolysis of Aqueous NaCl
The electrolysis of aqueous NaCl depends on the
concentration of electrolyte.
NaCl aqueous solution contains Na+ cation, Cl- anionand water molecules
On electrolysis,
the cathode attracts Na+ ion and H2O molecules
the anode attracts Cl- ion and H2O molecules
El t l f d l t d N Cl l t
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Cathode
2H2O (l) + 2e- H2 (g) + 2OH
- (aq) E0 = -0.83 V
Na+ (aq) + e- Na (s) E0 = -2.71 V
E0 for water molecules is more positive.
H2O easier to reduce.
Electrolysis of diluted NaCl solution
AnodeCl2 (g) + 2e
- 2Cl- (aq) E0 =+1.36 V
O2 (g) + 4H+ (aq) + 4e- 2H2O (l) E0 =+1.23 V
In dilute solution, water will be selected foroxidation because of its lower Eo.
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Reactions involved
2H2O (l) + 2e-
H2 (g) + 2OH-
(aq) E0
= -0.83 VCathode:
Anode:
Cellreaction:
6H2O(l) O2(g) + 2H2(g) + 4OH-(aq) + 4H+(aq)
E0cell = -2.06 V
2H2O (l) O2 (g) + 4H+ (aq) + 4e- E0 = -1.23 V
4H2O (l) + 4e- 2H2 (g) + 4OH
- (aq)
4 H2O
2H2O(l) O2(g) + 2H2(g)
El t l i f C t t d N Cl l ti
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Electrolysis ofConcentrated NaCl solutionCathode
2H2O (l) + 2e- H2 (g) + 2OH
- (aq) E0 = -0.83 V
Na+ (aq) + e- Na (s) E0 = -2.71 V
E0 for water molecules is more positive
H2O easier to be reduceAnode
Cl2 (g) + 2e- 2Cl- (aq) E0 =+1.36 V
O2 (g) + 4H+ (aq) + 4e- 2H2O (l) E0 =+1.23 V
In concentrated solution, chloride ions will beoxidised because of its high concentration.
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Reactions involved
2H2O (l) + 2e- H2 (g) + 2OH- (aq) E0 = -0.83 V
2Cl- (aq) Cl2 (g) + 2e-
E0 = -1.36 V
Cellreaction:
2H2O(l) + 2Cl- Cl2(g) + H2(g) + 2OH-(aq)
E0cell = -2.19 V
Cathode:
Anode:
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Na2SO4 aqueous solution contains Na+ ion, SO42- ionand water molecules
On electrolysis, the cathode attracts Na+ ion and H2O molecules
the anode attracts SO42- ion and H2O molecules
Exercise
Predict the electrolysis reaction whenNa2SO4 solution is electrolysed using platinum electrodes.
Solution
C h d
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Anode
E0 for water molecules is less positiveH
2
O easier to oxidise
S2O82- (aq) + 2e- 2SO4
2- (aq) E0 =+2.01 V
O2 (g) + 4H+ (aq) + 4e- 2H2O (l) E0 =+1.23 V
Cathode
2H2O (l) + 2e- H2 (g) + 2OH
- (aq) E0 = -0.83 V
Na+ (aq) + e- Na (s) E0 = -2.71 V
E0 for water molecules is more positiveH2O easier to reduce
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Cathode = H2 gas is produced and solution becomebasic at cathode because OH- ions are formed
Anode = O2 gas is produced and solution becomeacidic at anode because H+ ions are formed
Equation
Cathode:
Anode: E0 = -1.23 V
2H2O (l) + 2e- H2 (g) + 2OH
- (aq) E0 = -0.83 V
2H2O (l) O2 (g) + 4H+ (aq) + 4e-
Cell
Reaction:E0cell = -2.06 V2H2O(l) O2(g) + 2H2(g)
d f l l
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Faradays Law of Electrolysis
Describes the relationship between the amount ofelectricity passed through an electrolytic cell andthe amount of substances produced at electrode.
Faradays First LawStates that the quantity of substance formed at an
electrode is directly proportional to the quantity ofelectric charge supplied.
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Faradays 1st Law
m Q
Q= electric charge in coulombs (C)
m = mass of substance discharged
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The EndThe End
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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 = 96 500 C
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Example
An aqueous solution of CuSO4 is electrolysed using acurrent of 0.150 A for 5 hours. Calculate the massof 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 =2700C
1 mole of electron 1 F 96500C
No. of e- passed through =270096500
=0.028 mol
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Cu2+ (aq) + 2e- p Cu(s)
From equation:2 mol electrons p 1 mol Cu
0.028 mol electrons p 0.014 mol Cu
Mr for Cu=63.5Mass of Copper deposited =0.014 x 63.5
=0.889 g
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