Chapter 10 Electrochemistry
description
Transcript of Chapter 10 Electrochemistry
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Chapter 10
ElectrochemisElectrochemistrytry
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ElectrochemistryIs the study of the relationship between electricity
and chemical reaction
Chemical reactions involved in electrochemistry are :Chemical reactions involved in electrochemistry are :
OxidationReduction
REDOX REACTION
One type of reaction cannot One type of reaction cannot occur without the other.occur without the other.
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REDUCTION gain of electron Oxidation no. decrease Reaction at cathode
Remember…RED CATRED CAT
= REDREDuction at CATCAThode
Example:Cu2+ + 2e- Cu
Oxidation no.
OXIDATION loss of electron Oxidation no. increase Reaction at anode Example:Mg Mg2+ + 2e-
Oxidation no.
REDOX Reaction
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Reduction :
Oxidation :
Cu2+(aq) + 2e- Cu(s)
Zn(s) Zn2+(aq) + 2e-
Half-cellreaction
Cu2+(aq) + Zn(s) Cu(s) +
Zn2+(aq)
Example
Overall cellreaction :
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
ElectrochemicalCells
ElectrolyticCells
where chemical reaction produces electricity
Uses electricity to produce chemical reaction
ChemicalEnergy
ElectricalEnergy
ElectricalEnergy
ChemicalEnergy
Also called; Galvanic cell or Voltaic cell
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Component and Operation of Galvanic cell
Consists of :Consists of :1) Zn1) Zn metal in an aqueous solution of Zn2+
2) Cu 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.
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Zn electrode
Cu electrod
e
Salt bridge
Zn2+ Cu2+
Voltmeter
Galvanic cell
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) 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.
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Cu2+ (aq) + 2e- Cu (s) Copper is deposited. Reduction occurs at the Cu electrode. Cu is the +ve electrode cathode
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.
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Reactions Involved:
Anode :Cathode :
Zn (s) Zn2+ (aq) + 2e-
Cu2+ (aq) + 2e- Cu (s)Zn (s) + Cu2+ (aq) 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 charge to move from one half-cell to the other.
Salt bridgeAn inverted U tube containing a An inverted U tube containing a gel permeated with solution of an gel permeated with solution of an inert electrolyte such as electrolyte such as KCl, KCl, NaNa22SOSO44, NH, NH44NONO33..
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What happened if there is no salt bridge?
Zn Cu
eZn2+ Cu2+
V
ee
e
ee
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) Zn2+ (aq) + 2e- Cu2+ (aq) + 2e- 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 salt bridge move into Zn
half cellK+ ions from salt
bridge move into Cu half cell
Electrical neutrality is maintained
Zn (s) Zn2+ (aq) + 2e- Cu2+ (aq) + 2e- Cu (s)Zn2+ ions enter the
solution.Causing an overall excess
of +ve charge.
Cu 2+ 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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Salt bridge
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s)anode cathode
Also can be represented as:
Cell notation
Phase boundary
Cell notation
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ExerciseFor the cell below, write the reaction at anode and 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 2X 3
Cell notation
3 32 26e
6e-
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Salt Bridge
Galvanic cell
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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 PotentialA measure of the ability of a half-cell to attract electrons towards it.A measure of the ability of a half-cell to attract electrons towards it.
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-cell’s electrode potential, the stronger the attraction for electrons.
Tendency for reduction ↑ (cathode)
Zinc half-cell becomes anode.
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Cell Potential (Eocell) = Eo
catode — Eoanode
= +0.34 – (-0.76)
= +1.1 V
Cu2+(aq) + 2e → Cu(s) Eored = +0.34 V
Zn2+(aq) + 2e → Zn(s) Eored = -0.76 V
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Zn2+ (aq) + 2e- Zn (s) Cu2+ (aq) + 2e- Cu (s)
E0 = -0.76V
E0 = +0.34V
E0cell = E0
cathode - E0anode
= +0.34 – (-0.76)= +1.10 V
or E0cell = E0
red + E0oxd
= +0.34 + (+0.76)= +1.10 V Change the sign
Half-cell equation at:
Anode :Cathode :
Zn (s) Zn2+ (aq) + 2e-
Cu2+ (aq) + 2e- Cu (s)
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Standard Electrode Potentials (Eo)
at 25oC, the pressure is 1 atm (for gases), and the concentration of electrolyte is 1M.
A measure of the ability of half-cell to attract electrons towards it
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
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For example:
Cl2(g) + 2e- 2Cl-(aq) E0 = +1.36 V
½Cl2(g) + e- Cl-(aq) E0 = +1.36 V
Cl-(aq) ½Cl2(g) + e- E0 = -1.36 V
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Standard Hydrogen Electrode (SHE)
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
PtelectrodeH+ (aq)
1 M
H2 gasat 1 atm
The standard reduction of SHE is 0 V
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Standard reduction potential of Zinc half cell is
measured 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 = 0 E0 = +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 /Zn 0 2+
EZn /Zn = -0.76 V0 2+
Zn2+ + 2e- Zn E0 = - 0.76 V
E0 = EH /H - EZn /Zn cell0 0+ 2+
2
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Standard reduction potential of Copper half cell is measured by setting up the electrochemicalcell as below.
H2 (g) 25oC 1 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 /H 2+ +2
0 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 other half-cell connected on it.
The cell notation for SHE is either:
Pt(s) | H2(g) | H+ (aq) when it is anode
H+(aq) | H2(g) | Pt(s) when it is cathode
In either case, E0 of SHE remains 0
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Zn2+ (aq) + 2e- Zn (s) Cu2+ (aq) + 2e- Cu (s)
E0 = -0.76V
E0 = +0.34V
E0cell = E0
cathode - E0anode
= +0.34 – (-0.76)= +1.10 V
or E0cell = E0
red + E0ox
= +0.34 + (+0.76)= +1.10 V
Change the sign
Half-cell equation at:
Anode :Cathode :
Zn (s) Zn2+ (aq) + 2e-
Cu2+ (aq) + 2e- Cu (s)
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At standard-state condition
E0cell = E0
red + E0ox
E0cell = E0
cathode - E0anode
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) Co2+(aq) + 2e-
Ag+(aq) + e- Ag(aq) E0 = +0.80VE0
ox = +0.28V
E0cell = E0
cathode - E0anode
= +0.80 – (-0.28)= +1.08 V
Ag+(aq) + e- Ag(aq)Co2+(aq) + 2e- Co(s) E0 = -0.28V
E0 = +0.80V
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Oxidation agent → left of the half cell equationReduction agent → right of the half cell equationExample :
Oxidation agent
Reducing agent
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 V
Ni2+ (aq) + 2e- → Ni (s) E0 = -0.25 V Increase
strength asreducing
agent
The more +ve the value of E0 → the stronger the oxidizing agent
The more -ve the value of E0 → the stronger the reducing agent
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Arrange the 3 elements in order of increasing strength of reducing agents
X3+ + 3e- X E0 = -1.66 V
Y2+ + 2e- Y E0 = -2.87 V
L2+ + 2e- 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) Eθ = -2.38 V
Sn4+(ak) + 2e → Sn2+
(ak) Eθ= +0.15 V
Oxidation : Mg(p) → Mg2+(ak) + 2e Eo
ox = +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
Eθcell = E o red + E o ox
E0cell = Ecathode - Eanode
=+0.15- (-2.38)
=+2.53V
= +2.38 + 0.15= +2.53 V
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ExerciseA cell is set up between a chlorine electrode and a
hydrogen electrode
(a) Draw a diagram to show the apparatus and chemicalsused.
(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 process occur at anode and cathode
2. Overall reaction
- Half-cell reaction
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Answer Reduction (cathode)
Cl2 (g) + 2e- 2Cl- (aq) Oxidation(anode)
H2 (g) 2H+
(aq) + 2e-
Eocell =+1.36 V
E0 = 0
Eocell = Eo
cathode - E0anode
+1.36 = Eocathode – 0
E0cathode = +1.36 V
So the standard reduction potential for Cl2 is:Eo = +1.36 V
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Spontaneous & Non-Spontaneous reactionsSpontaneous & Non-Spontaneous reactions - Redox reaction is spontaneous when- Redox reaction is spontaneous when EEcellcell is +ve. is +ve.
- Non spontaneous is when E- Non spontaneous is when Ecellcell is –ve. is –ve.
Eθ cell = 0 The reaction is at equilibrium
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Sn4+/Sn2+
Predict whether the following reactions occur Predict whether the following reactions occur spontaneously or non-spontaneously under standard spontaneously or non-spontaneously under standard condition.condition.
Zn + SnZn + Sn4+ → 4+ → SnSn2+2+ + Zn + Zn2+ 2+
The two half-cells involved are:-The two half-cells involved are:-Anod : Zn Anod : Zn → → Zn Zn2+2+ + 2e E + 2e Eoo
ox ox = +0.76 V= +0.76 VCathode: SnCathode: Sn4+4+ + 2e + 2e → → Sn Sn2+ 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 non-Predict : 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 VEo
ox = -1.36
Eocell= Eo
red + Eo
ox
Reduction
Oxidation
= (-1.36) + (-0.13)
= -1.48 VNon-spontaneous No Reaction
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Example :
2Ag(s) 2Ag+(aq) + 2e Eθ
ox = - 0.80 V
Br2(aq) + 2e 2Br -(aq) Eθ
= +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 occur spontaneously :
2Ag(s) + Br2(aq) 2Ag+(aq) + 2Br-
(aq)
EAg /Ag = +0.8 V+0
2EBr /Br = +1.07 V0 -
standard reduction potential
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ExerciseA 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)
E0cell = E0
cathode - E0anode
= +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 = Eocell – 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]y log
Ecell = Eocell – 0.0592
n
log Q
n = no of e- that are involved Q = 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
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Eocell = Eo
red + Eoox @
= +0.34 V + 0.76 V
= +1.10 V
Eocell = Eo
cathode – 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
Eθcell = 0.0592
n log K
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Example 2Calculate the equilibrium constant (K) for the following 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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ElectrolysisElectrolysis 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.
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Anion Cation
Oxidation Reduction
Electrolytic Cell
Electrolyte (M+X-)
X-,OH- M+,H+
+ -
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Anode
Cathode
Positive electrode The electrode which is connected to the
positive terminal of the battery Oxidation takes place
Negative electrode The electrode which is connected to the
negative terminal of the battery Reduction takes place
Electrons flow from anode to cathode
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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 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)
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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 redox reaction releases energy
Oxidation occurs at anode, reduction occursat cathode
Anions move towards anode, cations move towards cathode. Electrons flow from anode to cathode in an external 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 atanode.
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 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 onthe 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 products of electrolysis
Factors influencing the products :1. Reduction/oxidation potential of the
species in electrolyte2. Concentrations of ions3. Types of electrodes used – active or
inert
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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- anion and water molecules. During electrolysis:- the cathode attracts Na+ ion and H2O molecules
the anode attracts Cl- ion and H2O molecules
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Cathode
2H2O (l) + 2e- H2 (g) + 2OH- (aq)E0 = - 0.83 VNa+ (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 VIn dilute solution, water will be selected for oxidation because of its lower Eo.
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Reactions involved
2H2O (l) + 2e- H2 (g) + 2OH- (aq)E0 = - 0.83 VCathode:
Anode:
Cell reaction: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 H2O2H2O(l) O2(g) + 2H2(g)
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Electrolysis of Concentrated NaCl solutionCathode
2H2O (l) + 2e- H2 (g) + 2OH- (aq)E0 = - 0.83 VNa+ (aq) + e- Na (s) E0 = - 2.71 V
E0 for water molecules is more positive H2O easier to be reduce
AnodeCl2 (g) + 2e- 2Cl- (aq) E0 = + 1.36 V
O2 (g) + 4H+ (aq) + 4e- 2H2O (l) E0 = + 1.23 VIn concentrated solution, chloride ions will be oxidised because of its high concentration.
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Reactions involved2H2O (l) + 2e- H2 (g) + 2OH- (aq)E0 = -0.83 V
2Cl- (aq) Cl2 (g) + 2e- E0 = -1.36 V
Cell reaction: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- ion
and water molecules On electrolysis,
the cathode attracts Na+ ion and H2O molecules the anode attracts SO4
2- ion and H2O molecules
ExercisePredict the electrolysis reaction when
Na2SO4 solution is electrolyzed using platinum electrodes.
Solution
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Anode
E0 for water molecules is less positive H2O easier to oxidise
S2O82- (aq) + 2e- 2SO4
2- (aq) E0 = + 2.01 VO2 (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 positive H2O 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
EquationCathode:
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: E0
cell = - 2.06 V2H2O(l) O2(g) + 2H2(g)
4H2O (l) + 4e- 2H2 (g) + 4OH- (aq)
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Faraday’s Law of Electrolysis
Describes the relationship between the amount ofelectricity passed through an electrolytic cell andthe amount of substances produced at electrode.
Faraday’s First LawStates that the quantity of substance formed at an electrode is directly proportional to the quantity of
electric charge supplied.
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Faraday’s 1st Law
m α QQ = electric charge in coulombs (C)
m = mass of substance discharged
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Q = ItQ = 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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ExampleAn aqueous solution of CuSO4 is electrolysed using a
current of 0.150 A for 5 hours. Calculate the massof copper deposited at the cathode.
AnswerElectric charge, Q = Current (I) x time (t)
Q = (0.150 A) x ( 5 x 60 x 60 )sQ = 2700 C
1 mole of electron Ξ 1 F Ξ 96 500 C
No. of e- passed through = 270096 500 = 0.028 mol
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Cu2+ (aq) + 2e- Cu (s)From equation:
2 mol electrons 1 mol Cu0.028 mol electrons 0.014 mol Cu
Mr for Cu = 63.5Mass of Copper deposited = 0.014 x 63.5
= 0.889 g