10 ELECTROCHEMISTRY

8/3/2019 10 ELECTROCHEMISTRY

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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

10 ELECTROCHEMISTRY

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