Chapter 10

ElectrochemistElectrochemistryry

ElectrochemistryIs the study of the relationship between electricity

and chemical reaction

Chemical reactions involved in electrochemistry are :Chemical reactions involved in electrochemistry are :

Oxidation

ReductionREDOX REACTION

One type of reaction cannot occur One type of reaction cannot occur without the other.without the other.

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

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’

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

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.

Zn electrode

Cu electrod

e

Salt bridge

Zn2+ Cu2+

Voltmeter

Galvanic cell

ZnSO4(aq)solution

CuSO4(aq)solution

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.

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.

Reactions Involved:

Anode :

Cathode :

Zn (s) Zn2+ (aq) + 2e-

Cu2+ (aq) + 2e- Cu (s)

Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)Overall cellreaction :

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 bridge

An inverted U tube containing a gel An inverted U tube containing a gel permeated with solution of an permeated with solution of an inert electrolyte such as KCl, Naelectrolyte such as KCl, Na22SOSO44, NH, NH44NONO33..

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

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

How does the cell maintains its electrical neutrality?

Left Cell Right Cell

Cl- ions from salt bridge move into Zn

half cell

K+ 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 tve charge.

Cu 2+ ions leave the solution.

Causing an overall excess of

-ve charge.

Electrochemical Cells

19.2

spontaneousredox reaction

anodeoxidation

cathodereduction

half

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 boundarySalt bridge

Cell notation

Exercise

For 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 2

X 3

Cell notation

3 3

2 26e

6e-

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

Electrode Potential

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.

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

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 Vor E0

cell = E0red + E0

ox

= +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)

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

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

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

Standard reduction potential of Zinc half cell is measured by setting up the electrochemicalcell 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- +

-+

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 V02+

Zn2+ + 2e- Zn E0 = -0.76 V

E0 = EH /H - EZn /Zn cell0 0

+ 2+2

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 + -

+-

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

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

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 Vor E0

cell = E0red + E0

ox

= +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)

At standard-state condition

E0cell = E0

red + E0ox

E0cell = E0

cathode - E0anode

or

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 = +0.80V

E0ox = +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

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

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

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

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

Answer

H2 (g),1 atm.

Pt

H+(aq), 1M

Pt

E0cell

=1.36V

- +

V- + Cl2 (g),1 atm.

Cl-(aq), 1M

1. Show the process occur at anode and cathode

2. Overall reaction

- Half-cell reaction

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

• Spontaneous & Non-Spontaneous reactions

- Redox reaction is spontaneous when

Ecell is +ve.

- Non spontaneous is when Ecell is –ve.

Eθ cell = 0 The reaction is at equilibrium

Sn4+/Sn2+

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 = +0.76 V

Cathode: Sn4+ + 2e → Sn2+ Eo = +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.

Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g)

Predict : Spontaneous or non-spontaneous?

Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(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 + Eo

ox

Reduction

Oxidation

= (-1.36) + (-0.13)

= -1.48 V

Non-spontaneous

No Reaction

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

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)

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]y

2.303 log

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 =

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

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

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

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

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

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.

Anion Cation

Oxidation Reduction

Electrolytic Cell

Electrolyte (M+X-)

X-,OH- M+,H+

+ -

Anode

Cathode

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

Electrons flow from anode to cathode

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)

Comparison between an electrochemical celland an electrolytic cell

CathodeAnode

- +e- e-

Anode Cathode

+ -

e- e-

+ -

Electrolytic Cell Electrochemical Cell

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:

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)

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’

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

Predicting the products of electrolysis

Factors influencing the products :

1. Reduction/oxidation potential of the species in electrolyte

2. Concentrations of ions

3. Types of electrodes used – active or inert

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

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 for oxidation because of its lower Eo.

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 H2O

2H2O(l) O2(g) + 2H2(g)

Electrolysis of Concentrated NaCl solution

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 be reduce

AnodeCl2 (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 be oxidised because of its high concentration.

Reactions involved

2H2O (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:

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

Exercise

Predict the electrolysis reaction when Na2SO4 solution is electrolysed using platinum electrodes.

Solution

Anode

E0 for water molecules is less positive H2O 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 positive H2O easier to reduce

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)

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.

Faraday’s 1st Law

m α Q

Q = electric charge in coulombs (C)

m = mass of substance discharged

The End

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

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.

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 96 500 C

No. of e- passed through = 270096 500

= 0.028 mol

Cu2+ (aq) + 2e- Cu (s)

From equation:2 mol electrons 1 mol Cu

0.028 mol electrons 0.014 mol Cu

Mr for Cu = 63.5

Mass of Copper deposited = 0.014 x 63.5

= 0.889 g