CHM3010_Electrochemistry

CHM 3010_Electrochemistry 1

ElectrochemistryElectrochemistry


Peringatan BersamaPeringatan Bersama

• Ingat Ibu Bapa Anda • Tanpa Mereka Kamu Tiada

• Jadikan Keluarga Sumber Kebahgiaan Anda


Electrolysis: Learning Electrolysis: Learning OutcomesOutcomes

Electrolyte conductivity, Faraday’s law,

1. Use Faraday’s law to calculate amounts of products formed, amounts of current passed, time elapsed and oxidation state

Electrochemical cell, standard electrode potential and redox reaction

2. Describe the differences between electrolytic cell and galvanic cell

3. Recognize oxidation and reduction half reactions

4. Write half reactions and overall cell reactions

5. Use standard reduction potential to calculate the standard cell potential, to identify the cathode and anode, to predict the spontaneity of a redox reaction


Electrolysis: Learning Electrolysis: Learning OutcomesOutcomes

Nernst equation

6. Use Nernst equation to relate electrode potentials and cell potentials to different concentrations and partial pressures

Application of electrochemistry in determination of equilibrium constant and Gibbs free energy

7. Relate standard cell potential to the standard Gibbs free energy change and the equilibrium constant


IntroductionIntroduction

• Discusses the relation between chemical processes and electrical energy.

Chemicalprocesses

Electrical energy

Battery orVoltaic cell

Electrolysis


ReferencesReferences• Whitten: Chapter 21• Brady: Chapter 21


ElectrochemistryElectrochemistry

• Electrochemistry deals with the chemical changes produced by electric current & with the production of electricity by chemical reactions

• All electrochemical reactions involve the transfer of electrons – oxidation-reduction reactions (redox reactions)

• In most applications the reacting system contained in a cell, and an electric current enters or exits by electrodes


ElectrochemistryElectrochemistry• The two parts of the reaction are physically

separated.– The oxidation reaction occurs in one cell.– The reduction reaction occurs in the other

cell.

• There are two types of electrochemical cells:1.Electrolytic cells are those in which electrical

energy from an external source causes nonspontaneous chemical reactions to occur

2.Voltaic or galvanic cells are those in which spontaneous chemical reactions produces electricity & supply it to an external circuit


Figure 21.3 General characteristics of voltaic and electrolytic cells

VOLTAIC CELLEnergy is released from

spontaneous redox reaction

Reduction half-reactionY++ e- Y

Oxidation half-reactionX X+ + e-

System does work on its surroundings

Overall (cell) reactionX + Y+ X+ + Y; G < 0

Energy is absorbed to drive a nonspontaneous redox reaction

ELECTROLYTIC CELL

Reduction half-reactionB++ e- B

Oxidation half-reactionA- A + e-

Surroundings(power supply)do work on system(cell)

Overall (cell) reactionA- + B+ A + B; G > 0


Electrical ConductionElectrical Conduction• Electric current represents transfer of

charge

• Charge can be conducted through metals & through pure liquid electrolytes (i.e. molten salts) or solutions containing electrolytes

• Metals conduct electric currents well in a process called metallic conduction.– Involves the electron flow with no atomic

motion & no obvious changes in the metal


ElectrodesElectrodes• Electrodes are surfaces on which oxidation

or reduction half-reactions occur

• They may or may not participate in the reactions

• Electrodes that do not react with the liquids or products of the electrochemical reactions are called inert electrodes

• Two examples of common inert electrodes are graphite and platinum.


ElectrodesElectrodes• The following identification for electrodes is

correct for either electrolytic or voltaic cells:

– The cathode is the electrode at which reduction occurs as electrons are gained by some species (CR)

– The cathode is negative in electrolytic cells and positive in voltaic cells.

– The anode is the electrode at which oxidation occurs as electrons are lost by some species (AO)

– The anode is positive in electrolytic cells and negative in voltaic cells.


Electrochemical processElectrochemical process• When a metal is immersed in a solution containing its

ion an equilibrium exists.M(s) M+(aq) + e

• Oxidation and reduction occur on the electrode surface where electron is transferred to or from the electrode.

• Electrical neutrality of every component of the cell must be maintained, therefore:

+-

+-

+i.

ii. When oxidation occurs on one electrode, reduction must occur on the other

Anode – electrode where oxidation occursCathode – electrode where reduction occurs


Voltaic or galvanic cellVoltaic or galvanic cell• A cell where electrical current is generated from

spontaneous chemical reaction on the electrode surface in the solution.

• Spontaneous chemical reaction occurs

• Cu2+ + Zn → Cu + Zn2+

• Energy is produced in form of heat (wasted) and cannot be benefited or used.

CuSO4

Zn

1 hour


Figure 21.4 The spontaneous reaction between zinc and copper(II) ion

Zn(s) + Cu2+(aq)

Zn2+(aq) + Cu(s)

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.


Electrochemical cellElectrochemical cell

• An electrochemical cell consists of at least two electrodes connected by a wire and an electrolyte.

Galvano/voltmeterG

electrolyte

electrodes

Connection wire


Voltaic or Galvanic CellsVoltaic or Galvanic Cells• Voltaic or galvanic cells are electrochemical cells in which a spontaneous (product-

favored) chemical redox reactions produce electrical energy.• The two halves of the redox reaction are separated, requiring electron transfer to

occur through an external circuit – useful electrical energy is obtained– electrons (from redox reaction) are forced to travel through wires and creating a potential

difference

• Examples of voltaic cells include:– Automobile batteries– Flashlights batteries– Computer & calculator batteries


A Galvanic Cell: The Copper - A Galvanic Cell: The Copper - Silver CellSilver Cell


Zn-Cu cellZn-Cu cell• Daniel cell – 2 part of the cell are separated and are

connected by wire.

• Zn metal is oxidised to Zn2+ while Cu2+ in the solution is reduced to Cu.Zn electrode Zn → Zn2+ + 2e [O] anodeCu electrode Cu2+ + 2e → Cu [R] cathode

• Electron is continuously taken from Cu thus become +ve electrode and given to Zn thus become –ve electrode.

• Electron flows from Zn (-ve) to Cu (+ve) giving electric current (energy).

CuZn

ZnSO4

CuSO4Porous pot

CuZn

ZnSO4 CuSO4

Salt bridge


The Construction of Simple The Construction of Simple Voltaic CellsVoltaic Cells

• Salt bridge– allows the movement of ions to keep the solutions

neutral– A tube filled with a solution composed of a salt of

ions not involved in the cell reaction – saturated salt/5% agar solution (sets to the consistency of firm gelatin)

– Common salt solutions: KNO3 or KCl

– Porous plugs are fitted at each end – prevents the salt solution from pouring out but at the same time allows ion exchanges with the solutions in half-cells


The Construction of Simple The Construction of Simple Voltaic CellsVoltaic Cells

• Salt bridge serves 3 functions:– Allows electrical contact between the two

solutions– Prevents mixing of the electrode solutions– Maintains the electrical neutrality in each half-

cell as ions flow into & out of the salt bridge

• A cell in which all reactants & products are in their thermodynamic standard states (1 M for dissolved species & 1 atm partial pressure for gases) is called a standard cell


Figure 21.5 A voltaic cell based on the zinc-copper reaction

Oxidation half-reactionZn(s) Zn2+(aq) + 2e-

Reduction half-reactionCu2+(aq) + 2e- Cu(s)

Overall (cell) reactionZn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)



Figure 21.6 A voltaic cell using inactive electrodes

Reduction half-reactionMnO4

-(aq) + 8H+(aq) + 5e-

Mn2+(aq) + 4H2O(l)

Oxidation half-reaction2I-(aq) I2(s) + 2e-

Overall (cell) reaction2MnO4

-(aq) + 16H+(aq) + 10I-(aq) 2Mn2+(aq) + 5I2(s) + 8H2O(l)


Another example


Sample Problem 21.2: Diagramming Voltaic Cells

PROBLEM: Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode.

PLAN:

SOLUTION:

Identify the oxidation and reduction reactions and write each half-reaction. Associate the (-)(Cr) pole with the anode (oxidation) and the (+) pole with the cathode (reduction).

Voltmeter

Oxidation half-reactionCr(s) Cr3+(aq) + 3e-

Reduction half-reactionAg+(aq) + e- Ag(s)

Overall (cell) reactionCr(s) + Ag+(aq) Cr3+(aq) + Ag(s)

Cr

Cr3+

Ag

Ag+

K+

NO3-

salt bridge

e-

Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s)


Cell notationCell notation

• Represent a cell in written form. The Zn-Cu cell can be represented as:

Zn(s)/Zn2+(aq)//Cu2+(aq)/Cu(s)or

Zn(s)/ZnSO4(aq)//CuSO4(aq)/Cu(s)

Anode – left, cathode – right. Metals (or conducting materials) for electrical terminals – at both ends. Phase separation - /. Salt bridge - //. If electrolyte contain > 1 participating in the reaction they are separated by a coma (,).

Examples:Cd(s)/Cd2+(aq)//H+(aq)/H2(g)/Pt

Ag/AgCl/Cl-(aq)//Fe3+(aq),Fe2+(ak)/Pt


Notation for a Voltaic Cell

components of anode compartment

(oxidation half-cell)

components of cathode compartment

(reduction half-cell)

phase of lower oxidation state

phase of higher oxidation state

phase of higher oxidation state

phase of lower oxidation state

phase boundary between half-cells

Examples: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu (s)

Zn(s) Zn2+(aq) + 2e- Cu2+(aq) + 2e- Cu(s)

graphite | I-(aq) | I2(s) || H+(aq), MnO4-(aq) | Mn2+(aq) | graphite

inert electrode


Why Does a Voltaic Cell Work?

The spontaneous reaction occurs as a result of the different abilities of materials (such as metals) to give up their electrons and the ability of the electrons to flow through the circuit.

Ecell > 0 for a spontaneous reaction

1 Volt (V) = 1 Joule (J)/ Coulomb (C)


Table 21.1 Voltages of Some Voltaic Cells

Voltaic Cell Voltage (V)

Common alkaline battery

Lead-acid car battery (6 cells = 12V)

Calculator battery (mercury)

Electric eel (~5000 cells in 6-ft eel = 750V)

Nerve of giant squid (across cell membrane)

2.0

1.5

1.3

0.15

0.070


Electromotive force (emf)Electromotive force (emf)

• A force is needed to move a charge. Total force to move electron(s) from a –ve electrode to a +ve electrode in an electrochemical cell is termed emf (in volts or V). Emf of a cell is often called cell potential or Ecell.

• emf depends on 3 factors:– Tendency of two half reactions to occur spontaneously.– Electrolyte concentration and gas pressure.– Temperature.

• Standard emf - Ecello measured at 25 oC, 1 atm of

gas pressure and 1.0 electrolyte concentration


Electrode and cell potentialsElectrode and cell potentials• Each ion has tendency to take up electron and undergoes

reduction. However the ability to be reduced is different and ion with higher ability will succeed while the other undergoing oxidation.

• Example – Zn-Cu cell, Zn2+ vs Cu2+. Cu2+ has higher ability to gain electron thus reduced and Zn2+ oxidised.

• This tendency is termed electrode (half cell) potential and the different (+ve value) between two electrode potentials is the cell potential.

• ThusEcell = Ecathode – Eanode

Ecell > 0 so Ecathode > Eanode

Based on cell notation Ecell = Eright - Eleft

Only cell potential can be measured experimentally. The electrode potential is calculated from the cell potential if the potential for the other half cell is known.

Ecathode = Ecell + Eanode or Eanode = Ecathode - Ecell


Standard hydrogen electrode and Standard hydrogen electrode and standard electrode potentialstandard electrode potential

• An electrode potential is calculated from a cell potential where the potential of one electrode is known. It is conventionally agreed that the potential of hydrogen electrode at 25 oC, [H+] = 1 M and PH2 = 1 atm is 0.0000 V.

• Example to determine the potential for Cu/Cu2+.

Pt/H2(g)/H+(aq)//Cu2+/Cu

• The cell potential (from meter) is 0.34 V

Ecello = ECu

o – EH2o = 0.34 V or ECu

o = 0.34 V

By this way a list of electrode (half cell) potentials are obtained. They are also referred to as standard reduction potentials. See the list in reference books.


The standard hydrogen electrode. Hydrogen gas at 1 atm is passed over finely divided platinum. The solution contains 1.00 M hydrogen ion. The reduction potential is exactly 0 V at 298 K (25oC).


• Using a hydrogen half-cell, other reduction potentials can be measured

A galvanic cell comprised of copper and hydrogen half-cells. The reaction is Cu2+(aq)+H2(g) Cu(s)+2H+(aq)

Cell notation:Pt(s), H2(g)|H+(aq)||Cu2+(aq)|Cu(s)


ReRedoxdox reactions reactions

• The reduction potential’s table can also be used to determine whether a redox reaction can occur spontaneously. The strength of a reduction or oxidation agent depend on the potential value. A substance or an ion with high reduction potential will be easily reduced thus is a strong oxidation agent and vice versa.

MnO4-, O3, F2 – strong oxidants

Li, K, Na – strong reductantsExample: Will the following mixture react spontaneously at standard state?Zn2+(aq) + Fe2+(aq) → ???

(a) Zn2+(aq) + 2e → Zn Eo = -0.76 V(b) Fe3+(aq) + e → Fe2+ Eo = 0.77 V

(a)-2(b); Zn2+(aq) + 2Fe2+(aq) → Zn(s) + 2Fe3+(aq)

The forward reaction is not spontaneous.

53.1)77.0(76.0 oreactionE


Free energy and EFree energy and Ecellcell

G is the maximum work that can be obtained or extracted from a system or process.

G = -wmax

In electrochemical cellWork (w) = - charge (Q) x emf

= - nF x emf= - nFEcell

Therefore G = - nFEcell

For standard stateGo = - nFEcell

o

Example: Zn + Cu2+ → Zn2+ + Cu Eo = 1.10 VGo = - 2 x 96500 x 1.10

= - 212300 J = - 212.3 kJ

Spontaneous reaction

G = -ve E = +ve

Nonspontaneous reaction

G = +ve E = -ve


Nernst’s equationNernst’s equation

For a general reactionaA + bB → eE + fF

The corresponding relation for Gibbs free energy is:

Therefore

Or

At 25 oC

If all species are at unit concentration log Q = 0 and Ecell = Ecello

ba

feo

BA

FERTEE

][][

][][ln

ba

feo

BA

FERTGG

][][

][][ln

ba

feo

BA

FERTnFEnFE

][][

][][ln

Nernst’s equation

ba

feo

BA

FE

nEE

][][

][][log

0592.0


= º - (0.059/n)log10(Q)

The Nernst Equation

, in analogy with G, indicates the direction of spontaneous change:

> 0 reaction goes left to right

= 0 no net reaction (equilibrium)

< 0 reaction goes right to left


Born: Briesen near Thorn (Torun) (Germany), 1864 Died: Ober-Zibelle near Bautzen (Germany), 1941

In 1883 Nernst became assistant of Wilhelm Ostwald in Leipzig, 1891 professor in Göttingen, 1905 in Berlin. He succeeded in explaining the phenomena connected with galvanic elements (Nernst´s equations 1889) and calculated the mobility of ions and the coefficient of diffusion. He received the Nobel Prize in 1920.

Walther Hermann Nernst


Note that the Nernst equation combines the tendency of both the oxidation reaction to proceed and the reduction reaction to proceed as a pair of coupled reactions. To establish a scale of reactivity we measure “half reactions” against a standard half reaction:

H2 2H+ + 2e

A Standard Half Reaction


The hydrogen electrode is the standard reference electrode for aqueous half-cell potentials. The half-cell reaction is the reduction

2 H+(aq) + 2e H2 (g)

The standard potential of this half-cell is set equal to zero. Potentials measured for a cell containing the standard hydrogen electrode will be equal to the half-cell potential of the other electrode. A table of standard half-cell potentials may be obtained from any of the references.

The Hydrogen Electrode


Conventions for (Half-cell Potentials)

1. Assign standard hydrogen electrode a potential of 0.0 V.

2. When half reactions are written as reductions, reactions that proceed as reductions more readily than H+/H2 are assigned positive voltages, those that proceed less readily are assigned negative voltages.

3. If the direction of the half reaction is reversed, the sign of the standard half cell potential is reversed. If the reaction is multiplied by some factor, the standard half cell potential remains unchanged.

4. The value of the half cell potential is a measure of the tendency of the reaction to proceed from left to right as written.


Usage of EUsage of Ecellcell

Determination of Equilibrium Constant

aA + bB ⇌ eE + fF

• The eq const can be calculated from or vice versa

ba

fe

eq BA

FEK

][][

][][

eqo K

nF

RTE ln

ba

feo

BA

FERTGG

][][

][][ln

0G and

ocellE


Figure 21.9

G0

E0cell K

G0 KReaction at

standard-state conditions

E0cell

The interrelationship of G0, E0, and K

< 0 spontaneous

at equilibrium

nonspontaneous

0

> 0

> 0

0

< 0

> 1

1

< 1

G0 = -RT lnKG0 = -nFEo

cell

E0cell = -RT lnK

nF

By substituting standard state values into E0

cell, we get

E0cell = (0.0592V/n) log K (at 250C)



ElectrolysisElectrolysis


• Electricity can be used to make nonspontaneous redox reactions to occur

• The process is called electrolysis • Electrolysis occurs in an

electrolysis or electrolytic cell • These cells require a source of

direct current, possibly one of the batteries just discussed, to provide electrical energy


In both types of cells oxidation occurs at the anode and reduction occurs at the cathode

• Electrolysis in aqueous solutions often involves water molecules

• This is often unintended and called a competing reaction

Electrolytic CellCathode is negative (reduction)Anode is positive (oxidation)

Galvanic CellCathode is positive (reduction)Anode is negative (oxidation)


Electrolytic cellElectrolytic cell

• Example: Electrolysis of fused NaCl

• Na+ is attracted to –ve electrode and reduced while Cl- attracted fo +ve electrode and oxidised.

Na+(l) + e → Na(s) anodeCl-(l) → 1/2Cl2(g) + e cathode

In term of electrode chargeelectrolytic cell is different from voltaic cell

Cl-

Na+

Battery

Anode Cathode

+ -

Cl2 released Na deposit

Cell Anode Cathode

Electrolytic

+ve -ve

Voltaic -ve +ve


Electrolysis from aqueous solutionElectrolysis from aqueous solution

• Consider electrolysis of aqueous NaCl. The process become complex in the presence of water as the water itself can be oxidised or reduced.

• Cathode – H2 is released• Anode – O2 is released for dilute NaCl

Cl2 is released for conc. NaClCathode : 2 competing reactions

2H2O + 2e → H2 + 2OH- Eo=-0.83 VNa+ + e → Na Eo=-2.71 V

H2O with higher red. potential is reduced.

Anode : 2 competing reactions2H2O → O2 + 4H+ + 4e Eo = -1.23 VCl- → Cl2 + 2e Eo = -1.36 V

The oxidation pot. for H20 is a little higher and since the conc of Cl- is low H2O is oxidised. However for high conc. NaCl, due to the concentration effect Cl- is oxidised.


Faraday’s LawFaraday’s Law

• Faraday relates the amount of substance obtained at the electrode with the amount of charge transferred during electrolysis.

Na+ + e → Na(s)Cl- → 1/2Cl2 + e

• ⇒ I mole of e is required to deposit 1 mole of Na and release half mole of Cl2

Charge of 1 e = 1.602 X 10-19 CCharge of 1 mole of e = 6.022 X 1023 X 1.602X 10-19

96500 CThis is well known as Faraday’s number, 1 F = 96500 C


Electrolysis of molten sodium chloride. The passage of an electric current through molten sodium chloride decomposes the material into molten sodium and chlorine gas. The products must be kept separated because they react on contact to re-form NaCl.


A microscopic view of the changes at the anode in the electrolysis of molten NaCl. The positive charge of the electrode attracts Cl- ions. At the surface of the electrode electrons are pulled from the ions yielding neutral Cl atoms which combine to form Cl2 molecules that rise to the surface as a gas.


• Sodium is a liquid at the melting point of sodium chloride (801 oC)

• Reaction at the cathode: Na+(l) + e- Na(l)

• Reaction at the anode: 2Cl-(l) Cl2(g) + 2e-

• Cell reaction: 2Na+(l) + 2Cl-(l) 2Na(l) + Cl2(g)

• The electrode polarities are reversed in electrolytic cells relative to those in galvanic cells


Electrolysis of an aqueous solution of potassium sulfate. The products of the electrolysis are H2 and O2 gas, not the “expected” products solid K and S2O8

2-. Why?

Look up the E0 values of all possible electrode reactions and work out how you may arrive at the statement given above.

Refer Chemistry: Matter and Its Changes by brady and Senese (4th Edn) Chp 21.


ExampleExample

• Calculate the mass of Cu (atomic mass = 64) deposited at cathode in the electrolysis of CuSO4 if 2 A current was flowed for 1 hour.

At cathode Cu2+ + 2e → Cu⇒ 2 F is required to deposit 1 mole Cu

Charge, Q = I X t = 2 C s-1 X 3600 s = 7200 C

Mole of Cu deposited = 7200/(2 X 96500) = 0.0373 mole

Mass of Cu = 0.0373 X 64 = 2.39 g


Table 21.4 Comparison of Voltaic and Electrolytic Cells

Cell Type G Ecell

Electrode

Name Process Sign

Voltaic

Voltaic

Electrolytic

Electrolytic

< 0

< 0

> 0

> 0

> 0

> 0

< 0

< 0

Anode

Anode

Cathode

Cathode

Oxidation

Oxidation

Reduction

Reduction

-

-

+

+


Types of CellsTypes of Cells• Look up uses


ExampleExample

• Calculate the mass of Cu (atomic mass = 64) deposited at cathode in the electrolysis of CuSO4 if 2 A current was flowed for 1 hour.

At cathode Cu2+ + 2e → Cu

2 F is required to deposit 1 mole Cu

Charge, Q = I X t = 2 C s-1 X 3600 s = 7200 C

Mole of Cu deposited = 7200/(2 X 96500) = 0.0373 mole

Mass of Cu = 0.0373 X 64 = 2.39 g


• A student added contains 50.00 mL solution of 0. 1216 M FeCl3 into an electrolytic cell. If a current of 0.310 A is passed through the cell for 40 minutes, how much (g) of Fe(s) would be deposited at the cathode? What is the concentration of Fe3+ (aq) in the cell at the end of the 40 minutes ?

• (One Faraday = 96500 Coulomb)


Group Exercise 1: Take HomeGroup Exercise 1: Take Home

• Use this list of half-reactions to answer the following six questions.

• MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) +1.51 V• Cr2O7

2-(aq) + 6e- 2Cr3+(aq) + 7H2O(l) +1.33 V• Pt2+(aq) + 2e- Pt(s) +1.20 V• Cu2+(aq) + 2e- Cu(s) +0.34 V• Pb2+(aq) + 2e- Pb(s) -0.13 V• Al3+(aq) + 3e- Al(s) -1.66 V • Using the half-reactions in the above table, write down

the cell notations for the four cells which can be constructed to give Eo

cell in excess of +1.90 V .



The value of Ecell at 25oC for the cell shown below is +1.27 V. Show how you would determine the value of Eocell.

Cd(s) | Cd2+(aq), 2.0 M || Ag+(aq), 2.0 M | Ag(s)

•Ans. 1.26 V



Calculate the mass of cobalt that will be deposited when a current of 2.00 A is passed through a solution of CoSO4 for 10.0 hours. (Co=58.93)

ANSWER: 22.0 g

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