Electrochemistry

An introduction

Electrochemistry is the study of the relationships that exist between chemical reactions & the flow of electricity……

Processes involving electrochemistry

BatteriesFuel cellsElectroplatingCorrosion

Redox ReactionsReactions involving electron transfer from one species to another.Reduction/oxidation.

OXIDATION OXIDATION — loss of electron(s) by a species; increase in oxidation — loss of electron(s) by a species; increase in oxidation number; increase in oxygen.number; increase in oxygen.

REDUCTIONREDUCTION — gain of electron(s); decrease in oxidation number; — gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen.decrease in oxygen; increase in hydrogen.

OXIDIZING AGENTOXIDIZING AGENT— electron acceptor; species is reduced.— electron acceptor; species is reduced.

REDUCING AGENT REDUCING AGENT — electron donor; species is oxidized.— electron donor; species is oxidized.

Remember …

OIL RIG

Electrochemical cells

Electrochemical cells consist of two electrodes in contact with an electrolyte

An oxidation reaction takes place at the anode and a reduction reaction takes place at the cathode oxidation reduction

Anode -ve Cathode +ve

Electron flow

Electrolyte solution

Electrochemical cells2 types :

- Galvanic Cell or Voltaic Cell the chemical rxn is spontaneous & produces electricity

- Electrolysis Cell the chemical rxn is non-spontaneous & is force by electricity from an external sources.

Spontaneous rxn that involves electron transfer can be used to generate electricity (e.g. : battery) Non-spontaneous rxn that involves electron transfer can be force proceed by the addition of an electric current (e.g. : electrolysis)

Galvanic Cell

An electrochemical cell that produces electricity as a result of a spontaneous reaction occurring inside it.

Electrodes contain a metallic conductor that acts as a source or sink for electrons.

Electrolyte solution – ionic conductor which may be a solution, a liquid or a solid.

Galvanic Cell

Electrode types :-

Redox Electrodes Metal / Metal ion

Metal / Insoluble-salt Gas Electrodes

Half-reactions

- A hypothetical reaction showing the gain or loss of electrons from either the oxidation or the reduction part of a redox equation.

- A redox reaction is the sum of

two half-reactions.

E.g.

Cu2+(aq) + 2e-

(aq) Cu(s) reduction

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

Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) redox reaction

Writing a balanced reaction from the half-reactions

Mg (s) + Cl2 (g) MgCl2 (s)

(Oxidized ) (Reduced)

1) Write Half-reactions Mg Mg2+ Cl2 2Cl-

2) Balance the equation by adding electrons

Mg Mg2+ + 2e- Cl2 + 2e- 2Cl-

3) Combine the Half-reactions

Mg + Cl2 + 2e- 2Cl- + Mg2+ + 2e-

Mg + Cl2 2Cl- + Mg2+

Na + Br2 NaBr

Try to balance this equation……

Cell Notation / Diagram

Diagram of cell with salt bridge

Diagram of cell with porous membrane

Cd(s)|CdCl2 (g)||NiCl 2 (aq)|Ni(s)

Denotes a phase boundary

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

zinc copper

ZnSO4 CuSO4

Denotes a salt bridge

salt bridge

porous membrane

Salt Bridge

Electron flow

+ve charged ions in solution are reduced, therefore concn of +ve ions decreases

+ve charged ions increase in solution due to oxidation

ANODE(negative electrode)

CATHODE(positive electrode)

reductionoxidation

K+Cl-

- The salt bridge consists of a glass U-tube filled with KCl containing Agar-Agar paste, which sets into a gel.

- When in a galvanic cell two solutions are kept in separate containers, an electrical contact between the two is needed to complete the circuit.- A salt bridge completes the circuit by allowing the migration of anions from one container into the salt bridge and from the salt bridge into the other container.- it also helps in maintaining the charge balance in the reactions taking place at the two containers by releasing counter ions into the solution. Other wise due to the accumulation of the respective charges in the two containers there will be no flow of electrons and the cell will stop functioning.

Function of aSalt Bridge

Redox Equation

Cell written with oxidation reaction on the left and reduction on the right (Right-hand side = Reduction)

Therefore the cell reaction occurring in electrochemical cell is expressed as

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

(aq)

Anode, -ve, oxidation Cathode, +ve, reduction

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

Electromotive Force (emf)

The potential difference between the anode and cathode in a cell is called the electromotive force (emf) & is measured in volts (V). The emf produced by a galvanic cell is called the cell potential, Ecell.

emf depends on concentrations of the ions, the

temperature & the partial pressure of any gases involved. When all ion concentrations are 1M, all partial pressures of gases are 1 atm & temp of the cell is 25oC Standard Cell Potential, Eo

cell.

1 V = 1 joule/coulomb

Standard Cell Potentials

Cu2+(aq) + 2e- Cu(s) E°= +0.34 V

Zn2+(aq) + 2e- Zn(s) E°= -0.76 V

Standard reduction potential of the positive electrode

Standard reduction potential of the negative electrode

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

(aq)

Eocell = Eo substance reduced - Eo substance oxidised

Refer Table of Standard Reduction Potentials Eo

cell = + 0.34 V – (- 0.76 V)= + 1.10 V

(cathode)

(anode)

Rxn with +ve E0 values = spontaneousRxn with –ve E0 values = non-spontaneous

Forward reactions

Standard Hydrogen Electrode (S.H.E)

In a galvanic cell, both electrodes are considered to make a contribution to the overall cell potential.

It is not possible to measure the potential of a single electrode.

The potential of one electrode is considered as a standard and its potential is set to zero.

The electrode selected for this is the standard hydrogen electrode.

Pt(s)|H2(g)|H+(aq) E° = 0 V

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

Standard Hydrogen Electrode (S.H.E)

As the hydrogen gas flows over the porous platinum, an equilibrium is set up between hydrogen molecules and hydrogen ions in solution. The reaction is catalysed by the platinum.

Standard conditionsThe hydrogen pressure = 1 bar temperature = 298K

A number of half reactions have been measured relative to the S.H.E Tables standard reduction potentials.

A +ve value of E° shows that the species has a greater tendency to be reduced and is therefore a good oxidising agent.

F2 + 2e- 2F- E° = +2.87 V

Ag+ + e- Ag(s) E° = +0.80 VCu2+ + 2e- Cu(s) E° = +0.34 VZn2+ + 2e- Zn(s) E° = -0.76 V

Decrease of

oxidising ability

Standard Hydrogen Electrode (S.H.E)

Table of Standard Reduction Potentials

S.H.E.

Calculating standard cell potentials

Calculate the following standard cell potential, label the anode and the cathode and write the net (redox) reaction for the following cell.

Zn(s)|Zn2+(aq)||Ni2+(aq)|Ni(s)

Zn2+(aq) + 2e- Zn(s) E°= -0.76 V Ni2+(aq) + 2e- Ni(s) E°= -0.25 V

ANODE (-ve) CATHODE (+ve)

Zn(s) + Ni2+(aq) Zn2+(aq) + Ni(s)

E°-

E°+

E °cell = E °+ - E °- = -0.25 –(-0.76) = +0.51 V = +ve spontaneous

Eocell = Eo substance reduced - Eo substance oxidised

cathode anode

Example : A mixture consisting of pieces of solid zinc and solid chromium in contact with a solution containing 1 M Zn2+ and 1 M Cr3+. Write the cell equation for this reaction and calculate the value of the E0

cell.

Cr3+ (aq) + 3e- Cr (s) E0 = - 0.74 V Zn2+ (aq) + 2e- Zn (s) E0 = - 0.76 V

From Table of Standard Reduction Potentials, Cr3+ is more easily reduced than Zn2+. So, the zinc half-reaction will occur as oxidation and chromium half-reaction will occur as reduction.

Cr3+ (aq) + 3e- Cr (s) (reduction)

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

2Cr3+ (aq) + 3Zn (s) 2Cr (s) + 2Zn2+ (aq)

2 x

3 x

Eocell = Eo substance reduced - Eo substance oxidised

= E0Cr

3+ICr - E0Zn

2+IZn

= ( - 0.74 V ) – ( - 0.76 V)= + 0.02 V

Cathode…???

Anode….??

E0cell = +ve spontaneous

Calculate the standard cell potential for an electrochemical cell in which this reaction takes place:

Zn(s) + Cd2+ (aq) Zn2+ (aq) + Cd (s)

Write the half-reactions

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

Refer table of Standard Reduction Potentials

Cd2+ (aq) + 2e- Cd (s) (reduction)

E0 Zn2+IZn = - 0.76 V E0 Cd

2+ICd = - 0.40 V

E0 cell = + 0.36 V E0

cell = +ve spontaneous

More calculations

From the following standard reduction potentials, write the equation for the spontaneous reaction and calculate the standard reduction potential.

Ag+(aq) + e- Ag(s) E°= +0.80 VNa+(aq) + e- Na(s) E°= -2.71 V

For a spontaneous reaction E°cell must be positive

Therefore E°cell = +0.80 –(-2.71) = +3.51 V

Overall reaction Ag+(aq) + Na(s) Ag(s) + Na+(aq)

(Red)

(Ox)

Nernst EquationReactions very rarely carried out under

standard conditions.

QvF

RTEE ln0

Gas constant temperature

Reaction quotient

Faraday constant (96 485 Coulombs mol-1)

Standard emf

Moles of electrons transferred

Emf under non-standard conditions

Reaction Quotient

For the reaction

aA(aq) + bB(aq) cC(aq) + dD(aq)

If one or more of the reactants are solids or pure liquids they do not enter into the expression.

The reaction will proceed until it has reached its equilibrium, at this point Q = K (the equilibrium constant) and the cell potential will be zero.

ba

dc

BA

DCQ

][][

][][

[X] =Concentration of X

Using the Nernst Equation

The spontaneous reaction under standard conditions :

Cu2+(aq) + Sn(s) Cu(s) + Sn2+(aq)

Cu2+ + 2e- Cu +0.34 VSn2+ + 2e- Sn -0.14 V

Then E°cell = +0.34 –(-0.14) = +0.48 V

What is the cell potential under the conditions [Cu2+] = 0.1M and [Sn2+] = 0.5M and 298K?

0.1

0.5ln

2F

298 x 8.3140.48lnQ

vF

RTEE 0

QvF

RTEE ln0

= 0.46 V

Calculating Equilibrium Constants

What is the equilibrium constant for this reaction at 298 K?

When the reaction reaches equilibrium the cell potential Ecell will fall to zero.

KvF

RTE

QvF

RTEE

ln

0ln

0

0

Therefore

So that

Calculation of equilibrium constant (cont)

161.54x10K

37.27Kln

Kln 2x96485

8.314x2980.48

Kln vF

RT0E

A large equilibrium constant indicates that the reaction goes almost to completion

Corrosion Corrosion or rusting of iron can be explained in

thermodynamic terms by comparing the following standard reduction potentials.

Corrosion can be defined as the deterioration of materials by chemical processes.

The special characteristic of most corrosion processes is that the oxidation and reduction steps occur at separate locations on the metal.

This is possible because metals are conductive, so the electrons can flow through the metal from the anodic to the cathodic regions.

The presence of water is necessary in order to transport ions to and from the metal but a thin film of adsorbed moisture can be sufficient.

Fe2+ (aq) + 2e- Fe(s) E°=-0.44 V

2H+ (aq) + 2e- H2(g) E° = 0 V

4H+ (aq) + O2 (g) + 4e- 2H2O (l) E° = +1.23 V

2H2O (l) + O2 (g) + 4e- 4OH-(aq) E° = +0.40 V

All of these standard reduction reactions are more positive than that for iron so all three of these could drive the oxidation of Fe Fe2+

Corrosion is a two-step process

Step 1

Fe Fe2+ + 2e-

- Fe atom at metal surface dissolves into moisture film, leaving negative charge in metal.

Step 2

- Corrosion continues as a depolarizers, removes electron from metal. - common depolarizers are :-

a) Oxygen O2 + 4H2O 4OH- + 4e-

b)Acid 2H+ + 2e- H2

c)Cation Metal Cu2+ + 2e- Cu

Schematic diagram of corrosion cells on iron

Protection against corrosionThe simplest way to protect iron against corrosion is to

cover the surface with paint- prevents oxygen from getting to the surface.

Sacrificial coatings

One way of supplying this negative charge is to apply a coating of a more active metal.

Common way of protecting steel from corrosion is to coat it with a thin layer of zinc; this process is known as galvanizing.

Cathodic protection

to maintain a continual negative electrical charge on a metal, so that its dissolution as positive ions is inhibited.