Download - Powerpoint in electrochemistry

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Dr. Paul CharlesworthMichigan Technological UniversityDr. Paul Charlesworth

Michigan Technological University

C h a pte rC h a pte r 1818ElectrochemistryElectrochemistry

Chemistry 4th EditionMcMurry/Fay

Chemistry 4th EditionMcMurry/Fay

Chapter 18 Slide 2Prentice Hall ©2004

Kinds of electrochemical cellsKinds of electrochemical cells

• Galvanic (voltaic)Spontaneous chemical reaction generates electricity

with which to do worke.g. batteries

• ElectrolyticElectricity used to drive nonspontaneous chemical

reaction


Redox Reactions 01Redox Reactions 01

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Redox reaction are those involving the oxidation and reduction of species.

OIL – Oxidation Is Loss of electrons.

RIG – Reduction Is Gain of electrons.

Oxidation and reduction must occur together.

They cannot exist alone.





• Oxidation Half-Reaction: Zn(s) → Zn2+(aq) + 2 e–.• The Zn loses two electrons to form Zn2+.


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• Reduction Half-Reaction: Cu2+(aq) + 2 e– → Cu(s)• The Cu2+ gains two electrons to form copper.


• Overall: Zn(s) + Cu2+(aq)→ Zn2+(aq) + Cu(s)

Galvanic cell, example 1Galvanic cell, example 1


Electrochemical Cells 01Electrochemical Cells 01

• Electrodes: are usually metal strips/wires connected by an electrically conducting wire.

• Salt Bridge: is a U-shaped tube that contains a gel permeated with a solution of an inert electrolyte.

• Anode: is the electrode where oxidation takes place.

• Cathode: is the electrode where reduction takes place.

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Al(s) + Pb2+(aq) Al3+(aq) + Pb(s)



Fe(s) + 2Fe3+(aq) 3Fe2+(aq)


Fe(s) + Fe3+(aq) 2Fe2+(aq)

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• Convention for expressing the cell:

Anode Half-Cell || Cathode Half-Cell

Electrode | Anode Soln || Cathode Soln | Electrode

Zn(s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu(s)

Fe(s) | Fe2+(aq) || Fe3+(aq), Fe2+(aq) | Pt(s)

• Electrons flow from anode to cathode. Anode is placed on left by convention.


Cell potential, Free-energy change


Driving force in electrochemical cells

Name: electromotive force (emf)cell potentialcell voltage

Symbol: E

SI units: Volt (V)

(1 J = 1 C x 1 V)

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Relationship between ∆G and ERelationship between ∆G and E

∆G = -nFE

F = 96,500 C/mol e-

faraday (Faraday contant)


Standard cell potential, EoStandard cell potential, Eo

Standard conditionsaqueous solutes: 1 Mgases: 1 atmsolids and liquids: pure

∆Go = -nFEo


Example Problem

The standard cell potential (Eo) for following reaction is 1.10 V at 2.5oC. What is ∆Go?

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

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Standard oxidation and reduction potentials

Eo = Eoox + Eo

red

Zn(s) Zn2+(aq) + 2e– Eoox = 0.76 V

2H+(aq) + 2e– H2(g) Eored = 0 V

Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) Eo = 0.76 V

Example:





• The standard half-cell potentials are determined from the difference between two electrodes.

• The reference point is called the standard hydrogen electrode (S.H.E.) and consists of a platinum electrode in contact with H2 gas (1 atm) and aqueous H+ ions (1 M).

• The standard hydrogen electrode is assigned an arbitrary value of exactly 0.00 V.

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H2(g) 2H+(aq) + 2e– Eoox =

Cu2+(aq) + 2e– Cu(s) Eored =

H2(g) + Cu2+( aq) 2H+(aq) + Cu(s) Eo = 0.34 V

Another example

0.34 V

0 V


Standard reduction potentialsStandard reduction potentials

Zn(s) Zn2+(aq) + 2e- Eoox = 0.76 V

Zn2+(aq) + 2e- Zn(s) Eored = -0.76 V

Eoox = -Eo

red



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Learning Goal

Determine Eo for a cell from a table of standard reduction potentials.



• When selecting two half-cell reactions the more negative value will form the oxidation half-cell.

• Consider the reaction between zinc and silver:Ag+(aq) + e– → Ag(s) E° = 0.80 VZn2+(aq) + 2 e– → Zn(s) E° = – 0.76 V

• Therefore, zinc forms the oxidation half-cell:Zn(s) → Zn2+(aq) + 2 e– E° = – (–0.76 V)



• What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2solution and a Cr electrode in a 1.0 M Cr(NO3)2solution?

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• What is the standard emf of an electrochemical cell made of a Mg electrode in a 1.0 M Mg(NO3)2solution and a Ag electrode in a 1.0 M AgNO3solution?


What is Eo for the following cell?

Fe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s)


Spontaneity of a Reaction 01Spontaneity of a Reaction 01

• The value of E˚ is related to the thermodynamic quantities of ∆G˚ and K.

∆G˚ = –nFE˚ and ∆G˚ = –RT ln K

combine to give:

E˚ = –(RT/nF) ln Kn = moles of electrons

RT/F = 0.0257 V (at 25oC)

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Equivalent forms of equation:

E˚cell = (0.0257/n) ln K

E˚cell = (0.0592/n) log K


Learning Goal

Determine equilibrium constant for a redox reaction from Eo.



• Calculate the standard free energy change (∆G˚) and the equilibrium constant (K) for the following reactions at 25°C:

1. Sn(s) + 2 Cu2+(aq) Sn2+(aq) + 2 Cu+(aq)

2. Fe2+(aq) + 2 Ag(s) Fe(s) + 2 Ag+(aq)

3. 4 Fe2+(aq) + O2(g) + 4 H+(aq) 4 Fe3+(aq) + 2 H2O(l)

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• Cell potentials can be modified by temperature and composition changes according to the equation:

∆G = ∆G° + RT lnQ

∴ –nFE = –nFE° + RT lnQ

• Giving:

The Nernst Equation 01The Nernst Equation 01

QnFRTEE o ln−=


Learning Goal

Determine a cell potential under non-standard-state conditions using Nernstequation (given: table of standard reduction potentials).


• Consider the reaction of metallic zinc with hydrochloric acid. Calculate the cell potential at 25°C when [H+] = 1.0 M, [Zn2+] = 0.0010 M, and PH2

= 0.10 atm.

The Nernst Equation 03The Nernst Equation 03

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What is E for the following cell?

Zn(s) | Zn2+(2.0 M) || Cu2+(0.010 M) | Cu(s)


Practical Use: pH Meter

What is pH of the solution in the anode half-cell below?Given: E is measured to be 0.78 V, and

Eo[Cu+(aq) Cu(s)] is 0.52 V.

Pt(s) | H2(g,1 atm) | H+(aq,? M) || Cu+(aq,1 M) | Cu(s)

Reference electrodeSolution with unknown pH


Actual pH meter

Ag(s) | AgCl(s) | Cl–(aq)

One electrode (“glass electrode”):

Reference electrode (“calomel electrode”):

Hg2Cl2(s) | Hg(l), Cl–(aq)

Ag(s) + Cl–(aq) AgCl(s) + e– Eoox = -0.22 V

Hg2Cl2(s) + 2e– 2Hg(s) + 2Cl–(aq) Eored = 0.28 V

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E depends on pH via potential across glass membrane, as well as on Eo

ox and Eored.