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ElectrochemistryElectrochemistry

An An electrochemicalelectrochemical cell produces cell produces electricity using a chemical reaction.electricity using a chemical reaction.

It consists of two half-cells connected via an It consists of two half-cells connected via an external wire with a salt bridge connecting external wire with a salt bridge connecting the solutionsthe solutions

An An electrolyticelectrolytic cell uses an external cell uses an external electricity source to produce a chemical electricity source to produce a chemical reaction. This is usually called electrolysis.reaction. This is usually called electrolysis.

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HistoricallyHistorically

Historically oxidation involved reaction with O2.i.e., Rusting

4 Fe(s) + 3O2 (g) Fe2O3 (s)

Another exampleZn(s) + Cu2+

(aq) Zn2+(aq) + Cu(s)

In this reaction:Zn(s) Zn2+

(aq) Oxidation

Cu2+(aq) Cu(s) Reduction

In a redox reaction, one process can’t occur without the other. Oxidation and Reduction reactions must simultaneously occur.

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Redox Between Zn and Cu Redox Between Zn and Cu

If Zn(s) and Cu2+(aq) is in

the same solution, then the electron is a transferred directly between the Zn and Cu.

No useful work is obtained. However if the reactants are separated No useful work is obtained. However if the reactants are separated and the electrons shuttle through an external path...and the electrons shuttle through an external path...

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Electrochemical CellsElectrochemical CellsVoltaic / Galvanic Cell Apparatus which produce

electricity

Electrolytic Cell Apparatus which consumes electricity

Consider: Initially there is a Initially there is a flow of e- flow of e- After a very short After a very short time the process time the process stopsstopsElectron transport Electron transport stops because of stops because of charge build up on charge build up on both sidesboth sides

Build up of Build up of positive chargepositive charge

Build up of negative charge

The charge separation will lead The charge separation will lead to process where it cost too to process where it cost too much energy to transfer much energy to transfer electron.electron.

ZnZn CuCu

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Completing the CircuitCompleting the CircuitElectron transfer can occur if the circuit is closedParts: Two conductorsElectrolyte solutionSalt Bridge / Porous membrane

Process that must happen if e- is to flow.A. e- transport through external circuitB. In the cell, ions must migrateC. Circuit must be closed using a salt bridge (no charge build up)

Anode (-)

Green

Negative electrode generates electron

Oxidation Occurs

Cathode (+)

Red

Positive electrode accepts electron

Reduction Occurs

A

B

C

Anode/Anion (-)

Cathode/Cation(+)

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Voltaic CellVoltaic CellElectron transfer can occur if the circuit is closedParts: Two conductorsElectrolyte solutionSalt Bridge / Porous membrane

3 process must happen if e- is to flow.A. e- transport through external circuitB. In the cell, ions a must migrateC. Circuit must be closed (no charge build up)

Anode (-)

Black

Negative electrode generates electron

Oxidation Occur

Cathode (+)

Red

Positive electrode accepts electron

Reduction Occur

Anode/Anion (-)

Cathode/Cation(+)

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Completing the Circuit: Salt Completing the Circuit: Salt BridgeBridge

In order for In order for electrons to move electrons to move through an through an external wire, external wire, charge must not charge must not build up at any build up at any cell. This is done cell. This is done by the salt bridge by the salt bridge in which ions in which ions migrate to different migrate to different compartments compartments neutralize any neutralize any charge build up.charge build up.

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Sign Convention of Voltaic CellSign Convention of Voltaic Cell

@ Anode: Negative Terminal (anions).

Source of electron then repels electrons. Oxidation occurs.

Zn(s) Zn+2(aq) + 2e- : Electron source

@ Cathode: Positive Terminal (cation)

Attracts electron and then consumes electron. Reduction occurs.

Electron target: 2e- + Cu+2(aq) Cu(s)

Overall:

Zn(s) + Cu+2(aq) Zn+2

(aq) + Cu(s) E° = 1.10 V

Note when the reaction is reverse: Cu(s) + Zn+2(aq) Cu+2

(aq) + Zn(s)

Sign of E ° is also reversed E° = -1.10 V

Oxidation: Zn(s) Zn+2(aq) E° = 0.76 V

Reduction: Cu+2(aq) Cu(s) E° = 0.34 V

1.10 V = E°CELL

or E°CELL = E°red (Red-cathode) - E°red (Oxid-anode)

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Another Voltaic CellAnother Voltaic CellZn(s) + 2H+

(aq) Zn+2(aq) + H2 (g) E° = 0.76 V

@ Anode: Negative Terminal (anions): ZnNegative Terminal (anions): Zn(s)(s) Zn Zn+2+2(aq)(aq) + 2e + 2e-- : :

Source of electron then repels electrons. Oxidation occurs.

@ Cathode: Positive Terminal (cation): 2ePositive Terminal (cation): 2e-- + 2H + 2H++(aq)(aq) HH2 (g)2 (g)

Attracts electron and then consumes electron. Reduction occurs.

Overall: ZnOverall: Zn(s)(s) + 2H + 2H++ (aq)(aq) Zn Zn2+2+ (aq)(aq) + H + H22 (g)(g)

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Another Voltaic CellAnother Voltaic CellZn(s) + 2H+

(aq) Zn+2(aq) + H2 (g) E° = 0.76 V

@ Anode: Negative Terminal (anions): ZnNegative Terminal (anions): Zn(s)(s) Zn Zn+2+2(aq)(aq) + 2e + 2e-- : :

Source of electron then repels electrons. Oxidation occurs.

@ Cathode: Positive Terminal (cation): 2ePositive Terminal (cation): 2e-- + 2H + 2H++(aq)(aq) HH2 (g)2 (g)

Attracts electron and then consumes electron. Reduction occurs.

Overall: ZnOverall: Zn(s)(s) + 2H + 2H++ (aq)(aq) Zn Zn2+2+ (aq)(aq) + H + H22 (g)(g)

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basically: metal | ionic solution || ionic solution | metal

This is a convenient way of representing cells1. Anode Cathode

[oxidation (-) ] [reduction (+)]

2. | represents a phase boundary

33. || represents the salt bridge

4. Concentration of component

Zn(s) ZnSO4 (aq,1.0M) CuSO4 (aq,1.0M) Cu(s)

Line Notation ConventionLine Notation Convention

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Line Notation ExamplesLine Notation ExamplesConsider : ZnZn(s)(s) + Cu + Cu+2+2

(aq)(aq) Zn Zn+2+2(aq)(aq) + Cu + Cu(s(s

Anode: Zn Zn+2 + 2e-

Cathode: Cu+2 + 2e- Cu

Shorthand “Line” notation

Zn (s) Zn+2 (aq)(1.0M) Cu+2(aq) (1.0M) Cu(s)

2nd Example : ZnZn(s)(s) + 2H + 2H+ + (aq)(aq) Zn Zn+2+2

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

Anode: Zn Zn+2 + 2e-

Cathode: 2H+ + 2e- H2 (g)

Shorthand “Line” notation

Zn (s) Zn+2 (aq)(1.0M) H+(aq) (1.0M), H2(g, 1atm) Pt(s)

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EMF - ElectroMotive ForceEMF - ElectroMotive Force

Potential energy of electron is higher at the anode. This is the driving force for the reaction (e- transfer)

eeAnode (-)

e- flow toward cathode(+)

Cathode

P.E. = V = J e - C

Larger the gap, the greater the potential (Voltage)

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ElectroMotive ForceElectroMotive ForceFor electrochemical cellsEmf – electromotive force

Potential energy difference between the two electrodesPotential energy difference between the two electrodesUnits of emf: V - Volts :

1V - 1 Joule / Coulomb

1 Joule of work per coulomb of charge transferred.

For electrolysisWhen a current of one amp flow for 1 second this equals one coulomb of charge passed.

Charge (C) = Current (A) x Time (s)

96,400 coulombs = 1 Faraday of charge1 Faraday is equal to 1 mole of electrons

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Standard Electrode Standard Electrode PotentialsPotentialsCell Potential is written as a reduction

equation. M+ + e- M E° = std red. potential

Wri

tten

as r

ed

ucti

on

Most spontaneous <Reduction occurs> Oxidizing Agent

Most non-spontaneous Spontaneous in the reverse direction. <Oxidation occurs> Reducing Agent

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Zoom of Std. Electrode Zoom of Std. Electrode PotentialsPotentialsCell Potential is written as a reduction

equation.

M+ + e- M E° =

F2 (g) + 2e- 2 F - (aq) 2.87 V

Ce4+ + e- Ce3+

(aq) 1.61 V

2H+ + 2e- H2 (g) 0.00 V

Li+(aq) + e- Li(s) -3.045 V

Wri

tten

as r

ed

ucti

on Most

spontaneous Reduction Oxidizing Agent

Most non-spontaneous Spontaneous in the reverse direction. Oxidation Reducing Agent

All reaction written as reduction reaction. But in electrochemistry, there can’t be just a reduction reaction. It must be coupled with an oxidation reaction.

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E°E°CellCell Evaluation EvaluationE°Cell Function of the reaction

Oxidation Process (Anode reaction)

Reduction Process (Cathode reaction)

E°Cell = E°red (cathode) - E°red (anode)

Simple to remember:

Eº(cell) = Eº(red) – Eº(ox)

E = E redox

The potential difference of the cell is equal to the electrode potential of the reduced component minus the electrode potential of the oxidised component.

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Standard Electrode PotentialStandard Electrode Potential

How is E°red (Cathode) and E°red (Anode)

determine.E° (EMF) - State Function; there is no absolute scaleAbsolute E° value can’t be measured experimentallyThe method of establishing a scale is to measure the difference in potential between two half-cells.Consider:

Zn Zn+2 + 2e- E°=?

Can’t determine because the reaction must be coupled

Question? - How can a scale of reduction potential be determined ?

Answer - Use a half reaction as reference and assign it a potential of zero.

Measure all other half cells in comparison to this reaction.

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Side-Bar: Relative ScaleSide-Bar: Relative Scale

Consider a baby whose weight is to be determine but will not remain still on top of a scale. How can the

parents determine the babies weight?

Carry the child in arms and weight both child and parent then subtract the weight of the parent from the total to yield the baby weight.

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Standard Reference Half CellStandard Reference Half CellSelected half reaction is:

H+ / H2 (g) couple half reaction: 2H+ (aq, 1.0M) + 2e- H2 (g,1atm)

by definition E° = 0.0 V, the reverse is also 0.0 V

H+/H2 couple - Standard Hydrogen Electrode (SHE)

To determine E° for a another half reaction, the reaction of interest needs to be connected to this SHE. The potential measured is then assigned to the half-reaction under investigation.

Example:

E°Cell = 0.76 V = E°red (Cat) - E°red

(Anode)

0.0 V - (?)

E°red (Anode) = - 0.76 V

Zn+2/Zn E° = -0.76 V reduction reaction

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Determining Other Half-Cell Determining Other Half-Cell PotentialsPotentials

Now consider the reaction:Zn(s)|Zn+2 (1.0 M)||Cu+2(1.0 M)|Cu(s)E°Cell = 1.10 V

E°Cell = E°red (Cat) - E°red (Anode)

recall, E° Zn+2/Zn = - 0.76 V

Therefore, E°Cell = E°Cu+2/Cu - E° Zn+2/Zn

1.10 V = (?) - (- 0.76 V)

E°Cu+2/Cu = + 0.34 V

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Example: Half-Cell PotentialExample: Half-Cell Potential

Example: For the reaction: Tl+3 + 2Cr2+ Tl+ + 2Cr3+ E°Cell = 1.19 V

i) Write both half reaction and balanceii) Calculate the E°Cell Tl+3 Tl+

iii) Sketch the voltaic cell and line notation

i) Tl+3 + 2e- Tl+

(Cr2+ 2Cr3+ + 2e- ) x 2 E° = 0.41 V

ii) E°Cell = 1.19 V = E°red (Cat) - E°red (Anode)

1.19 V= E°red (Cat) - 0.041 V

for Tl+3 + 2e- Tl+ : 1.19 V - 0.41 = E°red (Cat) = 0.78 V

Pt

Pt

Cr2+ Cr3+ Tl3+ Tl+

1.19 V

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Voltaic Vs. Electrolytic CellsVoltaic Vs. Electrolytic Cells

General characteristics of voltaic and electrolytic cells. A voltaic cell generates energy from a spontaneous reaction (G<0), whereas an electrolytic cell requires energy to drive a nonspontaneous reaction (G>0). In both types of cell, two external circuits provides the means or electrons to flow. Oxidation takes place all the anode, and reduction takes place at the cathode, but the relative electrode changes are opposite in the two cells.

Oxidation ReactionA- A + e-

Oxidation ReactionX X+ + e-

Reduction Reactione- + Y+ Y

Reduction Reactione- + B+ B

Overall (Cell) ReactionX + Y+ X+ + Y, G

= 0

Overall (Cell) ReactionA- + B+ A + B, G> 0

Voltaic CellEnergy is released from

spontaneous redox reaction

System does work on load (surroundings)

Electrolytic CellEnergy is absorbed to drive

nonspontaneous redox reaction

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

Anode (Oxidation)