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ElectrochemistryElectrochemistryElectrochemistry deals with the conversionsbetween electrical and chemical energy.In particular, the chemical reactions areoxidation-reduction reactions.

Oxidation reactions give off electrons and what is being oxidized is called

the reducing agent. Reduction reacts take in electrons and what is being reduced is called the oxidizing agent.

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Remember, LEO goes GER.

LEO is the loss of electrons - oxidation.

GER is the gain of electrons - reduction.

An electrochemical cell (also called a voltaicor galvanic cell) converts chemical energy

toelectrical energy.

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An Electrochemical CellAn Electrochemical Cell

V

voltmeter

salt bridge

Zn (anode/-)Cu (cathode/+)

electrolyte (NaNO3)

(Na+, NO3-)

ZnSO4 (Zn2+, SO42-)CuSO4 (Cu2+, SO4

2-)

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The following is what should be known about

this particular voltaic cell: The oxidation occurs in the anode compartment.

The anode reaction is Zn → Zn2+ + 2e-.

Because electrons are not good swimmers, they collect on the metal

strip which is called the anode (a negative electrode).

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The reduction occurs in the cathode compartment.

The cathode reaction is Cu2+ + 2e- → Cu.

The cathode is assigned a positive charge.

A voltmeter connected to the two electrodes will measure the electromotive force (emf) of the cell.

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Electromotive force (E) is really a misnomer because what is being measured is voltage.

Voltage really measures the electric potential energy in joules/coulomb, 1 V = J/C. Electrons will flow from negative to positive (anode to cathode) in the external circuit.

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Internally within the cell itself, the cations (positive ions) move to the cathode and the anions move to the anode. The salt bridge allows for ions to flow in and out of each cell compartment.

A salt is chosen so that it’s ions, in this case Na+ and NO3

-, will not react with the species in either compartment.

The salt bridge allows for electric neutrality.

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The Na+ provided by the salt bridge migrate to the cathode to compensate for the reduction of cations.

The NO3- migrate to the anode to

compensate for the increasing concentration of cations.

Standard state conditions for a voltaic cell exist when gases are at a pressure of 1 atm, all solutes are at a concentration of 1 M, and T = 25°C = 298K.

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All aqueous solutions must have a concentration of 1.0 M, a pressure of 1.0 atm and a temperature of 25°C. As previously stated, the term force is somewhat of a misnomer.

Because the anode is considered at a higher potential than the cathode,

what actually is being measured is electrical potential energy in volts,

1 V = 1 J1 C

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What Is The Voltmeter Measuring?What Is The Voltmeter Measuring?A voltmeter measures the cell emf whichmeasures the potential difference betweenthe anode and cathode.

The reference electrode which all other electrodes are compared to is the SHE (Standard Hydrogen Electrode).

2H+(aq) + 2e- → H2(g) E° = 0.00 V

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The cell potential for any combination of

electrodes is given by

E°cell = E°ox + E°red

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Predicting Spontaneous ReactionsPredicting Spontaneous Reactions

(a)Is the following reaction spontaneous? If it is spontaneous, determine the cell

emf and identify the anode and cathode.

Zn + 2H+ → Zn2+ + H2

Zn → Zn2+ + 2e- E°ox = 0.76 V

2H+ + 2e- → H2 E°red = 0.00 VZn(s) + 2H+(aq) → Zn2+(aq) + H2(g) E° = 0.76 V

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The reaction is spontaneous because E° > 0.

Zn is being oxidized making Zn the reducing agent or reductant.

H+ is being reduced making H+ the oxidizing agent or oxidant.

It did not occur in this problem but if you had to multiply the number of electrons

by a coefficient to balance them, you would not multiply the E° by the coefficient.

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The mass of the electrodes change as theoxidation-reduction reaction proceeds.

At the anode, zinc atoms are consumed

as they are oxidized into Zn2+ cations which decreases the mass of the Zn anode. At the cathode H+ cations are reduced to H2 molecules which increases the mass of the accumulated hydrogen.

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(b) Is the following reaction spontaneous? If

it is spontaneous, determine the cell emf

and identify the anode and cathode.

Li+ + Na → Na+ + Li

Na → Na+ + e- E° = 2.71 VLi+ + e- → Li E° = -3.04 V

E° = -0.33 V

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The reaction is not spontaneous because E° < 0.

The reverse direction is spontaneous and is given by Na+(aq) + Li(s) → Na(s) + Li+(aq) E° = 0.33 V Li is being oxidized making Li the reducing agent or reductant.

Na+ is being reduced making Na+ the oxidizing agent or oxidant.

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Sometimes a line convention is used toidentify the anode, salt bridge, and cathodeas shown below.

Zn|Zn2+(aq)||H+(aq)|H2

The anode components are written to the

left of the salt bridge symbol (||) and cathode components are written to the right.

The single vertical line indicates a phase boundary as in the case of (s) and (aq).

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Determine which of the following metals Ni,Cr, and Ca is the most active metal.

Ca2+(aq) + 2e- → Ca(s) E°red = -2.87 VCr3+(aq) + 3e- → Cr(s) E°red = -0.74 V Ni2+(aq) + 2e- → Ca(s) E°red = -0.28 V

The more active the metal, the more it wantsto give off electrons.

Reversing the reduction reactions givesE°ox = 2.87 V (TP!) corresponding to calciummaking it the most active metal.

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SRP SummarySRP SummaryThink Positive!The more positive the standard reductionpotential, E°red, the more the species wants

tobe reduced making it a stronger oxidizingagent.

The stronger the oxidizing agent the weaker the conjugate reducing agent. F2 is the strongest oxidizing agent because it is the most easily reduced, E°red = 2.87 V.

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If F2 is the easiest species to reduce, then F- must be the hardest to oxidize.

Although this logic may sound twisted, it is very similar to that of a strong acid-weak conjugate base pair. The stronger the reducing agent the weaker the conjugate oxidizing agent.

The more negative the standard reductionpotential, E°red, the more the species wants tobe oxidized making it a stronger reducingagent.

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The stronger the reducing agent the weaker the conjugate oxidizing agent. Li is the strongest reducing agent because it is the most easily oxidized, E°ox = 3.05 V. If Li is the easiest species to oxidize, then Li+ must be the hardest to

reduce.

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Spontaneity, Spontaneity, E°, and E°, and ΔΔG°G°

When there is a spontaneous redox reaction,

E° > 0.

The equation relating ΔG° (Gibb’s freeenergy) and cell potential is given by

ΔG° = -nFE°where

n is the number of moles of electronstransferred and F is Faraday’s constant

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and

Remember that the superscript ° indicates ifall the products are in their standard states,the P = 1 atm, the concentrations are 1.0 Mand T = 25°C = 298K.

Many times redox reactions will not be run at

standard state conditions and you will have to

use the Nernst equation given by

F =96500 JV • mol e-

.

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.

E =E° - 0.0591 Vn log Q

where E° is the cell potential at standard conditions, Q is the reaction quotient, n is the number of moles of electrons, and T = 25°C = 298K.

Remember, the concentrations of solids, liquids, and solvent do not appear in the reaction quotient.

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If a redox reaction reaches equilibrium,ΔG = 0 and E = 0 and at 298K the following

istrueE° = 0.0591 V

n log K

The emf is not converted to work with 100%efficiency.

There is always energy wasted in the form ofheat with ΔG° being the work attainable.

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Calculate Calculate ΔΔG in an Alkaline G in an Alkaline SolutionSolution

Calculate the standard free energy, ΔG, forthe following redox reaction in a basicsolution.

Ni2+(aq) + Cr(OH)3(s) → Ni(s) + CrO42-(aq)

Ni2+ + 2e- → Ni

Cr(OH)3 → CrO42-

-1-33

-2-86

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Cr(OH)3 → CrO42- + 3e-

Cr(OH)3 + H2O → CrO42- + 3e- + 5H+

3Ni2+ + 6e- → 3Ni2Cr(OH)3 + 2H2O → 2CrO4

2- + 6e- + 10H+3Ni2+ +2Cr(OH)3 + 2H2O → 3Ni + 2CrO4

2- + 10H+

10OH- 10OH-

3Ni2+(aq) +2Cr(OH)3(s) + 10OH-(aq) → 3Ni(s) + 2CrO4

2-(aq) + 8H2O(l)

E°cell = E°red + E°ox = -0.28 V + 0.13 V = -0.15 V

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ΔG° = -nFE°

ΔG° = - 6 mol e- ×

ΔG° = 8.7 x 104 J

The reaction is not spontaneous in theforward direction but is so is the reversedirection.

96500 C1 mol e- × -0.15 V

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The Nernst Equation RevisitedThe Nernst Equation Revisited

As an electrochemical cell operates, it isalways trying to achieve chemicalequilibrium.

When the cell potential is measured at theinstant the spontaneous redox reactionbegins, two conditions exist:

The cell potential is a maximum. The system is the furthest from equilibrium as it gets.

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For a given set of electrodes at 25°C = 298K,

cell potential is dependent on concentration. An increase in reactant and a decrease in product concentrations increases the cell potential.

At 25°C = 298K, the Nernst equation (whichwas presented on slides 23-25) calculates

thevoltage when the concentrations are not1.0 M.

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Using The Nernst EquationUsing The Nernst EquationFor the following reaction taking place in avoltaic cell at standard temperature,

2Al(s) + 3I2(s) → 2Al3+(aq) + 6I-(aq)

Determine the emf when [Al3+] = 4.0 x 10-3 M

and [I-] = 0.010 M.

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Al → Al3+ + 3e-

3I2 + 6e- → 6I-

2Al → 2Al3+ + 6e- E°ox = 1.66 V

3I2 + 6e- → 6I- E°red = 0.54 V2Al(s) + 3I2(s) → 2Al3+(aq) + 6I-(aq) E° = 2.20 V

E =E° - 0.0591 Vn log Q

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.

E = 2.20 V 0.0591 V6

log ((4.0 x 10-3)2 × 0.0106)

- ×

=2.37 V

As the cell continues to spontaneouslydischarge, the concentrations change and E° ofthe cell decreases.

Voltaic cells spontaneously discharge untilequilibrium is achieved, at which point the cellis dead.

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Electrolytic (Daniel) CellElectrolytic (Daniel) CellA nonspontaneous redox reaction can beforced to occur by connecting the electrodesto a source of direct current (DC) electricity.

An electrolytic cell converts electrical energy to chemical energy.

This process is called electrolysis.

An electrolytic cell is diagrammed on the nextslide.

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An Electrolytic CellAn Electrolytic Celldirect currentpower source

salt bridge

Zn (cathode/-)Cu (anode/+)

electrolyte (NaNO3)

(Na+, NO3-)

ZnSO4 (Zn2+, SO42-)CuSO4 (Cu2+, SO4

2-)

+ -

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Components of an Electrolytic CellComponents of an Electrolytic CellAt first glance, it is easy to confuse anelectrolytic cell with an electrochemical

cell.

For an electrolytic cell: A source of direct current replaces the voltmeter or electrical appliance. Electrons are forced onto the cathode making the cathode negative.

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Electrons are removed from the anode making it positive. The cathode being the negative electrode and the anode being the positive anode is the opposite of the electrode charge found in a voltaic

cell. The cathode is still the site of the reduction and the anode the site of

the oxidation.

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The reduction reaction is given by

Zn2+(aq) + 2e- → Zn(s)

and the anode reaction is given by

Cu(s) → Cu2+(aq) + 2e-

In an electrochemical cell using zinc and

copper electrodes, the E°cell = 1.10 V.

This requires of the direct current source to be greater than 1.10 V.

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Electrolytical Cell TerminologyElectrolytical Cell Terminology.

I = Qt

I is the current measured in amps (A), Q isthe magnitude of charge in coulombs (C), and t is the time in seconds (s).

1 A =1C1s

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The charge on 1 e- = 1.6 x 10-19 C

1 mol e- = 6.02 x 1023 e- ≈ 96500 C

The quantity of charge can also be expressedin faradays, F, where 1 F = 96500 C = 1 mol e-.

Whenever work is done on any system, thereis an increase of energy, whether it be kineticor potential.

In the case of an electrolytic cell, there will be an increase in electric potential energy.

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.

P = Wt = ΔE

t = Js = watts

1 J =1 watt-s or 3.6 x 106 J = 1 kWh

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Electrolytic ProblemsElectrolytic ProblemsAn acidic solution contains Pb2+ cations andan anion which is a spectator. When thesolution is electrolyzed, PbO2 is formed at theanode and Pb at the cathode.

(a)What reaction occurs at the anode?

The oxidation of Pb2+ occurs at the anode and the reaction is

Pb2+(aq) + 2H2O(l) → PbO2(s) + 4H+(aq) +

2e-

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(b) If a current of 1.25 A is applied to the cell

for 15.0 min, what mass of Pb is deposited

at the cathode?

I = 1.25 A t = 15.0 min 1 mol e- = 96500 Cm = 1.25 Cs × 15.0 min× 60 s

1 min ×

1 mol e-

96500 C × 1 mol Pb4+

2 mol e- ×

207.20 g Pb4+

1 mol Pb4+ =1.21 g Pb4+