Electrochemistry Full Slides

ELECTROCHEMISTRY

Electrochemistry

Oxidation-Reduction reactions.

LEORA (loose electron oxidation reducing agent)

GEROA (gain electron reduction oxidizing agent)

Cu2+ + Zn Zn2+ + Cu

Reduction: Cu2+ + 2e- Cu

Oxidation: Zn Zn2+ + 2e-

(Half-reactions of the redox process)

The two parts of the reaction can be physically

separated.

The oxidation reaction occurs in one cell .

The reduction reaction occurs in the other cell.

A cell is a compartment for the half-reaction

The cell must contain all physical forms of the species

involved

Reduction half-reaction cell contains aqueous Cu2+ and solid Cu

Oxidation half-reaction cell contains aqueous Zn2+ and solid Zn

The combination of two cells (reduction and

oxidation cell) is called a electrochemical cell

Electrochemistry

There are two kinds electrochemical cells.

1. Electrochemical cells containing

nonspontaneous chemical reactions are

called electrolytic cells.

2. Electrochemical cells containing

spontaneous chemical reactions are called

voltaic or galvanic cells.

Electrical Conduction

Metals conduct electric currents well in a process called metallic conduction.

In metallic conduction there is electron flow with no atomic motion.

In ionic or electrolytic conduction ionic motion transports the potential.

Positively charged ions, cations, move toward the negative electrode (cathode)

Negatively charged ions, anions, move toward the positive electrode (anode)

Electrodes

The surface in which oxidation or reduction half-reaction occurs is called the ELECTRODE. Electrode part of reaction: active electrode

Electrode not a part of reaction: INERT electrode

Convention for electrodes (correct for either electrolytic or voltaic cells): Cathode : reduction

Negative in electrolytic cells and positive in voltaic cells.

Anode : oxidation Positive in electrolytic cells and negative in voltaic cells.

Voltaic or Galvanic Cells

Electrochemical cells in which a spontaneous chemical reaction produces electrical energy.

Cell halves are physically separated so that electrons (from redox reaction) are forced to travel through wires and creating a potential difference.

Examples of voltaic cells include:

Construction of Simple Voltaic Cells

Voltaic cells consist of two half-cells which contain

the oxidized and reduced forms of an element (or

other chemical species) in contact with each other.

A simple half-cell consists of:

A piece of metal immersed in a solution of its ions.

A wire to connect the two half-cells.

And a salt bridge to complete the circuit, maintain

neutrality, and prevent solution mixing.

The Zinc-Copper Cell

Cell components for the Zn-Cu cell are:

1. A metallic Cu strip immersed in 1.0 M copper (II) sulfate.

2. A metallic Zn strip immersed in 1.0 M zinc (II) sulfate.

3. A wire and a salt bridge to complete circuit

The cells initial voltage is 1.10 volts

In all voltaic cells, electrons flow spontaneously from the negative electrode (anode) to the positive electrode (cathode).

The Zinc-Copper Cell

Short hand notation for voltaic cells.

The Zn-Cu cell provides a good example.

Zn | (1.0 M) Zn2+ || Cu2+ (1.0 M) | Cu

Reduction half-reactionOxidation half-reaction

Double

vertical

bar : Salt

bridge

Single vertical

bar: electrode

Single vertical

bar: electrode

The Copper - Silver Cell Cell components:

1. A Cu strip immersed in 1.0 M copper (II) sulfate.

2. A Ag strip immersed in 1.0 M silver (I) nitrate.

3. A wire and a salt bridge to complete the circuit.

The initial cell voltage is 0.46 volts.

The Copper - Silver Cell

Compare the Zn-Cu cell to the Cu-Ag cell The Cu electrode is the cathode in the Zn-Cu cell.

The Cu electrode is the anode in the Cu-Ag cell.

Whether a particular electrode behaves as an anode or as a cathode depends on what the other electrode of the cell is.

The Copper - Silver Cell

These experimental facts demonstrate that Cu2+ is a stronger oxidizing agent than Zn2+.

In other words Cu2+ oxidizes metallic Zn to Zn2+. Similarly, Ag+ is is a stronger oxidizing agent than Cu2+.

Because Ag+ oxidizes metallic Cu to Cu 2+. If we arrange these species in order of increasing strengths, we see

that:

Standard Electrode Potential

Potential in the chemical sense, means the tendency to transfer electrons

To measure relative electrode potentials, we must establish an arbitrary standard.

That standard is the Standard Hydrogen Electrode (SHE).

The SHE is assigned an arbitrary voltage of 0.00000000 V

To determine the ability of a reaction to transfer or accept electrons relative to the SHE

The Zinc-SHE Cell

For this cell the components are:

1. A Zn strip immersed in 1.0 M zinc (II) sulfate.

2. The other electrode is the Standard Hydrogen Electrode.



The cathode is the Standard Hydrogen Electrode.

In other words Zn reduces H+ to H2.

The anode is Zn metal.

Zn metal is oxidized to Zn2+ ions.

The Zinc-SHE Cell

The Copper-SHE Cell The cell components are:

1. A Cu strip immersed in 1.0 M copper (II) sulfate.

2. The other electrode is a Standard Hydrogen Electrode.



In this cell the SHE is the anode

The Cu2+ ions oxidize H2 to H+.

The Cu is the cathode.

The Cu2+ ions are reduced to Cu metal.

The Copper-SHE Cell

Uses of Standard Electrode Potentials

Electrodes that force the SHE to act as an anode are assigned

positive standard reduction potentials.

Electrodes that force the SHE to act as the cathode are assigned

negative standard reduction potentials.

Standard electrode (reduction) potentials tell us the tendencies of

half-reactions to occur as written.

Potassium (K0 ) has greater

tendency to give electrons (to

be OXidized relative to SHE)

Fluorine (F20 ) has greater

tendency to accept electrons

(to be REDuced relative to SHE)


Standard electrode potentials are used to predict

whether an electrochemical reaction (at standard

state conditions) will occur spontaneously.

Standard electrode potentials are used to predict

which redox couple (if there are many) will occur at

the specified standard conditions

Strongest

Oxidizing

Agent

Strongest

Reducing

Agent


1. Choose the appropriate half-reactions from a table of standard reduction potentials.

2. Write the equation for the half-reaction with the more positive E0 value first, along with its E0 value.

3. Write the equation for the other half-reaction as an oxidation with its oxidation potential, i.e. reverse the tabulated reduction half-reaction and reverse the sign of its E0.

4. Balance the electron transfer. Do not multiply the E0 values by the coefficient! (E0 values are INTENSIVE properties)

5. Add the reduction and oxidation half-reactions and their potentials. This produces the equation for the reaction for which E0cell is positive, which indicates that the forward reaction is spontaneous (galvanic). If the system has a negative E0cell , the system must be nonspontaneous(electrolytic).

Will permanganate ions, MnO4-, oxidize iron (II) ions to iron (III) ions, or

will iron (III) ions oxidize manganese(II) ions to permanganate ions in acidic solution? (Which is more spontaneous?)

Will silver ions (Ag+) oxidize metallic zinc (to Zn2+) or will zinc ions

(Zn2+) oxidize metallic Ag (to Ag+)? (Which is more spontaneous?)

Gibbs Free Energy and Electrical

Potential at Standard Conditions

For a redox reaction at standard states

G0 = -nFE0celln # of electrons involved in the balanced reaction

F 96 485 J/Vmol e- (Faradays Constant)

E0cell cell potential at standard conditions

G0 standard Gibbs Free Energy

Relationship of E0cell and Equilibrium constant Keq

G0 = -nFE0cell = -RTlnKeq

E0cell = RTlnKeq / nF or lnKeq = nFE0

cell / RT

kJ/mol 302.23-or rxn J/mol 302230- G

V 1.5662e mole V

J 485 96e mole 2- G

0

-

-0

=

+=

kJ/mol 285.60-or rxn J/mol 285595- G

V 0.74e mole V

J 485 96e mole 4- G

0

-

-0

=

+=

Consider the following reduction half-reactions involving water

V 0.83- OH2He2O2H

V 0.40 4OHe4O2HO

V 1.23 O2H 4e 4H O

-

(aq)2(g)

-

(l)2

-

(aq)

-

(l)22(g)

(l)2

-

(aq)2(g)

++

++

++ +

V 2.06 O2H 2H O

V 0.83 )e2OH2 OH22(H

V 1.23 O2H 4e 4H O

(i)2(g)22(g)

-

(l)2

-

(aq)2(g)

(l)2

-

(aq)2(g)

+

++

++ +

V 1.23 O2H 2H O

V 0.83 )e2OH2 OH22(H

V 0.40 4OH 4e O2H O

(i)2(g)22(g)

-

(l)2

-

(aq)2(g)

-

(aq)

-

(l)22(g)

+

++

++

More

spontaneous, but

cathode requires

H+ (will come from

H2 gas)

Less spontaneous,

but cathode

requires H2O

V 2.06 O2H 2H O

V 0.83 )e2OH2 OH22(H

V 1.23 O2H 4e 4H O

(l)2(g)22(g)

-

(l)2

-

(aq)2(g)

(l)2

-

(aq)2(g)

+

++

++ +

Latimer Diagrams

A convenient way of writing redox processes and

their corresponding potential

What is the potential needed to generate Fe(s) from Fe3+?

Iron Fe(s) has two stable oxidation states, Fe(III) and Fe(II).

G follows Hess Law, so

calculation of new E0 must

come from G0 obtained

from Hess Law summation

The values are all

standard reduction

potentials (Left-most has

the highest positive

oxidation state, then

right-most lowest

oxidation state)

Toward reduced state (reduction process)

Frost Diagrams

In general, anything which alters the free energy of the

system will change the redox potentials. The following

factors all affect the size of G:

(1) Concentration

(2) Temperature

(3) Other reagents which are not inert

(4) pH (a special case of (3))

Frost Diagrams

Another convenient way of writing redox

potential in graphical form

Latimer

Frost

Frost diagrams

are essentially

free energy

plotted against

oxidation state,

where G is

given in units of

nF (i.e. in Volts)

A Frost diagram relates the free energy of any given redox state to the energy of

the elemental form. We calculate the value nE for the overall redox couple

X(N)/X(0) and plot each couple against the oxidation state N. This means each

point in the plot is referenced against the elemental form.

For Manganese, we start by calculating

the two-electron process between Mn2+

and Mn, which works out to 2.36 V. For

the next step, we recognize that Mn+3 to

Mn2+ is an additional one-electron step,

so that +1.5 V must be added to the

previous step to get the value that relates

the energy of Mn3+ to that of elemental

manganese.

a) Identifying oxidizing and reducing

agents

Species lying high on the diagram are OA

(reduces) towards species on their left,

while species high on the diagram act as

RA (oxidizes) to other oxidizing agents

on their right. Another way to visualize

this is by considering the lines

connecting the higher and lower lying

species. If the line has a positive slope,

the higher-lying species is an OA. If the

line has a negative slope, the higher-

lying species is a RA.

b) Identifying strong and weak agents

Relative strength of RA and OA can be

determined by the steepness of the

slope of the lines. The steeper the

slope, the stronger the agent.

c) Identifying redox products (unreactive

redox states)

Species at the bottom of the graph have low

free-energy, thus little tendency to react. The

lowest species on the graph are the

thermodynamic final product(s) of the redox

reactions involving that element

d) Identifying species likely to undergo

disproportionation

If a species lies above the line connecting

its neighbors, it is thermodynamically

unstable towards disproportionation. This

has been described as a point lying along a

concave line.

e) Identifying species likely to undergo

comproportionation

Species likely to undergo

comproportionation to a third species

are located to left and right of a point

which lies below the line connecting the

two species

Disproportionate

Direction of

comproportionation

Which is the strongest oxidizing agent relative to N2 in acidic conditions?

Which is the strongest reducing agent relative to N2 in basic conditions?

Identify 2 species that will comproportionate in acidic conditions

Identify 4 species that will disproportionate in basic conditions

Effect of Concentrations (or Partial

Pressures) on Electrode Potentials

The Nernst Equation

Standard electrode potentials, those compiled

in appendices, are determined at

thermodynamic standard conditions.

The value of the cell potentials change if conditions are nonstandard.

The Nernst equation describes the electrode potentials at nonstandard conditions.

The Nernst Equation

The Nernst Equation

Substitution of the values of the constants into the

Nernst equation at 25o C gives:

The Nernst Equation

For this half-reaction:

The corresponding Nernst equation is:

The Nernst Equation

Substituting E0 into the above expression gives:

If [Cu2+] and [Cu+] are both 1.0 M, i.e. at standard conditions, then E = E0 because the concentration term equals zero.

The Nernst Equation

Calculate the potential for the Cu2+/Cu+ electrode at

250C when the Cu+ ion concentration is 1/3 of the Cu2+

ion concentration.

The Nernst Equation

Calculate the initial potential of a cell that consists of

an Fe3+/Fe2+ electrode in which [Fe3+]=1.0 x 10-2 M

and [Fe2+]=0.1 M connected to a Sn4+/Sn2+ electrode

in which [Sn4+]=1.0 M and [Sn2+]=0.10 M . A wire

and salt bridge complete the circuit.

Calculate the E0 cell by the usual procedure.

Substitute the ion concentrations into Q to calculate

Ecell.

Measurement of Equilibrium constants from Eocell

AgBr s Ag aq + Br aq E = 0.729V

E =0.0592

1log Ag Br =

0.0592

1logK

Ksp = ?

Cathode:

AgBr s + e Ag s + Br aq ! = 0.071V

Anode:

Consider the following half-reactions:

Ag aq + e Ag(s) != 0.800V

AgBr s + e Ag s + Br aq E = 0.071V

Ag s Ag aq + eE= 0.800V

AgBr s Ag aq + Br(aq)

Measurement of Equilibrium constants from Eocell

Consider the following half-reactions:

Au& aq + 3e Au(s) != 1.520V

[AuCl*] aq + 3e Au s + 4Cl aq ! = 1.002V

Au& aq + 4Cl(aq) [AuCl*] aq -. =?

Cathode:

Anode:

Au& aq + 3e Au(s) != 1.520V

Au s + 4Cl aq [AuCl*] + 3e aq != 1.002V

Au&(aq) + 4Cl aq [AuCl*](aq) != 0.518V

E =0.0592

3log

[AuCl*]

Au& Cl *=0.0592

3logK0

CONCENTRATION CELL

Measurement of Ksp using concentration cells

Ag 0.100M + e Ag(s)Ag saturated + e Ag(s)

!4566 = !4566 0.0592

1log

[78(9:;?)]

[78 = 0.100]

Ag 0.100M + e Ag(s)Cathode:

Anode: Ag s Ag saturated + e

Ag 0.100M Ag(saturated)

Electrolytic Cells Nonspontaneous electrochemical cells

E0cell is negative

G0 is positive [G0 = -nF(-E0cell value)= + value]

Eletrolysis means forcing a nonspontaneous redox reaction to

occur (at the electrodes) by applying potential

Spontaneous, releases 212.3

kJ/mole of energy

Nonspontaneous, requires

212.3 kJ/mole of energy

Electrolytic Cells

Electrolytic cells requires external potential source

In electrolytic cells, reduction still occurs at cathode, oxidation

at anode

Polarity is reversed

o cathode: negative

o anode: positive

Electrolysis can be done in 1-compartment or 2-compartment

cells

Two examples of commercial electrolytic reactions are:

Electroplating of jewelry and auto parts.

Electrolysis of chemical compounds.

Electrolytic Cell: 1-compartment

A container for the

reaction mixture.

Two electrodes

immersed in the

reaction mixture.

A voltage source of

direct current.

The motion of the ions through the solution = electric current.

Electrolytic conduction

o Positive ions migrate toward the negative electrode (cathode).

o Negative ions migrate toward positive electrode (anode).

Electrolytic Cell: Electrolysis of NaCl

Requires molten (melted) NaCl since ions conduct electricity Consists of Na+ and Cl- ions.

If direct current (greater than 3.852 V) is applied (by way of two inert electrodes) through the cell containing the molten NaCl, we observe the following: Pale green gas (Cl2) is formed in one electrode.

Molten, silvery Na forms at the other electrode. The Na particles floats on top of the molten NaCl.

Liquid Na is produced at the cathode (-).

2{Na+ + e Na(l)*}

Gaseous Cl2 is produced at the anode (+).

2Cl- Cl2 + 2e-

---------------------------------

2Na++2Cl-2Na(l)+Cl2(g)

*Na(l) Na(s)

Not possible since the set-up is at high temperatures (>801C)

Direct current (dc) source supplies energy to force the reaction forward.

AnodeCathode

Electrons are used in the cathode half-reaction (reduction) and produced in the anode half-reaction (oxidation).

Travel of e- : from ANODE(+) to CATHODE(-).

The dc source forces the e- to flow non-spontaneously from the positive electrode to the negative electrode.

Na and Cl2 must NOT be allowed to come into contact with each other because they react spontaneously, rapidly and explosively to form NaCl.

In Downs Cell, the liquid Na is drained off, cooled and cast into blocks, then stored in inert mineral oil to prevent reaction with atmosphere (O2) or H2O.

Downs Cell Industrial

production of Na(s)

Voltaic/Galvanic cellso Cathode (+): Most Positive E0redo Anode (-): Most Negative E0redo The MOST positive E0cell will

occur first

Electrolytic cellso Cathode (-): Least Negative E0redo Anode (+): Least Positive E0redo The LEAST negative E0cell will

occur first

In both cases, REDUCTION occurs at CATHODE, OXIDATION occurs at ANODE

(REDCAT, OXAN)

V 0.83- OH2He2O2H

V 0.40 4OHe4O2HO

V 1.23 O2H 4e 4H O

-

(aq)2(g)

-

(l)2

-

(aq)

-

(l)22(g)

(l)2

-

(aq)2(g)

++

++

++ +

V 2.06 O2H 2H O

V 0.83 )e2OH2 OH22(H

V 1.23 O2H 4e 4H O

(l)2(g)22(g)

-

(l)2

-

(aq)2(g)

(l)2

-

(aq)2(g)

+

++

++ +

V 1.23- 2H OO2H

V 0.40 - 4e O2H O4OH

V 0.83- )OH2He2OH22(

(g)22(g)(l)2

-

(l)22(g)

-

(aq)

-

(aq)2(g)

-

(l)2

+

++

++

Electrolytic conditionsVoltaic/Galvanic conditions

Electrolytic Cells: Electrolysis of Aqueous KI

K+ (aq) + e- K(s) -2.924 V

2H2O + 2e- 2H2 (g) + OH- -0.828 V

I2(s) + 2e- 2I- (aq) +0.535 V

O2 (g) + 4H+ +4e- 2H2O +1.229 V

Calculations of different redox combinations and the minimum applied voltages required for electrolysis:

E1 = E(K+/K) E(I2/I) = (-2.924 V) (+0.535 V)

= -3.459V

E2 = E(H2O/H2,OH) E(I-/I2) = (-0.828V) (0.535 V)

= -1.363V

E3 = E(H2O/H2,OH) E(H2O/O2, H+)=(-0.828V) (1.229)

= -2.057

2nd case has smallest V and hence, this

combination will occur first.

cathode anode

Electrolytic Cells: Electrolysis of Aqueous NaCl

Na+ (aq) + e- Na(s) -2.714 V

2H2O + 2e- 2H2 (g) + OH- -0.828 V

Cl2(s) + 2e- Cl- (aq) +1.360 V

O2 (g) + 4H+ +4e- 2H2O +1.229 V

Calculations of different redox combinations and the minimum applied voltages required for electrolysis:

E1 = E(Na+/Na) E(Cl-/Cl2) = (-2.714 V) (+1.360 V) = -4.074 V

E2 = E(H2O/H2,OH) E(Cl2/Cl-) = (-0.828V) (1.360 V) = -2.188 V

E3 = E(H2O/H2,OH) E(H2O/O2, H+)=(-0.828V) (1.229) = -2.057

The third case has the smallest V so we expect that O2(g) would be generated first before Cl2(g).

However, the overpotential is higher for the former (O2) which explains the experimental results.

Thus overpotential needs to be considered when deciding what will be the products of electrolysis.

E2 = E(H2O/H2,OH) E(Cl-/Cl2) = (-0.828V) (1.360 V) = -2.188 V

E3 = E(H2O/H2,OH) E(H2O/O2, H+)=(-0.828V) (1.229) = -2.057

OVERPOTENTIAL extra potential due to

interference of electron flow from electrode

to chemical species

Counting Electrons: Coulometry and

Faradays Law of Electrolysis

Faradays Law - The amount of substance undergoing chemical reaction at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the electrolytic cell.

A faraday is the amount of electricity that reduces one equivalent of a species at the cathode and oxidizes one equivalent of a species at the anode.

A coulomb is the amount of charge that passes a given point when a current of one ampere (A) flows for one second.

1 amp = 1 coulomb/second

@.ABACAD = E. FGG H @FGIJ = KELMNOPQRPSTU



Faradays Law states that during electrolysis, one faraday of electricity (96,485 coulombs) reduces and oxidizes, respectively, one equivalent of the oxidizing agent and the reducing agent.

This corresponds to the passage of one mole of electrons

through the electrolytic cell.



Calculate the mass of palladium produced by the reduction of palladium (II) ions during the passage of 3.20 amperes of current through a solution of palladium (II) sulfate for 30.0 minutes.



Calculate the volume of oxygen (measured at STP)

produced by the oxidation of water.

Commercial Applications of Electrolytic

Cells

Electrolytic Refining and Electroplating of Metals

Impure metallic copper can be purified

electrolytically to 100% pure Cu. The impurities commonly include some active metals plus

less active metals such as: Ag, Au, and Pt.

The cathode is a thin sheet of copper metal

connected to the negative terminal of a direct

current source.

The anode is large impure bars of copper.

Commercial Applications of Electrolytic Cells

The electrolytic solution is CuSO4 and H2SO4

The impure Cu dissolves to form Cu2+.

The Cu2+ ions are reduced to Cu at the cathode.

Any active metal impurities are oxidized to cations that are more difficult to reduce than Cu2+.

This effectively removes them from the Cu metal.

Corrosion Metallic corrosion is the oxidation-reduction

reactions of a metal with atmospheric components such as CO2, O2, and H2O.

( )points. exposedat rapidly occursreaction The

reaction overall OFe 2O 3+Fe 4 320

2

0

O2 + 4 e- + 2 H2O 4 OH

-

Fe Fe2+ + 2 e

4 Fe2+ + O2 4 Fe3+ + 2 O2

2e- + O2- + 2H2O 4OH-

Fe3+ + 3 H2O Fe(OH)3 + 3 H+

Fe(OH)3 FeO(OH) + H2O

2FeO(OH) Fe2O3 + H2O

Corrosion Protection Some examples of corrosion protection.

1. Plate a metal with a thin layer of a less active (less easily oxidized) metal.

2. Connect the metal to a sacrificial anode, a piece of

a more active metal.

Corrosion Protection3. Allow a protective film to form naturally.

4 Galvanizing, the coating of steel with zinc, provides a more active metal on the exterior.

Corrosion Protection

5. Paint or coat with a polymeric material such as

plastic or ceramic.

ceramic. with coated are bathtubs Steel

Battery an electrochemical cell

Primary cell non-rechargeable, non-reversible

spontaneous reaction involved (by design)

Secondary cell rechargeable, reversible

system involved (the design allows for

reversibility)

Flow battery or fuel cell materials can be

continuously passed through the battery, and

converts chemical energy to electrical energy

Primary Cells

-Leclanche battery

Side

reaction:

Alkaline

batteries:

Lead acid storage battery (secondary

battery)

Zn-Ag Button cell

Lithium Ion battery

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