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Electrochemistry

Oxidation-Reduction Reactions

Oxidation - refers to the loss of electrons by a molecule, atom or ion

Reduction - refers to the gain of electrons by an molecule, atom or ion

Oxidation number - the oxidation number of an atom is the charge that results when the electrons in a

covalent bond are assigned to the more electronegative atom; it is the charge an atom would possess if the

bonding were ionic

In oxidation numbers, the charge is written before the number (to distinguish them from actual

electronic charges - where the number is placed in front of the charge)

HCl

Hydrogen is less electronegative than Chlorine, therefore, Chlorine gets all the shared electrons in the HCl

bond.

In this compound, the H oxidation number is therefore +1, and the oxidation number of Cl is -1.

Chemical reactions in which the oxidation state of one or more substances changes are called oxidation-

reduction reactions (or redox reactions)

Keeping track of the oxidation numbers of the elements involved in a reaction will identify

whether the reaction is a redox reaction or not

Zn(s) + 2H+(aq) Zn

2+(aq) + H2(g)

The oxidation numbers for the above elements and ions are:

Zn = 0, Zn2+

= +2

H+ = +1, H2 = 0

Thus, the oxidation number of both the Zn(s) and H+(aq) change during the course of the reaction,

and so, this must be a redox reaction

In this reaction an unambiguous transfer of electrons occurs:

The elemental form of zinc (i.e. zinc metal) loses two electrons (and becomes soluble zinc ion)

Two H+ ions in solution gain two electrons and form a hydrogen (H2) molecule

In other types of reactions, we can identify changes in oxidation number, but it is not so clear that a

particular atom has gained or lost an electron:

Hydrogen has been oxidized and oxygen has been reduced in the above reaction involving the

production of water from elemental hydrogen and oxygen

However, the H-O bonds are not ionic, they are polar covalent, so there is not a complete transfer

of shared electrons

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Using oxidation numbers is just a convenient bookkeeping method to understand

reduction/oxidation types of reactions. It does not necessarily mean that complete electron transfer

has occurred.

In a redox reaction, both oxidation and reduction must occur. In other words, something must be oxidized

for something else to be reduced

The compound that contributes the electrons is called the reducing agent. The reducing agent

gives up electrons, causing another compound to be reduced (and is therefore oxidized in the

process)

The compound that accepts electrons is called the oxidizing agent. The oxidizing agent accepts

electrons from another compound, causing the other compound to be oxidized (in the process, the

oxidizing agent is reduced)

In the above reaction of hydrogen and oxygen:

Hydrogen is the reducing agent (it contributes electrons to oxygen, and is itself oxidized)

Oxygen is the oxidizing agent (it accepts electrons from hydrogen - oxidizing it, and is itself

reduced)

Balancing Oxidation-Reduction Reactions

In balancing redox reactions, gains and losses of electrons must be balanced

In many cases the equation can be balanced without explicitly considering the transfer of

electrons. For example:

Zn(s) + 2H+(aq) Zn

2+(aq) + H2(g)

However, many redox reactions are more complicated and cannot be balanced without taking into

account the number of electrons lost and gained during the reaction

Half-Reactions

Although oxidation and reduction must take place simultaneously (i.e. for something to be reduced,

something else must be oxidized) it is often convenient to consider them as separate processes

The following is the oxidation of Tin(II) by Iron(III)

Sn2+

(aq) + 2Fe3+

(aq) Sn4+

(aq) + 2Fe2+

(aq)

This reaction can be considered as two separate processes:

o The oxidation of Sn2+

o The reduction of Fe

3+

Oxidation: Sn2+

(aq) Sn4+

(aq) + 2e-

Reduction: 2Fe3+

(aq) + 2e-

2Fe

2+(aq)

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In the oxidation reaction the electrons are shown as products (i.e. the tin is oxidized and releases

electrons)

In the reduction reaction the electrons are shown as reactants (i.e. the iron ion combines with

electrons as it is reduced)

Equations that show either oxidation, or reduction, alone are called half-reactions

The overall electrons produced in the oxidation half-reaction must equal the number of electrons

consumed in the reduction half-reaction

o When the equations have been balanced in this way, they are combined to produce the

balanced overall redox reaction

Balancing Equations By The Method Of Half-Reactions

The use of half-reactions provides a general way to balance redox reactions

The reaction between permanganate ion (MnO4-) and oxalate ion (C2O4

2-) in acidic solution:

When these ions combine in an acidic solution, they react to produce manganese ion (Mn2+

) and

CO2 gas

Here is the unbalanced equation that describes this reaction:

MnO4- + C2O4

2- Mn

2+(aq) + CO2(g)

To balance this redox reaction using the method of half-reactions, begin by writing the incomplete

oxidation and reduction half-reactions:

MnO4- Mn

2+(aq)

C2O42-

CO2(g)

Which compound is being reduced and which one is being oxidized?

Now, balance the atoms undergoing oxidation or reduction by the appropriate addition of coefficients.

In the above half-reactions, the Mn appears to be balanced on both sides of its half-reaction

However, the carbon is not balanced in the second half-reaction - we need another carbon in the

products

MnO4- Mn

2+(aq)

C2O42-

2CO2(g)

If the reaction is done under acidic aqueous solution, H+ and H2O can be added to reactants or products to

balance H and O atoms. Likewise, if the reaction is done under basic aqueous solution, OH- and H2O can be

added to balance H and O atoms

In the manganese half-reaction we have 4 O atoms on the left and none on the right. We can add 4

H2O molecules on the right to take care of this:

MnO4- Mn

2+(aq) + 4H2O(l)

We now have 8 H atoms on the right that need to be balanced. This reaction is done under acidic

conditions, so we can balance the H atoms by adding H+(aq) ions on the left:

MnO4- + 8H

+ Mn

2+(aq) + 4H2O(l)

The atoms are now balanced, but the charges are not. We have a net charge of 7+ on the left, and

2+ on the right. We want to balance the charge without altering the number of atoms (they are

balanced already). We balance the charge by adding the appropriate number of electrons:

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MnO4- + 8H

+ + 5e

- Mn

2+(aq) + 4H2O(l)

(balanced atoms, balanced charge)

How about the other half-reaction?

C2O42-

2CO2(g)

The atoms are balanced, but the charges are not. We have a net charge of 2- on the left and 0 on

the right. We can balance this by adding the appropriate number of electrons:

C2O42-

2CO2(g) + 2e-

(balanced atoms, balanced charge)

Are we done? No. By the original definition above, in a redox reaction the overall electrons

produced in the oxidation half-reaction must equal the number of electrons consumed in the

reduction half-reaction. We have not yet identified which is the oxidation half-reaction and which

is the reduction half-reaction, but it is clear that the electrons are not balancing (we have one half-

reaction with 5e- and the other with 2e

-)

o The half-reactions must be multiplied by some factor, such that the electrons present in

the half-reactions balance each other

o For the above half-reactions, the lowest common multiple is 10. Thus, the half-reaction

with 5e- should be multiplied by 2, and the half-reaction with 2e

- should be multiplied by

5:

2MnO4- + 16H

+ + 10e

- 2Mn

2+(aq) + 8H2O(l)

5C2O42-

10CO2(g) + 10e-

We can now describe the redox reaction accurately:

o The oxidation half-reaction is:

5C2O42-

10CO2(g) + 10e-

o The reduction half-reaction is:

2MnO4- + 16H

+ + 10e

- 2Mn

2+(aq) + 8H2O(l)

In other words, the oxalate ion is oxidized by the permanganate ion (and the permanganate ion is reduced)

o The overall balanced redox reaction is the combination of the two balanced half-

reactions:

2MnO4- + 16H

+ + 10e

- 2Mn

2+(aq) + 8H2O(l)

5C2O42-

10CO2(g) + 10e-

2MnO4- + 16H

+ + 5C2O4

2- 10CO2(g) + 2Mn

2+(aq) + 8H2O(l)

Summary of balancing half-reactions in acidic solutions:

1. Divide reaction into two incomplete half-reactions

2. Balance each half-reaction by doing the following:

a. Balance all elements, except O and H

b. Balance O by adding H2O

c. Balance H by adding H+

d. Balance charge by adding e-as needed

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3. If the electrons in one half-reaction do not balance those in the other, then multiply each half-reaction to

get a common multiple

4. The overall reaction is the sum of the half-reactions

5. The oxidation half-reaction is the one that produces electrons as products, and the reduction half-reaction

is the one that uses electrons as reactants

Balancing Redox Reactions That Occur in Basic Solution

In a basic aqueous solution the half-reactions must be balanced using OH- and H2O (instead of H

+ and

H2O)

The procedure for balancing in this case is actually the same as for balancing in an acidic solution,

with one small twist:

o Since we are under basic conditions, we have to neutralize any H+ ions in solution

o We achieve this by adding the appropriate number of OH- ions

o We keep things balanced by adding the same number of OH- ions to both sides of the

half-reaction

Voltaic Cells

A spontaneous redox reaction involves the transfer of electrons

In principle, this transfer or movement of electrons can be used to do a type of work - specifically,

electrical work

A type of device that makes use of redox reactions to produce electron flow, and to allow

electrical work, is known as a Voltaic Cell (named after Count Alessandro Volta, 1745-1827, an

Italian physicist)

The development of the voltaic cell begins as follows...

A spontaneous redox reaction that we have previously discussed involves the metals Zinc (Zn) and Copper

(Cu)

If you recall the Activity Series for metals and hydrogen ion:

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When comparing Zinc with Copper in the activity series, we expect that Zinc should be easier to

oxidize (i.e. should give up electrons easier than Copper)

The spontaneity of the reaction between zinc and copper ion reflects the Activity Series

relationship:

Zn(s) + Cu2+

(aq) Zn2+

(aq) + Cu(s)

o The reaction as shown is spontaneous. The Zinc metal is oxidized, and the Copper ion is

reduced

o The reverse reaction, the oxidation of Copper and the reduction of Zinc ion, is not

spontaneous

The redox half-reactions for the above reaction would be:

Oxidation: Zn(s) Zn2+

(aq) + 2e-

Reduction: Cu2+

(aq) + 2e- Cu(s)

The oxidation of a Zinc atom releases 2 electrons

The reduction of a Copper ion is achieved by the acceptance of 2 electrons

Thus, there would appear to be a movement, or flow, of electrons from the Zinc metal to the Copper ions

According to the above equation, if Zinc metal is placed in an aqueous solution containing Cu2+

ions (e.g. a

solution of copper sulfate salt), the following will occur:

At the surface of the Zinc metal, the oxidation of Zinc atoms will be coupled to the reduction of

Cu2+

ions

Electrons from the oxidation of the zinc will reduce the copper ions to elemental copper on the

surface of the zinc

Over time, the zinc metal will dissolve (as solid zinc is oxidized to zinc ion) and zinc ion will

build up in the solution.

Also over time, the solution will no longer contain copper ions, an the copper will be present in

elemental form

In the above reaction, the zinc and copper metal/ions are in direct contact with each other and that is why

the reduced form of the copper (i.e. the metal) builds up on the surface of the zinc

What if we separated the Cu(s)/Cu2+

from the Zn(s)/Zn2+

?

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The reaction will not occur because there is no way for the electrons released by the oxidation of Zinc

metal to get over to the Copper ions and reduce them to Copper metal

What if we provide a path of conductance for the electrons released by the oxidation of the zinc to

get over to the copper?

Now that the electrons have a path to the Cu/Cu2+ side it would appear that the reaction could proceed

However, we now have another problem ...

At the beginning of the redox reaction we have a neutral aqueous salt solution, e.g. ZnSO4 (i.e. the

concentration of cation equals the concentration of anion)

As the redox reaction proceeds we build up Zn2+

ions in the solution where the Zinc is being

oxidized. Conversely, we remove Cu2+

ions from solution where the Copper ion is being reduced.

The anion concentration (sulfate ion in this case) does not change.

Thus, we are building up a net positive charge in the zinc solution, and a net negative charge in the

copper solution

These charges will oppose the flow of electrons. The positive charge in the zinc solution will make

it harder for the negative electrons to leave. Likewise, the negative charge in the copper solution

will repel the electrons that are trying to come over from the zinc side

We need a way to neutralize the charge build-up in the solutions due to the change in soluble ion

concentration

What if we had a tube filled with aqueous solution that connected the two redox reactions?

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This would allow anion to move from the copper reduction side to the zinc oxidation side and

keep the charges in solution neutral

In turn, this would now allow the electrons to flow

However, as the reaction proceeds, we have yet another problem...

As the reaction proceeds, although movement of anion allows the overall charges in solution to remain

neutral, the net movement of anion produces a concentration gradient across the two solutions. In other

words, after a while the net concentration of ions in the zinc oxidation side will be greater than in the

copper reduction side. This concentration gradient will oppose movement of anion

Actually, this is not really a problem that we will see. What it means is that we have to realize that charge

neutralization can occur by either anions moving to the left or cations moving to the right

In consideration of keeping the overall concentration of ions in balance between the two sides,

cations will also be moving to the right:

Thus, in the connecting tube of solution we have net movement of both types of ions:

o anions are going into the oxidation side

o cations going into the reduction side

The connecting tube of solution is called a Salt Bridge

Summary of the movement of ions, electrons and the redox half-reactions in a voltaic cell:

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The two solid metals in the different half-reactions are called electrodes

The metal in the half-reaction where oxidation is occurring is called the Anode

The metal in the half-reaction where reduction is occurring is called the Cathode

The cathode is often labeled with a "+"; "this electrode attracts electrons"

The anode is often labeled with a "-"; "this electrode repels electrons"

Cell EMF

Here is a summary of a voltaic cell:

A key question is: what causes the electrons to flow from the anode to the cathode?

The electrons flow from the anode to the cathode because of a difference in the potential energy of

the electrons at the anode compared to the cathode

In particular, the potential energy of the electrons is higher at the anode than at the cathode

Like water molecules in a waterfall, the electrons move in a direction from high potential energy

to low potential energy (i.e. "downhill" in an energy landscape)

The potential difference between two electrodes is measured in Volts (V)

One volt is the potential difference, between an anode and a cathode, that would be required in

order to impart 1 Joule of energy to a charge of 1 coulomb (note: 1 coulomb is the charge of a

collection of 6.24 x 1018

electrons)

1 V = 1 J/C (i.e. energy per electron)

The potential difference between two electrodes is called the electromotive force, or emf

The emf of a voltaic cell is called the cell potential, or the cell voltage

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The cell voltage of a voltaic cell will be positive value

The cell voltage will depend upon the specific half-reactions, concentrations of ions, and temperature

Common tabulations of cell voltages are performed at 25°C and standard conditions of

concentrations of liquid (i.e. 1M) and gas components (i.e. 1 atm)

Under standard conditions, the emf of a voltaic cell is called the standard emf or the standard cell

potential (E0

cell)

By convention, the potential associate with an electrode is the potential for reduction

The standard cell potential is given by the standard reduction potential of the cathode, minus the

standard reduction potential of the anode

E0cell = E

0red(cathode) - E

0red(anode)

If E0

red(cathode) > E0

red(anode) then the electron flow from anode to cathode is spontaneous

Standard Reduction Potentials

The cell potential of a voltaic cell depends upon the two redox half-reactions that comprise the functioning

cell

There are many different combinations of redox half-reactions. Do we need to tabulate all possible

unique combinations?

As you might have guessed, the answer is no, we don't. In principle, we could assign a unique

standard potential to each half-reaction and by comparing these values determine the cell potential

for any combination of half-reactions

There is one problem, however: we have seen that an isolated half-reaction doesn't result in reduction or

oxidation or electron flow. Therefore, how are we going to assign a standard potential to a half-reaction?

We are going to use a generally agreed upon reduction reaction as the reference half-reaction

against which to compare and tabulate all possible half-reactions

The reference reaction is the reduction of H+(aq) ions to produce H2(g):

2H+(aq, 1M) + 2e

- H2(g, 1 atm) E

0red = 0.0V

This is a Standard Hydrogen Electrode (SHE)

How do we use this SHE to determine the standard reduction potential, E

0red, for some redox half-reaction?

Let's say we are interested in determining the standard reduction potential for Zn2+

We would set up a voltaic cell using a SHE as one half-cell, and a Zinc electrode as the other half-

cell

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The half-cell reactions are:

Zn(s) Zn2+

(aq) + 2e- (Zinc electrode)

2H+(aq) + 2e

- H2(g) (SHE)

The overall cell reaction:

Zn(s) + 2H+(aq) Zn

2+(aq) + H2(g)

Note: this is performed under Standard Conditions, where concentrations = 1.0M, and p = 1 atm

For the reaction as shown, the zinc electrode is the anode (oxidation), and the SHE is the cathode

(reduction)

o The voltage for the overall cell is measured at 0.76 Volts (i.e. this is the standard cell

potential)

o From the previous definition of the standard cell potential:

E0cell = E

0red(cathode) - E

0red(anode)

o We know the value for the standard cell potential, it is +0.76V. We also know the value

for the reduction potential for the SHE, it is 0V. Thus,

0.76V = E0red(cathode) - E

0red(anode)

0.76V = 0 - E0red(anode)

E0red(anode) = -0.76V

In other words, in reference to the SHE, the standard reduction potential for the zinc electrode is

determined to be -0.76V.

This may seem a little strange because the cell reads +0.76V and yet the determined reduction potential for

the zinc electrode is -0.76V.

The key here is that the original definition of the standard cell potential is set up such that each

cell is referenced as a reduction reaction.

In other words what we have actually determined is the half-cell potential (in reference to the

SHE) for the reduction of zinc:

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Zn2+

(aq) + 2e- Zn(s) E

0red = -0.76V

Whenever we assign a potential to a half-reaction using the equation for the standard cell potential, it is

always done in reference to a reduction reaction

Thus, in reference to the SHE, the oxidation of Zn(s) is associated with a half-cell potential of

+0.76V (this is the spontaneous direction for the reaction; the reduction of Zn2+

and the oxidation

of H2(g) is not spontaneous)

The standard cell potential (Voltage) describes the potential per unit charge that the reaction produces

This value of the potential per unit charge (i.e. electron) is a constant that is a property of the two

half-reactions that are reacting in the cell. It is independent of the stoichiometry or amount of the

reactants in the cell

The more positive the value of E0

red the greater the driving force for reduction

In reference to H+/H2(g) , Ag

+ will preferentially be reduced to Ag(s) and H2(g) will be oxidized to

H+(aq). In reference to H

+/H2(g) , Zn(s) will preferentially be oxidized and H

+ will be reduced to

H2(g).

The cathode is the electrode where reduction occurs, therefore, in order for a cell to operate

spontaneously, the half-reaction at the cathode must have a more positive value of E0red than the

half-reaction at the anode

o The greater driving force for reduction at the cathode can force the reaction at the anode

to operate "in reverse" - i.e. to undergo the oxidation reaction

o The difference between the standard reduction potentials of the two reactions, E0

red

(cathode) - E0red (anode), is the excess potential that can be used to drive electrons

through the cell, E0

cell

Oxidizing and Reducing Agents

In the above discussion of reduction potentials and their relationship to one another in half-reactions, we

can predict which direction current will flow in a voltaic cell. In other words, we can predict which reactant

will be oxidized and which will be reduced

The same information can be used for redox reactions where the compounds are in direct contact,

and not separated into half-cells.

For example, from the example above, we would predict that silver ion will oxidize lithium metal

as a spontaneous redox reaction. Alternatively, we would predict that the reduction of lithium ion

by silver metal is not a spontaneous reaction (and would require the input of energy to drive)

The more positive the E0

red value of a half-reaction, the greater the tendency of the compound to be reduced

- and therefore to oxidize another compound (i.e. the greater the tendency to act as an oxidizing agent)

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Substances that are strong oxidizing agents are subsequently weak reducing agents. Likewise,

substances that are strong reducing agents are weak oxidizing agents.

Spontaneity of Redox Reactions

A positive voltage that forms across the electrodes of a voltaic cell indicates that the oxidation-reduction

reaction is a spontaneous reaction for reduction at the cathode and oxidation at the anode.

Conversely, if the potentials of the half-cells are known, then it is possible to predict whether a

given redox reaction will be spontaneous (i.e. result in a positive voltage)

A negative voltage indicates that the reverse reaction is spontaneous (i.e. oxidation at the cathode,

and reduction at the anode; by convention you would need to swap the labels on the electrodes)

Basically, any current flow indicates that there is a spontaneous redox reaction occurring in a

voltaic cell. The sign of the voltage indicates at which electrode the reduction or oxidation is

occurring. (electrons flow towards the half-cell where reduction is occurring - by convention, the

cathode)

As long as we can identify the actual reduction and oxidation processes that will occur in a redox reaction,

the general description of the standard reduction potential for any redox reaction (and not just one

occurring in a voltaic cell) would be:

E0 = E

0red (reduction process) - E

0red (oxidation process)

Thus, E0 will be positive for the case where the reaction is spontaneous

E0 will be zero for a redox reaction at equilibrium

E0 will be negative for the case where the reaction is spontaneous in the reverse direction

Note that there is no reference here to what is the cathode and what is the anode

Activity Series of Metals

The activity series of metals lists the metals in decreasing order of their relative ease of oxidation:

The stronger the reduction potential, the more difficult it is to oxidize the compound.

The weaker the reduction potential, the easier it is to oxidize the compound

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EMF and Free Energy Change

Free energy change, ΔG, is a measure of the spontaneity of a chemical reaction or process. Likewise, the

standard reduction potential (E0

cell) for a redox reaction is also a measure of the spontaneity of a redox

reaction. What is the relationship between these two values?

ΔG has units of J/mole

Reduction potential (electromotive force, E) has units of Volts. 1 Volt describes the potential

difference necessary to impart 1J of energy to a charge of 1 coulomb (6.24 x 1018

electrons)

o We need some way to relate charge in Coulombs (a collection of electrons) to moles of

electrons (another collection of electrons)

ΔG = -nFE

n = number of moles of electrons transferred in the reaction

F = the quantity of electrical charge (in coulombs) that is contained in 1 mole of electrons (this is

the Faraday constant). One Faraday is equal to 96,500 coulombs/mole of e-. Since 1 volt =

1Joule/coulomb, one Faraday also equals 96,500 J/volt*mole e-

Joules/mol = -(electrons)* (coulombs/mol of electrons) * (Joules/coulomb)

F, and n, are positive values. Therefore, a positive value of E (which indicates spontaneity) is a

negative value for ΔG (which also indicates spontaneity)

If everything is in the standard state:

ΔG0 = -nFE

0

Effect of Concentration on Cell EMF

The EMF of a redox reaction in a voltaic cell is determined not only by the type of redox reaction, but also

the concentrations of the reactants and products (i.e. the reducing agent and oxidizing agent)

The EMF of the cell will fall as the reactants are used up and products increase in concentration

At equilibrium concentrations of reactants and products, the EMF = 0. Electrons flow

spontaneously in a redox reaction because the system is attempting to achieve equilibrium. When

equilibrium is achieved, net electron flow is zero.

The Nernst Equation

Walther Hermann Nernst (1864 - 1941) was a German chemist who came up with an equation that related

the EMF of a redox reaction on the concentration of reactants and products

Free energy and the equilibrium constant (this is from the section on thermodynamics)

ΔG = ΔG0 + RT lnQ

where Q is the reaction quotient

Relationship between ΔG and the electromotive force (from that genius Faraday)

ΔG = -nFE

where n = moles of electrons transferred, F = 96,500 J/V mole e-, E = EMF in volts

Substituting into the free energy equation:

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-nFE = -nFE0 + RT lnQ

solving for E yields:

A common form of the equation does away with the natural log and puts the equation in the form

of log10:

Yet another simplification occurs if the temperature of the reaction is known. For example, 298K

is a common reference temperature for redox reactions (this is room temperature). At 298K the

value of 2.303*RT/F = 0.0592 V mole:

This is the Nernst Equation. It allows us to do the following for redox reactions:

1. If we know the value of E0 and we measure the EMF of the cell (under non-standard conditions)

we can determine the value of Q (and therefore the concentrations of reactants and products)

2. If we know the value of E0 and the concentrations of reactants and products, then we can

determine the resulting EMF of the cell

Equilibrium Constants for Redox Reactions

What happens when the redox reaction achieves equilibrium concentrations of reactants and products?

From thermodynamics, ΔG = 0

If ΔG = 0, then from Faraday's analysis of the EMF, E = 0 (because ΔG = -nFE, and n and F are

positive and non-zero, so E must be zero)

In other words, if the redox reaction is at equilibrium, there is no longer any net reduction or

oxidation reaction that drives an electromotive force (the cell is "dead")

or, rearranging to solve for K:

What does this equation tell us?

This says that the equilibrium constant is directly related to the standard EMF for the redox

reaction (conversely, the standard EMF for the reaction is directly related to the value of the

equilibrium constant)

A redox reaction where the equilibrium lies far to the right, will have a large value for K, and a large value

for the standard EMF

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If we can calculate the standard EMF for a reaction (by comparing the relative difference between

the two half-reactions) we can derive the equilibrium constant for the reaction

Commercial Voltaic Cells

Voltaic cells provide a convenient, safe and portable supply of electrical energy.

The Industrial Revolution marked the development of heat engines, and other devices, that utilized

the energy released from combustion reactions in the form of heat

The Digital Age (Electronic Revolution?) revolves around devices that require energy in the form

of electrical energy. Such devices require voltaic cells and redox chemistry (as opposed to

gasoline and combustion chemistry)

Technology in transition

The automobile industry has relied upon combustion reactions (the internal combustion engine).

But these pollute, causing acid rain and global warming. Also, fossil fuels are a limited resource.

Electric cars promise lower pollution, and are essentially silent (an important feature in a

population-dense world). A few electric cars have been developed, but they have problems:

o Range is limited

o The electrical systems are heavy

o They are expensive

All of the above drawbacks are related primarily to the voltaic cells

Hybrid combustion/voltaic systems have been developed and can be purchased

o These designs combine a small internal combustion (IC) engine with an electric motor

and voltaic cells

o The IC engine may power an electric generator for the electric subsystem. The IC engine

can therefore run at its most efficient rpm at all times (even when the car is not moving).

Alternatively, the IC engine can drive the wheels directly during situations when extra

power is needed (starting, passing, going uphill, etc.). All other times, the car is powered

by the electric motor.

o These cars still have problems related to the size/weight/storage capacity of voltaic cells

Lead-Storage Battery

The redox half-reactions in a lead-storage (lead acid) battery are as follows:

Cathode:

PbO2(s) + SO42-

(aq) + 4H+(aq) + 2e

- PbSO4(s) + 2H2O(l)

Anode:

Pb(s) + SO42-

(aq) PbSO4(s) + 2e-

Overall redox reaction:

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PbO2(s) + Pb(s) + 4H+(aq) + 2SO4

2-(aq) 2PbSO4(s) + 2H2O(l)

Sulfuric acid provides the protons and sulfate ions: 2H2SO4(aq) 2SO42-

(aq) + 4H+(aq)

The solid electrodes (Pb and PbO2) do not react in their respective redox reactions to produce

soluble ions. In both cases, PbSO4(s) is formed and remains attached as a solid to the electrode(s).

Thus, ions do not diffuse from one half-cell to the other. Therefore, the two electrodes can be

placed in the same container of acid.

Water is produced and sulfuric acid is consumed during the reaction.

The EMF per "cell" under standard conditions is:

E0cell = E

0red (cathode) - E

0red (anode) = (+1.685 V) - (-0.356 V) = 2.041 V

o 6 cells can be combined end-to-end (i.e. in series) to produce about 12 V (what you find

in a typical car)

This is a reversible reaction. If electrical current is applied in the opposite direction (this is the job

of a generator or alternator in your car) the electrodes are regenerated

2PbSO4(s) + 2H2O(l) Pb(s) + PbO2(s) + 4H+(aq) + 2SO4

2-(aq)

Dry Cell

These are your basic (not alkaline) type battery. The 6V battery in your emergency flashlight is most likely

a dry cell type. It was invented in 1866. The reactions are curiously rather complex. A simple version of the

half-reactions is as follows:

Cathode:

2NH4+(aq) + 2MnO2(s) + 2e

- Mn2O3(s) + 2NH3(aq) + H2O(l)

Anode:

Zn(s) Zn2+

(aq) + 2e-

The construction consists of a zinc electrode (for the anode). The cathode is a bit weird - it is an

inert support of graphite is immersed in a paste of ammonium chloride and manganese dioxide.

The dimanganese trioxide solid precipitates out on the surface of the inert graphite

This is not reversible, so the battery cannot be recharged (the electrode reaction products diffuse

throughout the cell)

In an "alkaline" type battery the ammonium chloride is replace by potassium hydroxide (KOH).

This provides more useable voltage and greater capacity than the typical dry cell

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Nickel Cadmium (Nicad, or Cadnica cell)

Cathode:

NiO2(s) + 2H2O(l) + 2e- Ni(OH)2(s) + 2OH

-(aq)

Anode:

Cd(s) + 2OH-(aq) Cd(OH)2(s) + 2e

-

This type of cell uses a cadmium anode and a nickel dioxide cathode

The solid products of the respective electrode reactions adhere to the electrodes and do not diffuse

throughout the cell. Thus, the redox reaction is reversible (i.e. like the lead acid cell, the nickel

cadmium cell is reversible)

No gases are produced, so the cell can be sealed

Fuel Cells

Combustion reactions produce heat that, in turn, can be used to produce electricity. However, typically less

than 40% of the heat energy is converted to electrical energy - the rest is "wasted" as heat.

Combustion reactions are actually redox reactions: diatomic oxygen (0 oxidation number) is reduced to

carbon dioxide (-2 oxidation number) or water (-2 oxidation number). Direct production of electricity from

redox chemistry, instead of combustion, for these reactions could result in higher efficiency of production

of electrical energy. Voltaic cells that perform this type of redox reaction for conventional fuels (such as

hydrogen or methane) are called fuel cells

A common reaction being utilized in fuel cells is the reduction of oxygen by hydrogen

o Cathode:

O2(g) + 2H2O(l) + 4e- 4OH

-(aq)

o Anode:

2H2(g) + 4OH-(aq) 4H2O(l) + 4e

-

o Overall reaction:

2H2(g) + O2(g) 2H2O(l)

This is currently an very expensive way to generate energy, but is extremely efficient and compact. Its

broadest application to date has been to provide electricity (and drinking water) for spacecraft.

Electrolysis

Voltaic cells are based upon the spontaneity of a particular redox reaction

This in turn is based upon the relative reduction potentials for the two half-reactions

The half-reaction with the greater reduction potential will drive the other half-reaction in the

direction of oxidation

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The relative difference between the two reduction potentials provides a quantitative description of

the electromotive force of the voltaic cell

However, we have seen that some commercial voltaic cells can be "recharged"

This is achieved by the application of current in the opposite direction to that of the spontaneous

direction of current flow in the voltaic cell

This application of current in the opposite direction will cause the redox reaction to proceed in the

opposite direction

o The half cell where oxidation occurred will now undergo a reduction reaction

o The half cell where reduction occurred will now undergo an oxidation reaction

Sodium metal and chlorine gas spontaneously react to form the ionic compound Sodium Chloride:

2Na(s) + Cl2(g) 2NaCl(s)

This reaction is a redox reaction:

Anode: 2Na(s) 2Na+(s) + 2e

-

Cathode: Cl2(g) + 2e- 2Cl

-(s)

The reverse reaction, the decomposition of ionic NaCl into Na(s) and Cl2(g), can be achieved by

the application of an EMF (i.e. external current)

2NaCl(l) 2Na(l) + Cl2(g)

Redox reactions that are driven by an external EMF are called electrolysis reactions and take place in

electrolytic cells

The electrolytic cell contains two electrodes and either a molten salt solution (as with NaCl) or some other

type of solution (e.g. aqueous)

The external EMF is provided by a source of direct current (sometimes another voltaic cell, or

battery)

This external current acts as an electron pump to push electrons onto one electrode (and drive a

reduction reaction) and to withdraw electrons from the other electrode (and drive an oxidation

reaction)

o The electrode where oxidation occurs is still called the anode. However, for an

electrolytic cell it is labeled as "+" to indicate that this is the electrode where electrons

are being withdrawn by the external EMF

o The electrode where reduction occurs is stall called the cathode. However, for an

electrolytic cell it is labeled as "-" to indicate that this is the electrode where electrons are

being pumped into by the external EMF

Electrolysis of Aqueous Solutions

With aqueous solutions of salts it must be considered that the electrolysis might end up driving the

reduction or oxidation of H2O(l)

H2O(l) can be oxidized to produce O2(g)

H2O(l) can be reduced to produce H2(g)

If a salt is present, the cation of the salt may or may not be preferentially reduced compared to

H2O(l). Likewise, the anion may or may not be preferentially oxidized compared to H2O(l)

Here are the reduction potential data for the reduction of H2O(l) and Na+(aq):

2H2O(l) + 2e- H2(g) + 2OH

-(aq) E

0red = -0.83 V

Na+(aq) + e

-

0red = -2.71 V

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When comparing any two reduction reactions the reaction that is favored is the one with the more

positive reduction potential. In this case, the reduction of H2O(l) is the favored reduction reaction

Here are the reduction potential data for the oxidation of H2O(l) and Cl-(aq). Remember that the reduction

potential provided is referencing the reduction (i.e. reverse) reaction

2H2O(l) +(aq) + O2(g) + 4e

- E

0red = 1.23 V

2Cl-(aq) Cl2(g) + 2e

- E

0red = 1.36 V

When comparing the reduction potentials, the reduction of Cl- is favored. However, we are

interested in the reverse reactions, i.e. the oxidation reactions. In this case, remember that a

substance that is readily reduced is difficult to oxidize, and vice versa. Thus, if the Cl- is favored

for reduction, then the water reaction is favored for oxidation

What this analysis suggests is that electrolysis of an aqueous solution of Na+(aq) and Cl

-(aq) ions

will not result in the reduction of Na+ or oxidation of Cl

-, but rather, will result in the reduction of

H2O to produce hydrogen, and the oxidation of H2O to produce oxygen.

Note: although this analysis is correct, there is another effect that can happen, known as the overvoltage

effect. Additional voltage is sometimes required, beyond the voltage predicted by the E0cell. This may be

due to kinetic rather than thermodynamic considerations. In other words, although the voltage applied is

thermodynamically sufficient to drive electrolysis, the rate is so slow that to make the process proceed in a

reasonable time frame we must increase the voltage of the external source (hence, overvoltage).

In the case of the oxidation of H2O versus Cl-(aq), the oxidation of H2O to produce O2(g) is

thermodynamically favored, but kinetically quite slow. Thus we could try to crank up the voltage. However,

the oxidation of chlorine occurs at a voltage not much higher than that for water. Thus, in practical terms,

oxidation of chlorine will preferentially take place at the anode.

The redox reactions for the electrolysis of an aqueous solution of NaCl are therefore:

(reduction) 2H2O(l) + 2e- H2(g) + 2OH

-(aq)

(oxidation) 2Cl-(aq) 2(g) + 2e

-

And the overall reaction is:

2H2O(l) + 2Cl-(aq) 2(g) + H2(g) + 2OH

-(aq)

Thus, while electrolysis of molten NaCl produces chlorine gas and sodium metal, the electrolysis of an

aqueous solution of NaCl produces chlorine gas, hydrogen gas and hydroxide ion

What external EMF is required do drive this particular electrolysis?

(reduction; cathode) 2H2O(l) + 2e- H2(g) + 2OH

-(aq) E

0red = -0.83 V

(oxidation; anode) 2Cl-(aq) 2(g) + 2e

- E

0red = 1.36 V

E0cell = E

0(cathode) - E

0(anode) = -0.83 V - (1.36 V) = -2.19 V

This value is negative, indicating that the reaction is not spontaneous as written and must be driven by an

external EMF

Electrolysis with Active Electrodes

Inert electrodes do not enter into the chemical reaction, and just serve as a surface upon which

electron transfers can occur (and be conducted to the other half-cell)

Active electrodes chemically participate in the redox reaction

If an aqueous solution contains solid metal electrodes, the electrodes may be oxidized to the ionic form if

the metal is easier to oxidize than water. Since we think in terms of reduction potentials and not ease of

21

oxidation, this means that the metal electrodes will be oxidized to ionic form if the reduction potential of

the metal is less than that of water

Ni(s) Ni2+

(aq) + 2e- E

0red = -0.28 V

2H2O(l) 4H+(aq) + O2(g) + 4e

- E

0red = 1.23 V

In this case the reduction potential of H2O(l) is greater than that for Nickel metal. Thus, H2O is

easier to reduce, and therefore more difficult to oxidize, than Nickel. Thus, at the anode, Nickel

will preferentially be oxidized compared to H2O(l) if we provide an external EMF that "sucks"

electrons out at this electrode

What about the cathode (reduction) reaction involving Nickel ion and water?

Ni2+

(aq) + 2e- Ni(s) E

0red = -0.28 V

2H2O(l) + 2e- H2(g) + 2OH

-(aq) E

0red = -0.83 V

In this case, the half-reaction with the greater reduction potential is the reduction of Nickel.

Therefore, at the cathode, Nickel will preferentially be reduced if we provide an external EMF that

forces electrons onto this electrode

Thus, a metallic cathode may be coated with solid nickel metal in a process known as

"electroplating"

Quantitative Aspects of Electrolysis

Reduction half-reactions provide information about the stoichiometric requirements for the production of

the elemental form of the metal. In particular, the number of electrons needed per molar basis of metal

reduced

Na+ + e

- Na

Cu2+

+ 2e- Cu

Al3+

+ 3e- Al

From the above information, 1 mole of electrons will allow the production (i.e. reduction) of 1

mole of sodium metal, 0.5 mole of copper metal, and 0.333 mole of aluminum metal

For any half-reaction the amount of substance reduced is proportional to the number of electrons that pass

into the cell

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We can quantitate the number of electrons by keeping track of the charge that passes into a cell

Charge is measured in Coulombs

1 mole of electrons has a charge of 96,500 Coulombs (C). This number of Coulombic charge is

equal to 1 Faraday (F)

Charge of 1 mole of electrons = 96,500 C = 1F

The word, current, refers to the rate of flow of electricity

One ampere (amp) is equal to a rate of flow of charge of 1 Coulomb every second

amps = Coulombs/second

and

Coulombs = amps * seconds

In summary…

Amps = charge / unit of time

Charge the number of electrons

The number of electrons the number of moles of a compound reduced

Electrical Work

Relationship between E and ΔG:

A positive value for E (electromotive force) is associated with a spontaneous redox reaction

Likewise, a negative value for ΔG is associated with a spontaneous reaction or process

+E or -ΔG means a spontaneous reaction

The relationship between ΔG and E:

ΔG = -nFE

Useful work

For any spontaneous process, ΔG is a measure of the maximum useful work (wmax)that can be done

by the process.

Therefore, -nFE also is a measure of the maximum useful work that can be done by a redox

reaction:

wmax = -nFE

wmax = moles * (Coulombs/moles) * (Joules/Coulomb) = Joules

Note: work done by the system on the surroundings is indicated by a negative sign for w

Non-spontaneous redox reactions

Electrolysis describes the application of an external EMF to drive a non-spontaneous process

For a non-spontaneous process, ΔG is positive and E will be negative

To force the redox reaction to proceed, the external potential must be larger in magnitude than the

EMF of the cell (i.e. Ecell)

Eexternal - Ecell

When an external potential is applied, the surroundings do work on the system. Therefore, we

expect the sign of the w term to be positive

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wmax = +nFEexternal

Units of power

A watt is a unit of energy expenditure, i.e. the amount of energy consumed or produced per unit of

time. It has units of Joules/second:

1 watt = 1 Joule/second

and

1 Joule = 1 watt * second

How does the Power Company charge you for your electric usage? It charges you for the

watt*hours that you consume, or kilowatt*hours

1 k watt hour = (1000 watt)*(1 hour) (360 sec/hour) ((1 J/s)/watt) = 3.6 x 106 Joules

Corrosion

Corrosion reactions are redox reactions in which the elemental form of a metal is oxidized to produce an

unwanted compound or a water soluble ion

Oxidized forms of a metal may not have the strength and conductive properties of the metal in the

elemental form

Diatomic oxygen in the atmosphere has a relatively high reduction potential - higher than many metals

For metals exposed to air, oxygen (in the presence of H+) is reduced to H2O, and the metal is

oxidized, in a spontaneous redox reaction

For some metals, however, the initial oxidation on the surface protects the underlying metal against further

oxidation. This is because the metal oxide layer is impervious to both oxygen and H2O (a source of H+)

This is the case for aluminum, which although quite reactive upon exposure to oxygen and water,

develops a metal oxide surface layer that protects the metal underneath

Corrosion of Iron

The corrosion of iron (i.e. rusting) costs the U.S. economy about $70 billion annually

Formation of rust requires both O2 and H2O

Anode reaction:

Fe(s) Fe2+

(aq) + 2e- E

0red = -0.44 V

Cathode reaction:

O2(g) + 4H+(aq) + 4e

- 2H2O(l) E

0red = 1.23 V

Thus, the reduction of oxygen is favored over the reduction of iron. In other words, the iron will

preferentially oxidize.

The Fe2+

that is formed eventually is oxidized further to Fe3+

which forms a hydrated iron (III)

oxide (i.e. rust)

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Prevention of Corrosion

If iron is painted, then oxygen and water are prevented from contacting the metal, and corrosion is avoided

Sometimes the iron is coated with a thin layer of another metal, such as Tin (the layer of Tin keeps oxygen

and water away from the iron)

2000 Dr. Michael Blaber