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Chapter 20 - Electrochemistry
Homework:
9, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21,
22, 23, 24, 25, 27, 31, 33, 34, 36, 37,
39, 41, 42, 43, 44, 45, 46, 47, 49, 51,
51, 52, 53, 55, 57, 58, 60, 61, 63, 64,
83, 85, 87, 88, 89
20.1 - Oxidation States

How do we tell whether a given reaction
is an oxidation-reduction reaction?


We do so by keeping track of the oxidation
states of all the elements involved in the
reaction
This procedure tells us which elements (if
any) are changing oxidation state
Changing Oxidation State

Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)


We can say that Zn is oxidized and H+ is
reduced because electrons are transferred
from zinc to hydrogen
The transfer of electrons that occur
produces energy in the form of heat

The reaction is thermodynamically favored, and
proceeds spontaneously
Changing Oxidation?
We can write the previous reaction in
terms of oxidation numbers
 Zn(s) + 2H+(aq) 
Zn2+(aq) + H2(g)
0
+1
+2
0
 By writing the oxidation state for each
element above or below the equation,
we can see the oxidation state changes
that happen

In some reactions there is no clear transfer of
electrons, but oxidation states do change
 Consider the combustion of hydrogen

2 H2(g) + O2(g) 
2H2O(g)
0
0
+1 -2
0
+1 -2
-2
 In this reaction, H2 is oxidized from the 0 to
+1 state and oxygen is reduced from the 0 to
-2 state

 Since oxidation states changes, this is an
oxidation-reduction reaction

In the previous example



2 H2(g) +
O2(g) 
2H2O(g)
Water is not ionic, therefore there is not a
complete transfer of electrons from hydrogen
to oxygen
Using oxidation states is a convenient
bookkeeping method


However, do not equate the oxidation state of an
atom with its actual charge in a chemical
compound
DO think of the oxidation number as if we had an
ionic compound, what would its charge be?

In any redox reaction, both oxidation and reduction
must occur



The substance that makes it possible for another
substance to be oxidized is called either the
oxidizing agent or the oxidant



In other words: If one substance is oxidized, then another
must be reduced
The electrons MUST go from one place to another
Removes electrons from another substance
It itself becomes reduced
The substance that gives up electrons is called the
reducing agent or the reductant

It itself is oxidized
In Other Words

Oxidizing Agent



Gains electrons
Is reduced
Reducing Agent


Loses electrons
Is oxidized
Using Our Examples

Zn(s) + 2H+(aq) 

Zn2+(aq) + H2(g)
Zn(s) is the reducing agent and H+(aq) is
the oxidizing agent


Since H+ received the electrons, it is therefore
oxidizing agent
Making Zn(s) the reducing agent

But remember: H+ itself was reduced while Zn(s) was
oxidized
20.2 - Balancing OxidationReduction Equations


Whenever we balance an equation, we MUST
obey the law of conservation of mass
As we balance an oxidation-reduction
reaction, we add another rule




The gains and losses of electrons must be
balanced
This means if a substance loses a certain number
of electrons during a reaction, some other
substance must gain that same number of
electrons
Some redox reactions do this pretty much
automatically
Others do not.
Half-Reactions


Although oxidation and reduction take place at
the same time, often easier to consider them
separate processes
Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2 Fe2+(aq)



We can think of this as two separate process
The oxidation of Sn2+ and the reduction of Fe3+
Oxidation:


Reduction:


Sn2+(aq)  Sn4+(aq) + 2e2Fe3+(aq) + 2e-  2Fe2+(aq)
Note: During oxidation electrons are a product, while
during reduction electrons are a reactant

Equations that show either oxidation or
reduction alone are called half-reactions


In the overall redox reaction, the number of
electrons lost in the oxidation half-reaction must
equal the number of electrons gained in the
reduction half-reaction
When this condition is met and each half-reaction
is balanced, the electrons on the two sides cancel
when the two half-reactions are then added to give
the overall, balanced equation
Balancing Equations by
Method of Half-Reactions
Assign oxidation states to see which
atoms are gaining electrons and which
are losing electrons
Divide the equation into two halfreactions
1.
2.

One for oxidation and the other for
reduction
Balance each half-reaction
3.
a)
b)
c)
d)

Balance the elements other than H and O
Balance the O atoms by adding H2O as needed
Balance the H atoms by adding H+ as needed
Balance the charge by adding e- as needed
check number of electrons in each half-reaction
corresponds to the changes in oxidation state
found in step #1
4.
5.
6.
7.
Multiply the half-reactions by integers if
necessary so that the number of electrons
lost in one half-reaction equals the number
gained in the other
Add the two half-reactions and simplify
whenever possible
Check to make sure the number of atoms
on each side of the reaction equal
Check to make sure the total charge on the
left is the same as the total charge on the
right
Example

Consider reaction between
permangante ion (MnO4-) and oxalate
ion (C2O42-) in an acid aqueous solution

MnO4-(aq) + C2O42-(aq)  Mn2+(aq) +
CO2(aq)


Note: Experiments show that H+ is consumed and
H2O is produced in this reaction
We will now balance this by half-reaction

MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g)

Mn goes from the 7+ oxidation state in MnO4- to the 2+
oxidation state in Mn2+



Which means that the Mn is reduced by 5 electrons
Which means it GAINS five electrons
Carbon goes from the 3+ oxidation state in C2O42- to
the 4+ oxidation state in CO2.


So it is oxidized by one electron
Which means each carbon loses one electrons
Half-Reactions

MnO4-(aq)  Mn2+(aq)
C2O42- (aq)  CO2(g)

Now we balance these


Start with the top one
MnO4-(aq)  Mn2+(aq)


Mn balanced
Need four oxygens, we add four waters


We have 8 hydrogens, need to add to reactants


MnO4-(aq)  Mn2+(aq) + 4 H2O(l)
8H+(aq) + MnO4-(aq)  Mn2+(aq) + 4 H2O(l)
Add electrons, 5 are gained



How do I know? Oxidation numbers
8(1+) + 1- = 7+ for reactants
2+ + 4(0) = 2+ for products


So we lost 5 positive charges, in other word we added 5
electrons
5e-+ 8H+(aq) + MnO4-(aq)  Mn2+(aq) + 4 H2O(l)
C2O42- (aq)  CO2(g)

Balance carbon first



C2O42- (aq)  2CO2(g)
Which also balances oxygens
Balance electrons




2- charge in reactants
0 charge in products
So a gain of 2 negative charges, which means a
gain of 2 electrons
C2O42- (aq)  CO2(g) + 2e-


Now that we have balanced half-reactions,
we add them to find the overall balanced
reaction

In overall reaction, should be no free electrons

5e-+ 8H+(aq) + MnO4-(aq)  Mn2+(aq) + 4 H2O(l)

+ C2O42- (aq)  CO2(g) + 2e-
I can see that I will need to apply a multiplier
of 2 to the top equation and a multiplier of 5
to the bottom equation

To cancel out the electrons
New Equations


10e-+ 16H+(aq) + 2MnO4-(aq)  2Mn2+(aq) + 8
H2O(l)
5C2O42- (aq)  5CO2(g) + 10e
Electrons cancel

Leaving me with net equation of
16H+(aq) + 2MnO4-(aq) + 5C2O42-(aq) 
2Mn2+(aq) + 8H2O(l) + 10 CO2(g)

And we are done!

Basic Solution

If a redox reaction occurs in a basic solution, we
complete it using OH- and H2O to balance H and O



Rather than H+ and H2O in an acidic solution
This can be a little more difficult, because both OHand H2O contain hydrogen and oxygen atoms
To help, balance the half-reactions as if they occurred
in an acidic solution


Then count the H+ in each half-reaction and add the same
number of OH- to BOTH SIDES of the half-reaction
If you then have an H+ and an OH- on the same side, they
neutralize to form H2O, which can be canceled as needed.
Example

Complete and balance this redox reaction
that takes place in basic solution


CN-(aq) + MnO4-(aq)  CNO-(aq) + MnO2(s)
First assign oxidation states

Mn goes from 7+ to 4+


Reduced (gains 3 electrons)
CN- total must be -1


In CNO-, O has oxidation state of 2-, which tells us that
the CN portion must have a total of +1
Therefore CN- is oxidized by two electrons
Half-Reactions



CN-(aq)  CNO-(aq)
MnO4-(aq)  MnO2(s)
We now balance the half-reactions as if
they occurred in an acidic solution
CN-(aq)  CNO-(aq)


C and N balanced
Need another O


Now H needs balancing


CN-(aq) + H2O  CNO-(aq)
CN-(aq) + H2O  CNO-(aq) + 2H+
Now balance electrons


1- on reactants, 1+ on products
CN-(aq) + H2O  CNO-(aq) + 2H+ + 2e-
MnO4-(aq)  MnO2(s)


Mn already balanced
Need to balance O by adding water


Balance H now


MnO4-(aq)  MnO2(s) + 2H2O(l)
MnO4-(aq) + 4H+  MnO2(s) + 2H2O(l)
Balance electrons



Reactant: (1-) + 4(1+) = 3+
Products: 0 + 0
3e- + MnO4-(aq) + 4H+  MnO2(s) + 2H2O(l)

Now we account for the basic solution by
adding an OH- to both sides of each equation
for every H+

CN-(aq) + H2O + 2OH-  CNO-(aq) + 2H+ + 2e- +
2OH


CN-(aq) + H2O + 2OH-  CNO-(aq) + 2H2O(l) + 2eCN-(aq) + 2OH-  CNO-(aq) + H2O(l) + 2e-
3e- + MnO4-(aq) + 4H+ + 4OH- MnO2(s) +
2H2O(l) + 4OH

3e- + MnO4-(aq) + 4H2O(l)  MnO2(s) + 2H2O(l) + 4OH3e- + MnO4-(aq) + 2H2O(l)  MnO2(s) + 4OH-


CN-(aq) + 2OH-  CNO-(aq) + H2O(l) + 2e3e- + MnO4-(aq) + 2H2O(l)  MnO2(s) + 4OH



Now that we have our equations, we multiply each
so that the electrons balance
Multiply top equation by 3 and bottom by 2
3CN-(aq) + 6OH-  3CNO-(aq) + 3H2O(l) + 6e6e- + 2MnO4-(aq) + 4H2O(l)  2MnO2(s) +
8OH-

Now we add the equations together



3CN-(aq) + 6OH-  3CNO-(aq) + 3H2O(l) + 6e6e- + 2MnO4-(aq) + 4H2O(l)  2MnO2(s) + 8OH-
3CN-(aq) + H2O(l) + 2MnO4-(aq) 
3CNO-(aq) + 2MnO2(s) + 2OH-
20.3 - Voltaic Cells

The energy released in a spontaneous
redox reaction can be used to perform
electrical work

This is done through a voltaic (or
galvanic) cell


A device in which the transfer of electrons takes
place through an external pathway
Rather than directly between reactants

An example of this is when a strip of
zinc is placed in contact with a solution
containing Cu2+



As reaction proceeds, the blue color of the
Cu2+ ions fade and the copper metal
deposits onto the zinc.
At the same time, the zinc dissolves
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

Same basic
principle as last



But this time the
Zn(s) and Cu2+ are
not in direct contact
The Zn is placed in
contact with Zn2+ in
one part of the cell
The Cu is placed in
contact with the Cu2+
in another cell
How Does this Work?

The reduction of the Cu2+ can occur only by
the flow of electrons through an external
circuit


The wire connecting the Zn and Cu strips
By physically separating the reduction half of
the reaction from the oxidation half


We create a flow of electrons through an external
circuit
We have electricity!

The two solid metals that are connected
by the external circuit are called
electrodes


The electrode at which oxidation occurs is
called the anode
The electrode at which reduction occurs is
called the cathode


Remember that oxidation and anode both start
with a vowel
While reduction and cathode both start with a
consonant

As in this example, the electrodes can
be made of materials that participate in
the reaction


But typically the electrodes are made of a
conducting material that does not gain or
lose mass during the reaction
Just acts as a surface at which electrons
are transferred

Such as platinum or graphite

Each compartment in a voltaic cell is
called a half-cell.


One half-cell is where the oxidation halfreaction takes place
One half-cell is where the reduction halfreaction takes place
Zn(s) + Cu2+(aq)  Zn2+(aq) +
Cu(s)

Continuing with our present example...




Anode:
Zn(s)  Zn2+(aq) + 2eCathode: Cu2+(aq) + 2e-  Cu(s)
Electrons become available as the zinc metal is
oxidized at the anode



Zn is oxidized while Cu2+ is reduced
They flow through the external circuit to the cathode
There they are consumed as Cu2+(aq) is reduced
Because Zn(s) is oxidized in the cell, the zinc
electrode loses mass, and the concentration of the
Zn2+ solution increases

Similarly, the Cu electrode gains mass and the Cu2+ solution
becomes less concentrated as Cu2+ is reduced to Cu(s)

For a voltaic cell to work


The solutions in the two half-cells must
remain electrically neutral
As Zn is oxidized, Zn2+ ions enter the
solution

So we need a way to for the positive ions
to leave the anode compartment or for
negative ions to enter

To keep solution electrically neutral
Similarly

The reduction of Cu2+ at the cathode
removes positive charge from the
solution


Leaving an excess of negative charge in
that half-cell
Therefore, positive ions must migrate
into the compartment or negative ions
must leave
How do we do this?

A salt bridge allows ions to move from
one solution to the other



Keeping each solution electrically neutral
Maintaining the electron flow
A salt bridge is simply a tube that
contains an electrolyte that will not react
with other ions in the cell or with the
electrode material

As oxidation and reduction proceed at
the electrodes, ions from the salt bridge
migrate to neutralize the charge in each
cell

No matter how this works, anions will
always migrate toward the anode and
cations toward the cathode

Note the direction of ions in the solution

Notice also that in any voltaic cell, the
electrons will flow from the anode through the
external circuit to the cathode.


Because the negative charged electrons flow from
the anode to the cathode, the anode in a voltaic
cell is labeled with a negative sign
And the cathode is labeled with a positive sign

Think of the electrons as being attracted to the positive
cathode from the negative anode through the external
circuit
20.4 - Cell EMF Under
Standard Conditions

Why do electrons transfer spontaneously
from a Zn atom to a Cu2+ ion?


In this section we will look at the main driving force
that pushes the electrons through the external
circuit in a voltaic cell
The chemical processes that make up any
voltaic cell are spontaneous


Electrons flow from the anode of a voltaic cell to
the cathode because of a difference in potential
energy
The potential energy of the electrons is higher in
the anode than in the cathode

So they spontaneously move to lower their energy

The difference in potential energy per
electrical charge (called potential
difference) between two electrodes is
measured in volts

One volt (V) is the potential energy
required to impart 1 J of energy to a charge
of 1 coulomb (C)

Remember that an electron has a charge of
1.60x10-19 C.

The potential difference between the two
electrodes in a voltaic cell provides the
driving force to push the electrons through
the external circuit

We will therefore call this potential difference the
electromotive force, or emf


electromotive means to
cause electron motion
The emf of a cell (Ecell) is also called the cell
potential.


Because Ecell is measured in volts, we often refer to this
as the cell voltage.
For any cell reaction that proceeds spontaneously (like in
a voltaic cell), the cell potential will be positive.

The actual emf of a particular cell
depends on the



specific reactions that take place at the
cathode and anode
concentrations of the reactants and
products
temperature (which we will generally
assume to be 25 C) unless told otherwise

We will focus on this temperature, and refer to
this as standard conditions
At standard conditions

If we have 1M concentrations for
reactants and products and 1 atm
pressure for gases and 25 C

Then the emf for these standard conditions
is called the standard emf or the standard
cell potential


E
cell
For example, for the Zn-Cu voltaic cell
seen earlier, the standard cell potential is
+1.10 V
Standard Reduction (HalfCell) Potentials

The emf of a voltaic cell depends on the
particular cathode and anode half-cell
involved



We COULD tabulate the standard
potentials for all possible combinations
However, that is not necessary
We assign a standard potential to each
individual half-cell, and use these half-cells
to determine E cell

The cell potential is the difference between
the two electrode potentials


One from the cathode, one from the anode
Due to convention, the potential associated
with each electrode is the potential for
reduction to occur at that electrode


Thus, standard electrode potentials are found for
reduction reactions
Called standard reduction potentials, E red
Calculating E

cell
The cell potential (E cell) is given by the
standard reduction potential of the
cathode, E red(cathode), minus the
standard reduction potential of the
anode, E red(anode)

E
cell
=E
red(cathode)
-E
red(anode)

Since every voltaic cell involves two
half-cells, it is impossible to measure
the standard reduction potential directly


All based off of a certain reference halfreaction
The reference half-reaction is the reduction
of H+(aq) to H2(g) under standard
conditions


2H+(aq, 1M) + 2e-  H2(g, 1 atm)
E red =
0V
The standard reduction potential of this reaction
is defined to be 0 V
2H+(aq, 1M) + 2e-  H2(g, 1 atm)
V

E
red
=0
An electrode designed to produce this
half-reaction is called the standard
hydrogen electrode (SHE) or the
normal hydrogen electrode (NHE)

Various standard reduction potentials
for half-reactions are found on pg. 863
(table 20.1) or in Appendix E

This is also information given to you during
the AP test (we just have not seen it yet)

Because electrical potential measures
potential energy per electrical charge,
standard reduction potentials are intensive
properties



This means that if we increased the amount of
chemicals in the redox reaction, we would
increase both energy and charges involved
But the ratio of the energy and charge would
remain constant
Therefore, changing the stoichiometric
coefficients in a half-reaction does not affect
the value of the standard reduction potential
Example

Given the standard reduction potential
of Zn2+ to Zn(s) is -0.76 V, calculate the
E red for the reduction of Cu2+ to Cu for
the following voltaic cell


Zn-Cu2+ cell
Zn(s) + Cu2+(aq, 1M)  Zn2+(aq, 1M) +
Cu(s)


E cell = 1.10 V
Cu2+(aq, 1M) + 2e-  Cu(s)

E




cell
=E
red(cathode)
-E
red(anode)
Zn is oxidized, and is therefore the anode
We were also given the E cell
1.10 V = E red(cathode) -(-0.76)
E red(cathode) = 1.10 V - 0.76 = 0.34 V
Example 2


Using the standard reduction potentials
provided, calculate the standard emf for
the voltaic cell given by this reaction
Cr2O72-(aq) + 14H+(aq) + 6I-(aq)  2 Cr3+(aq) + 3I2(s) +
7H2O(l)
Cr2O72-(aq) + 14H+(aq) + 6I-(aq)  2
Cr3+(aq) + 3I2(s) + 7H2O(l)

First we need to identify the half-reactions
involved

Cr2O72-(aq) + 14H+(aq) + 6e-  2Cr3+(aq) + 7H2O(l)

Cathode
E red(cathode) = 1.33 V
6I-(aq)  3I2(s) + 6e


Anode

E
red(anode)
= 0.54 V

Now we plug into equation

E
cell
=E
red(cathode)
-E
red(anode)
E
E

E
= E red(cathode) red(anode)
cell
cell
= 1.33 V - 0.54 V = 0.79 V
Analyzing the Equation

For each of the half-cells in the voltaic cell,
the standard reduction potential gives a
measure of the driving force for the reaction
to occur


This means that the more positive the value of
E red, the greater the driving force for reduction
In all voltaic cells the reaction at the cathode has a
more positive value of E red than the reaction at
the anode.


So the greater driving force of the cathode half-reaction
is used to force the anode reaction to occur in reverse
(oxidation)
Since we can think of oxidation as the reverse of
reduction, while all values are given for oxidation

We know that E cell is the difference
between the standard reduction
potential of the cathode reaction and the
standard reduction potential of the
anode

We can think of E cell as the net driving
force that pushes the electrons through the
external circuit
Example

A voltaic cell is based on the following
two half-reactions:




Cd2+(aq) + 2e-  Cd(s)
Sn2+(aq) + 2e-  Sn(s)
Find the half-reactions that occur at the
anode and cathode
Find the standard cell potential
Finding the Half-Reactions

Look up the E
reactions


E
E



red
for the two half-
2+/Cd) = -0.403 V
(Cd
red
2+/Sn) = -0.136 V
(Sn
red
Since the standard reduction potential for Sn2+
is more positive (less negative) the reduction of
Sn2+ occurs at the cathode
Cathode: Sn2+(aq) + 2e-  Sn(s)
Anode: Cd(s)  Cd2+(aq) + 2e-
Determine the standard cell
potential

E

cell
E
=E
cell
red(cathode)
-E
red(anode)
= (-0.136 V) - (-0.403 V) = 0.267 V
Strengths of Oxidizing and
Reducing Agents

Generally, the more positive the E red
value for a half-reaction, the greater the
tendency for the reactant of the halfreaction to be reduced

Which means a greater chance of it to
oxidize another species
Good oxidizing agents

Among the most often used oxidizing
agents we find the halogens, O2 and
oxyanions


Such as MnO4-, Cr2O72- and NO3All have a central atom with a large,
positive oxidation state
Good Reducing Agents


Like acid-base strength, the strongest
oxidizers are the weakest reducing
agents
Which also means the strongest
reducing agent is the weakest oxidizer


So the half-reaction with the smallest
reduction potential is the most easily
reversed as an oxidizer
Means they most easily give up electrons
to other species

Common good reducing agents include:


H2, and active metals such as the alkali
metals and the alkaline earth metals
Other metals whose cations have negative
E red values are also used


Such as Zn and Fe
Often difficult to store solutions of
reducing agents because of the
presence of O2 (a strong oxidizing
agent) in the air
Table 20.14, pg. 867


Table like this given on AP test
The list orders the ability of substances
to act as an oxidizing or reducing agent

Remember, the more positive the value of
E red, the strong oxidizing agent it is

Which means it will tend to be reduced
20.5 - Free Energy and Redox
Reactions

Voltaic cells use redox reaction to
proceed spontaneously


Any reaction that occurs in a voltaic cell to
produce a positive emf must be
spontaneous
So we can determine spontaneity of a
redox reaction by deciding if it produces a
positive emf

We start by applying a previous
equation to all redox reactions, not just
limiting it to voltaic cells

E

Becomes

E = E red(reducing process) - E
process)
cell
=E
red(cathode)
-E
red(anode)
red(oxidation

E = E red(reducing process) E red(oxidation process)




A positive value of E tells us this is a spontaneous
process
While a negative value of E indicates a
nonspontaneous one
Use E to represent emf under nonstandard
conditions
Use E to represent emf under standard
conditions
Example

Using standard reduction potentials,
determine whether or not the following
reactions are spontaneous under
standard conditions


Cu(s) + 2H+(aq)  Cu2+(aq) + H2(g)
Cl2(g) + 2I-(aq)  2Cl-(aq) + I2(s)
Cu(s) + 2H+(aq)  Cu2+(aq) +
H2(g)

In this reaction, Cu is oxidized to Cu2+, while
H+ is reduced to H2


Half-reactions would be
Reduction:


Cu(s)  Cu2+(aq) + 2e-
Therefore, using E = E
E


E
red =
0V
Oxidation:


2H+(aq) + 2e-  H2(g)
red(oxidation
E
red =
red(reducing
0.34 V
process) -
process)
E = (0 V) - (0.34 V) = -0.34 V
Because E is negative, the reaction is not spontaneous
in this direction
Cl2(g) + 2I (aq)  2Cl (aq) +
I2(s)


Cl2 is reduced, while I-is oxidized
Reduction:


E
red =
1.36
Oxidation:


Cl2(g) + 2e-  2Cl-(aq)
V
E

2I-(aq)  I2(s) + 2eV
E
red =
0.54
= (1.36 V) - (0.54 V) = 0.82 V
This reaction would be spontaneous, and
could be used to build a voltaic cell
Activity Series of Metals

Remember, when dealing with single and double
replacement reactions that you consult the
activity series to determine if one metal will
replace another


If metal A is above metal B on the table, then A will
replace B
Any metal in the activity series will be oxidized by
the ions of any metal below it


Activity series really shows the oxidation reactions of
the metals, ordered from strongest reducing agent at
the top to the weakest reducing agent at the bottom
So we can use the relative strength of oxidizers to
predict the results of displacement reactions with
metals
EMF and ΔG

The change in Gibbs free energy, ΔG
measures the spontaneity of a process at constant
temperature and pressure

Because EMF, E, of a redox reaction is spontaneous, there is a
relationship between these two terms
Relationship

ΔG = -nFE

n is a positive number without units


F is called Faraday’s constant


Represents the number of electrons transferred in the reaction
Equal to 96,485 J/V-mol
The units of ΔG is J/mol

The /mol means per mole of reaction as written
ΔG = -nFE

Both n and F are positive numbers

So a positive value of E means a negative
ΔG


So a positive value of E and a negative value for ΔG both indicate a
spontaneous reaction
Now, since ΔG = -RT(lnK)

We can now relate the standard emf to the equilibrium
constant for the reaction
Example

Use the standard reduction potentials to
calculate the standard free-energy
change, ΔG and the equilibrium constant, K, at
room temperature (T = 298 K) for the reaction

4 Ag(s) + O2(g) + 4H+(aq)  4 Ag+(aq) + 2 H2O(l)
Find E

First
To find E we need to break into halfreactions and obtain E red values

Reduction:



O2(g) + 4H+(aq) + 4 e-  2 H2O(l)
E red = +1.23 V
Oxidation:


4 Ag(s)  4Ag+(aq) + 4eE red = +0.80 V


E = 1.23 V - 0.80 V = 0.43 V
Now we use ΔG = -nFE


ΔG = -(4)(96485)(0.43)
= -1.7x105 J/mol = -170 kJ/mol
Now solve for K



ΔG = -RT(lnK)
K = e-ΔG /RT = e-1.7x10^5/(8.314x298)
K = 9x1029



Why would we do this?
Such a large K value very hard to measure
However, the voltage measured is easy to measure.
20.6 - Cell EMF Under
Nonstandard Conditions

As a voltaic cell is discharged, the reactants
are consumed and products generated


Which changes the concentrations
The emf keeps dropping until E = 0, at which point
we say the cell is dead



At this point the concentrations of products and reactants
are cease to change
They are at equilibrium
This section will look at how emf changes
with concentration, which will show
nonstandard conditions
The Nernst Equation

The dependence of cell emf on concentration
is found from the dependence of free-energy
change on concentration

ΔG = ΔG + RT(lnQ)

Remember, Q is the reaction quotient

Substitute ΔG = -nFE and
-nFE = -nFE - RT(lnQ)
Solving this for E gives the Nernst equation

E = E - (RT/nF)(lnQ)



E = E - (RT/nF)(lnQ)



Commonly expressed as a base 10 logarithm
E = E - (2.303RT/nF)logQ
At room temperature (T = 298 K)


E = E - (0.0592/n)logQ
E = E - (0.0592/n)logQ

Consider the reaction

Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

In this case, n = 2




Two electrons transferred from Zn to Cu2+
Standard emf for this cell is 1.10 V
Therefore at 298 K, the Nerst equation gives us
E = 1.10 V -(0.0592/2)log([Zn2+]/[Cu2+])




Where did the solids go?
Now if we know the concentrations of these chemicals,
we can find the emf of the cell
At [Cu2+] = 5.0 M and [Zn2+] = 0.050 M we get
E = 1.10 V -(0.0592/2)log([0.050/5.0) = 1.16 V

So by changing the concentrations, we get a different emf.
In General

If the concentrations of reactants
increase relative to the concentrations
of the products, the emf increases

Whereas if the concentrations of products
increase relative to the reactants, the emf
decreases

Which is what happens when a batter goes
dead!
Example

Calculate the emf at 298 K generated by the
following cell





Cr2O72-(aq) + 14H+(aq) + 6I-(aq) 
2Cr3+(aq) + 3I2(s) + 7H2O(l)
[Cr2O72-] = 2.0 M, [H+] = 1.0 M, [I-] = 1.0 M
and [Cr3+] = 1.0x10-5 M
Standard emf for this cell is 0.79 V
As we have shown earlier, there are 6
electrons transferred in this reaction

First find Q

Cr 
Q
Cr O H  I 
3 2
2
2
 14
 6
7
Q
1.0 10 
5 2
2.01.0 1.0
14
6
 5.0 10
11
Now plug into Nernst equation

E = E - (0.0592/n)logQ



= 0.79 - (0.0592/6)log(5.0x10-11)
= 0.79 + 0.10 = 0.89 V
We should expect this result.

We increased the concentration of a
reactant (5M) and kept everything else at
standard molarities (1M0, so we should
expect emf to increase
Example 2

If the voltage of a Zn-H+ cell (like we
have seen before) is 0.45 V at 25 C
when [Zn2+] = 1.0 M and PH2 = 1.0 atm,
what is the concentration of H+?

Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
Plan




Find E for the reaction first
Determine n from the reaction equation
Solve Nernst equation for Q
Use equation for the cell to write
expression for Q that contains [H+] to
solve for it
Find E


Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
E = E red(reduction) - E red(oxidation)

= 0 V - (-0.76 V) = +0.76 V
Finding n

Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)


Each zinc loses 2 electrons
So n = 2
Use Nernst Equation

E = E - (0.0592/n)logQ



log Q = (0.76 - 0.45)(2/0.0592)
log Q = 10.47
Q = 1010.47 = 3.0x1010

Zn P
Q
H 
2
H2
 2
H 
 2

1.01.0
10

 3.0 10
H 
 2
1.0
11

 3.3 10
10
3.0 10
H  

3.3 10
11
6
 5.8 10 M
Concentration Cells

In each of the voltaic cells we have seen, the
reactive species at the anode has been
difference from the one at the cathode


Cell emf depends on concentration
Which means that a voltaic cell can be made
using the same species at both the anode and
cathode


As long a the concentrations are different
A cell based only on the emf generated because of
difference in concentration is called a concentration cell
Features of a Concentration
Cell

Still requires two vessels, connected by
a salt bridge, and an external wire
Calculations

When calculating emf for a
concentration cell, the following rules
must be observed
1.
2.
Standard emf is zero (E
Use Nernst equation


= 0)
Q = [dilute]/[concentrated]
When concentration is equal in both
cells, Q = 1 and E = 0
Picture Example

What is the emf provided by the
concentration cell in this picture?

E = E - (0.0592/n)logQ






Cu2+
So 
E =
E =
E =
E =
 Cu(s), or vice
= 2
0 0 0 - (-0.0592)
0.0592 V
20.9 - Electrolysis

Voltaic cells are based on spontaneous redox
reactions

It is possible to use electrical energy to cause
nonspontaneous redox reactions to occur




For example, we can use electricity to decompose
molten sodium chloride into its component elements
2 NaCl(l)  2 Na(l) + Cl2(g)
A nonspontaneous process driven by outside
electrical energy is called an electrolysis
reaction
These reactions take place is electrolytic cells
Electrolytic Cell

An electrolytic cell is made of two
electrodes in molten salt or a solution

A battery or some other source of direct
electrical current acts as an electron pump


Pushing electrons into one electrode and
pulling them from the other
Just like in a voltaic cell, the electrode at
which reduction occurs is called the
cathode (and where oxidation occurs is the
anode)
Things to notice

In a voltaic cell (or any other source of direct
current)


So the electrode of the electrolytic cell that is
connected to the negative terminal of the
voltage source is the cathode


electrons move from the negative terminal
It receives the electrons that are used to reduce
the substance
The electrons that are removed during the
oxidation process at the anode travel to the
positive terminal of the voltage source
Uses

Electroplating

Electroplating uses electrolysis to deposit a
thin layer of one metal on another metal



Metal in solution becomes deposited onto
cathode
Only requires a small voltage source to
accomplish
pg. 884
Electrical Work

Remember

A positive value of E is associated with a
negative value of free-energy change


And thus spontaneous processes
We also know that for any spontaneous
process ΔG measures the maximum useful work, wmax for
the process

ΔG = wmax

Since ΔG = -nFE


wmax = -nFE
The cell emf for a voltaic cell is positive, so wmax will be
negative

Negative work means work is being done by the system on the
surroundings

In an electrolytic cell, we use external energy
to bring about a nonspontaneous
electrochemical reaction



Therefore ΔG is positive and Ecell is negative
To force the process to occur, we need to apply an external potential, Eext,
whose magnitude is larger than Ecell
When Eext is applied to a cell, the surroundings are doing work on the
system, and therefore


w = nFEext
So we can now calculate maximum work from a voltaic cell, and the minimum
work needed for electrolysis
A few final notes...

Electrical work is expressed in terms of
watts times time



1 W = 1 J/s
Therefore, a watt-second = 1 J
The unit power companies use is the
kilowatt-hour (kWh), which is equal to
3.6x106J