1
1.
2.
3.
4. Be able to balance redox equations using both the half-reaction method and the oxidation number method. (19-2)
Chemistry 30
Chapter 19 - Oxidation-Reduction Reactions
Objectives
Assign oxidation numbers to each atom in an element or compound. (19-1)
Explain what an oxidation-reduction (redox) reaction is. (19-1)
Be able to identify the species oxidized and reduced in a redox reaction. (19-1)
5. Relate chemical activity to oxidizing and reducing strength. (19-3)
8.
9.
6.
7.
Explain the concept of auto-oxidation. (19-3)
Explain what is required for an electrochemical cell. (19-4)
Describe the nature of voltaic cells. (19-4)
Explain the process of electroplating. (19-4)
10.
11.
Describe the chemistry of a rechargeable cell. (19-4)
Calculate cell potentials from a table of standard electrode potentials. (19-4)
Vocabulary half-reaction oxidation oxidizing agent electrochemical cell electrolysis reduction potential oxidation-reduction redox reducing agent electrochemistry electrolytic cell oxidized reduction anode electrode electroplating standard electrode potential voltaic cell reduced autooxidation cathode electrode potential half-cell

2
Oxidation and Reduction
Oxidation and reduction are common reactions; we call them burning, rusting and other things. Oxidation and reduction are the basis of batteries of all kinds and fuel cells, among other things. Oxidation and reduction are the reactions involved in both photosynthesis and respiration.
Oxidation and reduction involve the exchange of electrons (acid-base reactions involve protons).
Oxidation is the loss of an electron by an atom or ion. Charge of the atom or ion becomes more positive.
The term oxidation comes from an early belief that it always involved oxygen.
Reduction is the gain of an electron by an atom or ion. Charge of the atom or ion becomes more negative.
The term reduction comes from early work with metals and their ores; if you have 100 kg of iron ore (Fe you get roughly 70 kg of iron metal. The mass of the substance is reduced.
2
O
3
)
Just like acid-base pairs, oxidation cannot happen without reduction; an atom or ion cannot give away an electron without a species to give the electron to. That is why this phenomenon is often called REDOX.
Oxidation Numbers
We have to be able to tell whether oxidation or reduction has occurred. We thus have to assign a charge, or
oxidation number, to every atom in every element, compound or ion we see. This applies even to molecular compounds, where we don’t usually consider charge.
To assign oxidation numbers we apply the following rules:
1.
2.
The oxidation number of any substance in its elemental state is zero (0).
The oxidation number of any simple ion is the charge of the ion. (Ca 2+
 Group 1 ions always have an oxidation number of 1+
= 2+; P 3 = 3-)
 Group 2 ions always have an oxidation number of 2+
 Aluminum ions always have an oxidation number of 3+
 Group 17 ions most often have an oxidation number of 1-
3. Oxygen in a compound has an oxidation number of 2-
 The exception is in peroxides; a combination of oxygen with group 1 or 2 metals where oxygen must have an oxidation number of 1- (H
2
O
2
, MgO
2
)
4. Hydrogen in a compound has an oxidation number of 1+
 The exception is in metal hydrides; a combination of hydrogen with another metal, where the oxidation number of the hydrogen is 1- (NaH, CaH
2
, BH
3
)

5.
3
The sum of the oxidation numbers in any compound or complex ion is equal to the charge of the compound or ion.
H
 This allows the determination of the oxidation number of transition metals, lanthanides, actinides and some non-metals:
Example 1:
2
SO
4
-
-
- hydrogen generally has an O.N. of 1+ oxygen generally has an O.N. of 2- sulfur is the problem, we get the answer mathematically:
2(H) + 1(S) + 4(O) = 0 (the substance has no total charge)
2(1+) + S + 4(2-) = 0
(2+) + S + (8-) = 0
S = 0 - (2+) - (8-)
S = 6+
The oxidation number of S is 6+
Example 2:
Cr
2
O
7
2-
-
(the dichromate ion) oxygen is usually 2-
2(Cr) + 7(O) = 2-
2Cr + 7(2-) = 2-
2Cr + (14-) = 2-
2Cr = 2- - (14-)
2Cr = 12+
Cr = 6+

4
Recognizing Oxidation and Reduction in Chemical Reaction Equations
Consider the following balanced chemical equation:
3 Cu 2+ + 2 Al → 3 Cu + 2 Al 3+
Let’s apply what we know about oxidation numbers to this equation:
3 Cu 2+ + 2 Al → 3 Cu + 2 Al
O.N. 2+ 0 0
3+
3+
In this reaction the copper goes from 2+ on the reactant side to 0 on the product side; it’s charge got lower, so it is reduced lost electrons):
O.N.
(it gained electrons). The aluminum goes from 0 to 3+; it’s charge got greater, so it is reduction (gained 2 electrons each)
┌──────────┐
3 Cu 2+ + 2 Al → 3 Cu + 2 Al
2+ 0 0 3+
└─────────┘
3+ oxidation (lost 3 electrons each) notice there are 3 coppers in the balanced equation; the total electrons gained is 3(2 e oxidized
1) = 6 electrons.
(it there are 2 aluminums in the equation; the total electrons lost is 2(3 e
1) = 6 electrons.
A bit of terminology; the substance that is oxidized can be said to have caused the other substance to be reduced. It is the reducing agent.
substance to be oxidized. It is the
Our equation now looks like this:
The substance that is reduced can be said to have caused the other oxidizing agent .
O.N.
reduction (gained 2 electrons each)
┌──────────┐
3 Cu 2+ + 2 Al → 3 Cu + 2 Al 3+
2+ 0 0 3+
└─────────┘ oxidation (lost 3 electrons each)

5
Part B on your worksheet asks you to do this for each of 5 equations. The first equation would look like this:
O.N.
reduction (gained 1 electron each)
┌────────────┐
O
2
+ 2 H
2
O → 2 H
2
O
2
0 1+ 2- 1+ 1-
└───────┘ oxidation (lost 1 electron each)
Note that oxygen is both oxidized and reduced in this equation (this happens sometimes); the oxygen in H a peroxide. Nothing happened to the hydrogen; it is basically around for the ride.
Repeat this process for the rest of part B of the worksheet.
Balancing Redox Equations
2
O
2
is
1) i) ii) iii)
In addition to recognizing oxidation numbers and being able to determine which species is oxidized and which has been reduced, you also must be able to balance chemical equations. To do this you must do two things:
-
- balance for the number of atoms (just like you are used to) balance for the number of electrons (the electrons lost must equal the electrons gained)
There are two methods for doing this. You must be able to use both. iv) v) vi) vii)
The oxidation number method uses the following steps: balance the equation for atoms other than hydrogen and oxygen. assign oxidation numbers to each atom in the chemical reaction. determine which species is oxidized and which is reduced and how many electrons are gained and lost.
Take into account the number of atoms of the oxidized and reduced species.
the species on each side of the reaction equation by a factor required so that the number of electrons gained and lost are the same. balance the equation for oxygen by adding water molecules (H
2
O). balance the equation for hydrogen by adding hydrogen ions (H check the final answer. It should be balanced for atoms and the charge on each side of the equation should be equal.
1+ ).

6
2) i) ii) iii)
The half-reaction method uses the following steps: write the equation as two half-reactions; one for oxidation and the other for reduction. balance each equation for atoms other than hydrogen and oxygen. balance each equation for oxygen by adding water molecules (H
2
O). iv) v) vi) balance each equation for hydrogen by adding hydrogen ions (H balance equation for charge by adding electrons (e
each equation by a factor required so that the number of electrons gained and lost are the
Balancing for basic conditions
1).
1+ ). vii) same. add the two equations together. Cancel out like terms. The electrons must cancel out. viii) check the final answer. It should be balanced for atoms and the charge on each side of the equation should be equal. iii) iv)
Both methods of balancing assume acidic conditions; that is why we add H best under basic conditions.
Balancing is very straightforward: i) ii) balance as you did previously. when done, add as many OH 1-
1+ ions. Some reactions proceed
ions to each side of the reaction equation as you have H 1+ ions. one side now has H if you have H
2
1+ and OH 1 ions. They combine to form H
2
O.
O on each side of the equation, cancel terms appropriately.

7
Half-Reaction Method
ClO
3
1 + MnO
2
 Cl 1 + MnO
4
1ii) iii) i)
1x
2x vii) vi) v) iv)
Write the half-reactions:
ClO
3
MnO
1-
2
 Cl
Balance for O by adding H
2
O:
1-
 MnO
4
1-
Balance for atoms; both Cl and Mn are already balanced.
Balance for charge by adding e
6 e
2 H
 MnO
Multiply to equalize number of electrons:
6 e
4 H
1-
2
1-
2
ClO
3
2 H
Balance for H by adding H
6 H
+ 6 H
2
1-
O + MnO
1+
2 H
2
1+
O + MnO
+ 6 H 1+
 Cl
2
 MnO
O + 2 MnO
2
1+ :
+ ClO
+ ClO
+ ClO
3
1 + 3 H
2
2
O
4
1-
1 Cl 1 + 3 H
2
O
O + MnO
2
1-
3
1-
:
4
1-
 MnO
 Cl 1-
+ 4 H
4
1 + 4 H 1+
+ 3 H
1+ + 3 e

3
1 Cl
2 MnO
4
1-
1 + 3 H
+ 8 H 1+
2
+ 6 e
O
1-
Add the equations together, cancelling like terms:
2
1-
O
6 e 1-
4 H
2
H
2
+ 6 H 1+
O + 2 MnO
O + ClO
3
1-
+ ClO
+ 2 MnO
3
1-
2
 2 MnO
2
 Cl 1-

4
1-
Cl 1-
+ 3 H
+ 8 H 1+ + 6 e
+ 2 MnO
2
1-
O
4
1 + 2 H 1+ viii) Check work; balanced for atoms, balanced for charge.

8 vi) v) vii)
Oxidation Number Method
ClO
3
1 + MnO
2
 Cl 1 + MnO
4
1i) ii)
Balance for atoms; already balanced for Cl and Mn.
Determine oxidation number of each species:
ClO
3
1 + MnO
2
 Cl 1 + MnO
4
1-
5+ 2- 4+ 2- 1- 7+ 2- iii) Determine which is oxidized and which is reduced, and the number of electrons:
ClO
3
1 + MnO
2
 Cl 1 + MnO
4
1-
5+ 2- 4+ 2- 1- 7+ 2- iv)
- Cl goes from 5+ to 1-
-
-
-
- it is reduced it gains 6 electrons
- Mn goes from 4+ to 7+ it is oxidized it loses 3 electrons
Multiply by a factor to equalize electrons gained and lost:
- multiply the Mn by 2, to give them 6 electrons lost, to equal the 6 electrons gained by the Cl
ClO
Balance for O by adding H
H
Balance for H by adding H
H
2
O + ClO
2
3
1 + 2 MnO
2
 Cl
O + ClO
3
1 + 2 MnO
3
1 + 2 MnO
2
2
2
O
 Cl

1+
Cl
1 + 2 MnO
1-
1-
+ 2 MnO
+ 2 MnO
4
1-
4
1-
4
1-
+ 2 H 1+
Check balancing of atoms and charge; balanced.

9
Balancing for basic conditions
ClO
3
1 + MnO
2
 Cl 1 + MnO
4
1i) ii) iii) balance the equation using either method: add OH
H
1-
2
O + ClO
3
1-
to both sides in equal number to the H
2 OH make water:
1 + H
2
+ 2 MnO
2
O + ClO
3
1-
 Cl 1 + 2 MnO
4
1-
1+
+ 2 H 1+
:
+ 2 MnO
2
 Cl 1 + 2 MnO
4
1 + 2 H 1+ + 2 OH 1iv)
2 OH
2 OH
1 + H cancel like terms
2
O + ClO
1 + ClO
3
1-
3
1 + 2 MnO
2
+ 2 MnO
2
 Cl
 Cl 1 + 2 MnO
1 + 2 MnO
4
1 + H
2
4
1-
O
+ 2 H
2
O
You should be prepared to balance redox word equations as well as the reaction equations we have just practiced:
The dichromate ion and mercury metal react in a battery to produce the chromium (III) ion and the mercury (II) ion. Write and balance the equation for this redox reaction:
Half-Reaction Method:
1x
3x
14 H
3
1+
(2) e
+ Cr
1-
2
O
+ 3 Hg  3 Hg
7
2 2 Cr
2+
3+ + 7 H
2
O + 6 e 1-
14 H
14 H 1+
1+
Oxidation-Number Method
+ Cr
2
+ Cr
O
7
2-
2
O
+
7
2-
3
+ 3 Hg  2 Cr 3+
┌─────────────┐
Hg  2 Cr 3+
+ 3 Hg
reduction (gained 3 electrons each) x 2 atoms = 6 e
+ 3 Hg 2+
6+ 2- 0 3+ 2+
└────────────┘
+
2+ + 7 H
2
7 H
2
O
1-
O
gained
1oxidation (lost 2 electron each) x 1 atom = 2 e lost x 3 = 6 e 1 lost
When you do problems they should look like the last two examples (you don’t have to write the equations over and over like the first examples)
Complete the worksheet problems as instructed in class.

10
Voltaic Cells
The energy released in a spontaneous redox reaction may be used to perform electrical work. Voltaic, or
galvanic cells, are devices in which electron transfer occurs via an external circuit.
If a strip of Zn is placed in a solution of CuSO
4
, Cu is deposited on the Zn and the Zn dissolves by forming Zn 2+ .
Zn
( s )
+ Cu 2+
( aq )
 Zn 2+
( aq )
+ Cu
( s )
Zn is spontaneously oxidized to Zn
The Cu 2+
2+
by Cu
2+
is spontaneously reduced to Cu 0
.
by Zn.
The entire process is spontaneous.
Each of the two compartments of a voltaic cell is called a half-cell.
This voltaic cells consists of:
An oxidation half-reaction: Zn( s )
Oxidation takes place at the anode.
 Zn 2+ ( aq ) + 2e –
–  Cu( s ) A reduction half-reaction: Cu 2+ ( aq ) + 2e
Reduction takes place at the cathode.
A salt bridge is used to complete the electrical circuit:
- Cations move from anode to cathode.
- Anions move from cathode to anode.

11
•
The two solid metals are the electrodes (cathode and anode).
As oxidation occurs, Zn is converted to Zn 2+ and 2e – .
The electrons flow toward the cathode where they are used in the reduction reaction.
We expect the Zn electrode to lose mass and the Cu electrode to gain mass.
Electrons flow from the anode to the cathode.
-
-
-
-
-
Therefore, the anode is negative and the cathode is positive.
Electrons cannot flow through the solution; they have to be transported through an external wire.
Anions and cations move through a porous barrier or a salt bridge.
Cations move into the cathodic compartment to neutralize the excess of negatively charged ions
(Cathode: Cu oxidation.
2+ + 2e –  Cu, so the counter ion of Cu is in excess).
Anions move into the anodic compartment to neutralize the excess Zn
A Molecular View of the Electrode Process
2+ ions formed by
“Rules” of voltaic cells:
• At the anode electrons are products.
•
•
•
•
•
Oxidation occurs at the anode.
At the cathode electrons are reagents.
Reduction occurs at the cathode.
The flow of electrons from anode to cathode requires an external wire.
The transfer of ions through a salt bridge maintains overall charge balance for the two compartments.

12
Standard Reduction Potential
This is related to the activity series of elements you studied in grade 11 (see text, p.266). The elements highest on the table were able to replace compounds of the lower substances on the table; the higher elements were more active than ones lower down.
If we look at one of these reactions:
CuCl
2 (aq)
+ Al
Since Al is higher on the activity table than Cu, the Al will replace the Cu in the compound in a single replacement reaction:
(s)

3 CuCl
2 (aq)
+ 2 Al
(s)
 3 Cu
(s)
+ 2 AlCl
The reverse reaction:
3 (aq)
3 Cu
(s)
+ 2 AlCl
3 (aq)
 No Reaction will not happen because Cu is lower on the chart than Al, so it is incapable of replacing it.
If we look at the reaction again we see it as a redox reaction:
3 CuCl
2 (aq)
+ 2 Al
(s)
 3 Cu
(s)
+ 2 AlCl
3 (aq)
2+ 1- 0 0 3+ 1-
The Cu goes from 2+ to 0; it is reduced.
The Al goes from 0 to 3+; it is oxidized.
Cell EMF Under Standard Conditions
The flow of electrons from anode to cathode is spontaneous, but what is the “driving force”?
Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than the anode.
• Potential difference is the difference in electrical potential.
• The potential difference is measured in volts.
• One volt (V) is the potential difference required to impart one joule (J) of energy to a charge of one coulomb (C):
1 V

1
J
C
• Electromotive force (emf) is the force required to push electrons through the external circuit.
• Cell potential:
•
• E cell
E cell
is the emf of a cell.
This is known as the cell voltage .
is > 0 for a spontaneous reaction.

13
For 1 M solutions or 1 atm pressure for gases at 25°C (standard conditions), the standard emf (standard cell
potential) is called E° cell
.
• For example, for the reaction:
E°
Zn
( s )
+ Cu 2+
( aq )
 Zn 2+
( aq )
+ Cu
( s )
• cell
= +1.10 V
We can now use the standard reduction potential table. It ranks a number of chemical reactions in order of their standard emf, based on their potential to be reduced (every equation is written as a half-reaction in the direction of gaining electrons) relative to the hydrogen electrode. The higher on the table, the more likely the reduction will occur, the lower the less likely.
We can use the standard reduction potential table both to balance redox equations and to predict whether a particular reaction is likely to occur.
For example:
ClO
4
1 + MnO
2
 Cl 1 + MnO
4
1-
If we divide this redox reaction into 2 half reactions:
ClO
4
MnO
1 Cl 1-
2
 MnO
4
1-
Now look at the table of Standard Reduction Potentials, and find these reactions. Complete each half reaction based on the equations you find on the table. Include the electrochemical potential:
8 H 1+
4 OH
+ 8
1e 1 + ClO
+ MnO
2

4
1 Cl
MnO
4
1-
1 + 4 H
2
O
+ 2 H
2
O + 3 e 1-
Eº
+1.37 V
-0.59 V
Note: the sign of the electrochemical potential of the second equation is reversed, because the equation is reversed.
To balance this equation simply multiply each equation so as to arrive at the lowest common multiple for the electrons:
3x 8 H 1+
8x 4 OH
+ 8
1e 1 + ClO
+ MnO
2

4
1-
MnO
 Cl 1 + 4 H
4
1 + 2 H
2
2
O
O + 3 e 1-
Eº
+1.37 V
-0.59 V
Note: Do not multiply the Eº value. The potential of a cell is determined only by the identity of the reactants,
Gives
24 H 1+ + 24
32 OH 1 + 8 MnO
24 H 1+
The H
+ 32 OH
1+ not their amount e 1-
and OH
+ 3 ClO
1-
1-
2

4
1-
8 MnO
+ 3 ClO
 3 Cl
4
1-
4
1-
1 + 12 H
+ 16 H
+ 8 MnO
2
2
2
O
O + 24 e
 8 MnO
1-
4
1 + 3 Cl 1 + 28 H
2
O
become water, so the final form of this equation is:
8 OH 1 + 3 ClO
4
1 + 8 MnO
2
 8 MnO
4
1 + 3 Cl 1 + 4 H
2
O
Eº
+1.37 V
-0.59 V
+0.78 V
+0.78 V

14
The electrochemical potential tells us if this reaction will actually happen:
-
- if the potential is positive, the reaction will occur if the potential is negative, the reaction will not occur without help.
Applications of Oxidation / Reduction Chemistry
1. Batteries and Fuel Cells
• A battery is a portable, self-contained electrochemical power source consisting of one or more voltaic cells. Each cell has a given potential; cells are attached in series to increase the potential of the battery. For instance, alkaline batteries have a potential of 1.5 V. Four batteries in series would have a potential of
4 (1.5 V) = 6.0 V.
Lead-Acid Battery
• A 12 V car battery consists of six cathode/anode pairs each producing 2 V:
• Cathode: PbO
2
on a metal grid in sulfuric acid:
PbO
2( s )
+ HSO
4
• Anode: Pb:
1-
( aq )
+ 3H 1+
( aq )
+ 2e 1 PbSO
4( s )
+ 2H
2
O
( l )
Pb
( s )
+ HSO
4
1-
( aq )
 PbSO
4( s )
+ H 1+
• The overall electrochemical reaction is
PbO
2( s )
+ Pb
( s )
+ 2HSO
4
1-
( aq )
+ 2H 1+
( aq ) aq )
+ 2e
 2PbSO
4( s )
1-
+ 2H
2
O
( l )

15
• The cell potential for this reaction is:
E° = E° (cathode) - E° (anode)
= (+1.685 V) - -0.356 V)
= +2.041 V
•
•
Wood or glass-fiber spacers are used to prevent the electrodes from touching.
An advantage of these cells is that they can be recharged.
• An external source of energy is used to reverse the process (the car alternator, which generates electricity while the car is running to both charge the battery and power the car’s electrical devices).
Alkaline Battery
• The most common nonrechargeable battery is the alkaline battery.
• Powdered zinc metal is immobilized in a gel in contact with a concentrated solution of KOH.
• Thus, these batteries are
• The reaction at the anode is:
•
Zn
( s )
+ 2OH 1-
(aq) alkaline

.
Zn(OH)
2
( aq ) + 2e 1-
The reaction at the cathode is the reduction of MnO
2
:
•
2MnO
2( s )
+ 2H
2
O
( l )
+ 2e 1 H)
( s )
+ 2OH 1-
( aq )
The cell potential of these batteries is 1.55 V at room temperature.
Nickel-Cadmium, Nickel-Metal-Hydride, and Lithium-Ion Batteries
•
•
A common rechargeable battery is the nickel-cadmium (NiCad) battery.
The reaction at the cathode is:
•
2NiO(OH)
( s )
+ 2H
2
O
( l )
+ 2e 1 2Ni(OH)
2( s )
+2OH 1-
( aq )
The reaction at the anode is:
Cd
( s )
+ 2OH 1-
( aq )
 Cd(OH)
2( s )
+ 2e 1-
•
•
•
The cell potential of this battery is about 1.30 V at room temperature.
Cadmium is a toxic heavy metal.
• There are environmental concerns to be addressed regarding the disposal of such batteries.
Other rechargeable batteries have been developed.
• NiMH batteries (nickel-metal-hydride).
• Li-ion batteries (lithium-ion batteries).

16
Hydrogen Fuel Cells
•
•
• Direct production of electricity from fuels occurs in a fuel cell.
• An example is a hydrogen fuel cell.
• At the cathode:
2H
2
O
( l )
+ O
• At the anode:
2( g )
+ 4e 1 4OH 1-
( aq )
•
2H
2( g )
+ 4OH 1-
( aq )
 4H
2
O
( l )
+ 4e 1-
This cell is known as a PEM fuel cell (proton exchange membrane).
The anode and cathode are separated by a polymer membrane that is permeable to protons but not electrons.
It acts as a salt bridge.
Direct Methanol Fuel Cells
• These fuel cells are similar to the PEM cell but use methanol as a reactant instead of hydrogen gas.
Corrosion
• An example of an undesirable redox reaction is the corrosion of metals.
• Metal is attacked by a substance in the environment and converted to an unwanted compound.
•
•
•
•
•
Corrosion of Iron
Consider the rusting of iron:
• Since E° (Fe 2+ ) < E° (O
• Cathode: O
• Anode: Fe
(
2( g )
+ 4H s )
 Fe
1+
2+
( aq )
2
), iron can be oxidized by oxygen.
( aq )
+ 4e 1-
+ 2e 1-
 2H
2
O
( l )
E°
E° = 1.23 V.
= - 0.44 V.
Dissolved oxygen in water usually causes the oxidation of iron.
The Fe 2+ initially formed can be further oxidized to Fe 3+ , which forms rust, Fe
Oxidation occurs at the site with the greatest concentration of O
2
O
3
.
x H
2
O
( s )
.
2
.
Other factors to consider are the pH, presence of salts, stress on the iron, and contact with other metals.

17
Preventing the Corrosion of Iron
• Corrosion can be prevented by coating the iron with paint or another metal.
• This prevents oxygen and water from reacting at the surface of the iron.
• Galvanized iron is coated with a thin layer of zinc.
• Zinc protects the iron since Zn is the anode and Fe is the cathode:
Zn 2+
Fe 2+
( aq )
+2e 1-
( aq )
+ 2e 1-
 Zn
(
 Fe s )
( s )
E°
E°
= - 0.76 V
= - 0.44 V
• The standard reduction potentials indicate that Zn is easier to oxidize than Fe.
• This process is cathodic protection (the sacrificial anode is destroyed).
• We can use something similar to protect underground pipelines.
• Often, Mg is used as a sacrificial anode:
Mg 2+
Fe 2+
( aq )
( aq )
+2e 1-
+ 2e 1-
 Mg
( s )
 Fe
( s )
E° = - 2.37 V
E° = - 0.44 V

18
Electrolysis
• Electrolysis reactions are nonspontaneous reactions that require an external current in order to force the reaction to proceed.
• They take place in electrolytic cells.
• In voltaic and electrolytic cells, reduction occurs at the cathode and oxidation occurs at the anode.
• However, in electrolytic cells, electrons are forced to flow from the anode to the cathode.
• In electrolytic cells the anode is positive and the cathode is negative.
• In voltaic cells the anode is negative and the cathode is positive.
• Example: The decomposition of molten NaCl.
• Cathode: 2Na 1+
( l )
+ 2e 1 2Na
( l )
• Anode: 2Cl 1-
( l )

2 ( g )
+ 2e 1-
• Industrially, electrolysis is used to produce metals like Al.
• Electrolysis of high-melting ionic substances requires very high temperatures.
• Do we get the same products if we electrolyze an aqueous solution of the salt?
• Water complicates the issue!
• Example: Consider the electrolysis of NaF(aq):
Na 1+
( aq )
+ e 1-  Na
( s )
2H
2
O
( l )
+ 2e 1-  H
2( g )
+ 2OH 1-
( aq )
• Thus, water is more easily reduced than the sodium ion.
E° = - 2.71 V
E° = - 0.83 V
2F
2H
1-
(
2 aq )
 F
O
( l )
2( g )
 O
+ 2e
2( g )
1-
+ 4H 1+
( aq )
+ 4e 1-
• Thus, it is easier to oxidize water than the fluoride ion.
E°
E°
= +2.87 V
= +1.23 V
• Electrolysis does not always involve inert
• An example is electroplating.
electrodes.
• Active electrodes are electrodes that take part in electrolysis.
•
•
• Consider an active Ni electrode and another metallic electrode (steel) placed in an aqueous solution of
NiSO
4
:
• Anode (nickel strip):
• Cathode (steel strip):
Ni
Ni
( s )
2+
(
 Ni aq )
2+
(
+ 2e aq )
1-
+ 2e 1-
 Ni
( s )
Ni plates on the inert electrode.
Electroplating is important in protecting objects from corrosion.

19
Quantitative Aspects of Electrolysis
• We want to know how much material we obtain with electrolysis.
• Consider the reduction of Cu 2+ to Cu.
Cu 2+
( aq )
+ 2e 1-  Cu
( s ).
• 2 mol of electrons will plate 1 mol of Cu.
• The charge of one mol of electrons is 96,500 C (1
• A coulomb is the amount of charge passing a point in one second when the current is one ampere.
• The amount of Cu can be calculated from the current (
Q
F ).
= I t
I ) and time ( t ) required to plate.
Electrical Work
• Free energy is a measure of the maximum amount of useful work that can be obtained from a system:
• We know:
• and:
G = w max
• thus:
G = - nFE w max
= - nFE w max
will be negative. • If E cell
is positive,
• Work is done by the system on the surroundings.

20
The emf can be thought of as being a measure of the driving force for a redox process.
• In an electrolytic cell an external source of energy is required to force the reaction to proceed. w = nFE external
• In order to drive the nonspontaneous reaction, the external emf must be greater than
• From physics we know that work is measured in units of watts:
1 W = 1 J/s
• Electric utilities use units of kilowatt-hours:
E cell
.
1 kWh


1000 W

1 hour




3600s
1 hour




1 J/s
1 W



3.6

10
6
J