Chapter 19: Redox & Electrochemistry
1. Oxidation-Reduction Reactions
Definitions
Oxidation - refers to the _____ of electrons by a molecule, atom or ion
Reduction - refers to the _____ of electrons by an molecule, atom or ion
Chemical reactions in which the oxidation state of one or more substances
changes are called oxidation-reduction reactions (or ______ reactions)
L
E
O
goes
G E R
or
R
O
L
R
The # of electrons lost ____ the # of electrons gained (they are transferred!)
Oxidation ___________ - a means of determining whether the atom is
neutral, electron-rich, or electron-poor. Does not necessarily imply ionic charges.
1. Elements = ____.
2. All oxidation numbers for atoms in compounds add to ____.
All oxidation numbers for atoms in ions add to _______.
3. A fixed charge cation retains its charge. (i.e Ca2+ = ____
4. Hydrogen attached to a nonmetal (oxid # = ___)
Hydrogen attached to a metal (oxid # = ___)
5. Oxygen (usually has oxid # = ___) (unless it violates #2; e.g. peroxides)
6. Halide (F always -1; others usually has oxid # = ___) (unless it violates #2)
7. Determine remaining element oxidation # by following rule #2.
Example: Find the oxidation number of Cr in K2Cr2O7.
K2
Cr2
Note, multiply when going to
upper row; divide when going to
lower row.
Answer is always in bottom row.
O7
19-1
LP#1. Determine the oxidation number of phosphorus in PO43-.
Recognizing Redox Reactions - keeping track of oxidation #s
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
The oxidation numbers for the above elements and ions are:
Zn(s) = ___ → Zn2+ = ___
H+ = ___ → H2 = ___
For a reaction to be a redox reaction, oxidation numbers must change!
Note: All single replacement reactions are redox reactions, but
not all redox reactions are single replacement.
In the above reaction of hydrogen and oxygen:

Zn is ________________;
It is the _________________

H+ is ________________;
It is the ___________________
Oxidizing Agent:
Reducing Agent:
 causes oxidation
 causes reduction
 undergoes reduction
 undergoes oxidation
 gains electrons
 looses electrons
 becomes more (-) (or less +)
 becomes more (+) (or less -)
19-2
LP#2. Identify the oxidized reactant, the reduced reactant, the oxidizing agent
and the reducing agent in the following reaction:
2 Al(s)
+
Cr2O3(s)  2 Cr(s)
+ Al2O3(s)
Note: We can say that Cr is reduced, but we must include the entire compound and say that
Cr2O3 is the oxidizing agent. We must include the entire compound for solids and covalent
compounds and entire ions for polyatomic ions in agents.
Balancing Oxidation-Reduction Reactions
In balancing redox reactions:
the number of electrons gained must equal the number of electrons lost.

Some can be balanced by inspection. For example:
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

However, some require a more sophisticated process.
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
Equations that show only oxidation, or reduction, are called half-reactions

The following is the oxidation of Tin(II) by Iron(III)
Sn2+(aq) + 2 Fe3+(aq)  Sn4+(aq) + 2 Fe2+(aq)
Oxid. ½ Rxn:
Red. ½ Rxn:

Oxidation ½ reactions have electrons as ___________.

Reduction ½ reactions have electrons as ___________.
19-3
Balancing Equations By The Method Of Half-Reactions
The use of half-reactions provides a general way to balance redox reactions
The following is a general set of steps that can be followed to balance almost all
redox equations:
Steps to Balancing Redox Reactions
1. Write the unbalanced net ionic equation and check to see if the final solution
is acidic or basic.
2. Identify and label the oxidation state of each atom in every species.
3. Identify the oxidation and the reduction half reactions. Make a note of how
many e- are lost or gained by each species.
4. Write the separate half-cell reaction equations.
To balance the atom number:
a) Balance all atoms on each side (except O and H).
b) Balance the number of oxygen atoms by adding water to the opposite
side.
c) Balance the number of hydrogen atoms by adding H+ to the opposite
side.
To balance the charge:
d) Make the charges on each side of the equation the same by adding e-.
(NOTE: this is a good place to check; the number of e- lost or gained in
step 3 per atom should be the same as in the half-reactions.)
e) If the number of electrons in each half-reaction is not the same, multiply
by a factor so that both are the same. (remember, e- lost must = egained)
5. Combine half-reactions to make a full equation. Simplify by combining like
terms.
6. Check to make sure that the atom number and the total charge on each side is
the same.
7. If the solution is basic, add enough OH- to neutralize all the extra H+. These
turn into waters on that side. Also add the same number of OH- to the other
side of the equation. (Just like algebra; what you do to one side must be done
to the other.)
19-4
Balancing Redox Reactions That Occur in Basic Solution
The procedure for balancing in this case is actually the same as for balancing
in an acidic solution, with one small twist.
Even though the previous reaction does not occur in basic solution, we are going
to pretend that it does for the sake of demonstrating the procedures for balancing
redox reactions under basic conditions.
 Start with the balanced reaction in acidic solution.
 Neutralize any H+ by adding OH- to both sides of the equation.
 Combine H+ and OH- to form water.
 Reduce any replicate water molecules.
Finding Oxidation Numbers
Oxidation Number - a means of determining whether the atom is neutral,
electron-rich, or electron-poor. This does not imply ionic charges.
8.
9.
Elements = 0.
K, N2, Xe (oxid # = 0)
All oxidation numbers add to zero (or to ion charge for an ion).
10.
A fixed charge cation retains its charge.
Group #1 (oxid # = +1)
Group #2 (oxid. # = +2)
Ag ion = +1; Cd ion = +2; Zn ion = +2; Al ion = +3
11.
Hydrogen attached to a nonmetal (oxid # = +1)
Hydrogen attached to a metal (oxid # = -1)
12.
Oxygen (usually has oxidation # = -2) (unless it violates #2; e.g. peroxides)
13.
Halide (F always -1; others usually has oxidation # = -1) (unless it violates #2)
14.
Determine remaining element oxidation # by following rule #2.
19-5
Example: The reaction between MnO4- and H2C2O4 in acidic solution:
Here is the unbalanced equation that describes this reaction:
MnO 4 - + H 2 C 2 O 4
 Mn 2+ (aq) + CO 2 (g)
To balance this redox reaction using the method of half-reactions, begin by
writing the incomplete oxidation and reduction half-reactions:
Oxid:
Red:
 Balance everything except H and O.
Oxid:
H2C2O4  CO2
Red:
MnO4-  Mn2+
 Add H2O to balance any O
Oxid:
H2C2O4  2CO2
Red:
MnO4-  Mn2+
 Add H+ to balance any H
Oxid:
H2C2O4  2CO2
Red:
MnO4-  Mn2+ + 4H2O
 Balance charge with electrons
Oxid:
H2C2O4  2CO2 + 2H+
Red:
8H+ + MnO4-  Mn2+ + 4H2O
 Check to see if electrons agree with original predictions…_______.
 Make #e- lost = # e- gained.
(H2C2O4  2CO2 + 2 H+ + 2e-)
(5e- + 8H+ + MnO4-  Mn2+ + 4H2O)
H2C2O4 
CO2 +
e- +
H+ +
MnO4- 
 Add half reactions.
H+ +
eMn2+ + H2O
Reduce any occurrences of H+ or H2O on both sides of the equation: none here.
19-6
Balancing Redox Reactions That Occur in Basic Solution
The procedure for balancing in this case is actually the same as for balancing
in an acidic solution, with one small twist.
Now let’s pretend that this reaction occurred in basic solution for demo of
process.
 Start with the balanced reaction in acidic solution.
5H2C2O4 + 6H+ + 2MnO4-  10CO2 + 2Mn2+ + 8H2O
 Neutralize any H+ by adding OH- to both sides of the equation.
5H2C2O4 + 6H+ + 2MnO4-  10CO2 + 2Mn2+ + 8H2O
 Combine H+ and OH- to form water.
5H2C2O4 + 2MnO4-  10CO2 + 2Mn2+ + 8H2O + 6 OH Reduce any replicate water molecules.
5H2C2O4 + 2MnO4-  10CO2 + 2Mn2+ + 6 OH-
19-7
2.
Electrochemical Cells
Definition: An electrochemical cell is a system consisting of electrodes that dip into
an electrolyte in which a chemical reaction either uses or generates an electric
current.
Electrochemical cells are based on redox chemistry.
 electrochemical cells in which a spontaneous reaction generates an electric
current are known as:__________________
_____________________
A voltaic cell is named after Count Alessandro Volta, 1745-1827,
(an Italian physicist)
 Electrochemical cells in which an electric current drives an otherwise
non-spontaneous reaction are known as______________ .
Voltaic Cells
Suppose we created a cell using the reaction between zinc and copper.
The reaction is spontaneous. Zn(s) is oxidized and Cu2+(aq) is reduced.

The redox half-reactions for the above reaction would be:
Oxidation:
Reduction:

2 e- transferred
If zinc metal is put directly into the solution of copper ions:

We can’t capture the flow of electrons
(or the energy of the cell.)
If we separate the Cu(s)/Cu2+ from the Zn(s)/Zn2+?
19-8
What if we provide a path (wire) for electron flow?
However, we now have another problem ...

We start with a neutral soln. ([cations] = [anions])

In the zinc solution there is a build up of:

___________________
In the copper solution there is a build up of: __________________

This will cause the flow of electron to _______
We need a way to neutralize the charge build-up in the solutions due to the
change in soluble ion concentrations!
What if we had a tube filled with an electrolyte
that connected the two redox reactions?
This would allow:

- cations to move towards the:
_________
- anions to move towards the: __________
- keeping the charges in solution neutral

In turn, this would now allow the electrons to flow
The connecting tube of solution is called a _________________________
and is usually filled with a gel or solution of a strong electrolyte such as KCl.
Summary of the movement of ions, electrons and the redox half-reactions in a
voltaic cell:
19-9
Electrodes
The two solid metals in the different half-reactions are called: ___________
 The electrode in the oxidation half-reaction is called the: __________
 The electrode in the reduction half-reaction is called the: __________
 Electrons always flow from: ________________ .
In voltaic (aka galvanic) cells:
 The cathode attracts electrons and is labeled with a:
 The anode produces electrons and is labeled with a:
_____
_____
Inert (aka Passive) vs Active electrodes.


Inert electrodes _________ 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).
 The material of the electrode is not important as long as it does not
enter into the overall reaction.
Active electrodes chemically participate in the redox reaction
Question: In the previous example, are the Zn and Cu active or passive electrodes?
 The copper electrode in the previous example is an ___________
electrode. (The Cu could deposit on any metal!)
 The zinc electrode in the previous example is an _____________
electrode.(The electrode must be made of Zn to create Zn2+ ions!)
Shorthand Notation for Cells
It is convenient to have a shortcut way of designating particular voltaic cells.
 The cell just discussed can be written as;
 The anode (oxidation half cell) is written on the left.
 The cathode (reduction half-cell) is written on the right.
 The connecting salt bridge is denoted by two vertical bars.
 The cell terminals are at the extreme ends of the cell notation.
 A single vertical bar indicates a phase change boundary.
 It follows the path of the electron flow.
19-10
 When the half cell involves a gas, an inert
material such as platinum serves as a terminal and
an electrode surface on which the reaction occurs.
 This figure shows a hydrogen electrode
The cathode half-reaction is:
The notation for the hydrogen electrode as a
anode is: ___________________
cathode is: __________________
A complete voltaic cell notation includes concentrations of solutions and pressure
of gases written in parentheses. If not noted, concentrations are assumed to be
standard conditions of 1.0 M concentration and pressure = 1.0 atm.
i.e.
Zn(s)Zn2+(1.0M)H+(1.0M)H2(1.0 atm)Pt
LP#3. Give the overall cell reaction for the voltaic cell
Cd(s)Cd2+(1.0M)H+(1.0M)H2(1.0 atm)Pt
LP#4. Write the shorthand notation for a galvanic cell that uses the reaction:
Fe(s) + Sn2+(aq)  Fe2+(aq) + Sn(s)
Where the concentration of iron (II) is 0.5M and the concentration of tin (II) is 0.1M.
3. Electromotive Force (EMF)
What causes the electrons to flow from the anode to the cathode?
• The movement of electrons is analogous to the movement of water over a
waterfall.
– Water moves from a point of ________ potential energy
to a point of ________ potential energy.
Thus, a PE difference is required.
19-11
– The work that can be done by the water depends on
the amount of water (rate x time) and the PE difference.
• An electric charge also moves from a point of high electrical potential to
one of lower electrical potential.
– Current is the rate of flow of electrons.
– The work that can be done by an electrical charge moving through a
conductor depends on the amount of charge (e.g. # of e-)
[current x time] and the potential difference __________.
• Potential difference is the difference in electric potential (PE) between two
points.
– Measured in: _______.
– The volt, V, is the SI unit of potential difference
equivalent to 1 joule (J) of energy per coulomb (C) of charge.
– 1V = 1J
1C
OR
1 Joule = 1 Coulomb x 1 Volt
• Charge is measured in Coulombs (C)
– The Coulomb is the SI derived unit of electric charge.
– A Coulomb is the charge carried by 6.242 x 1018 electrons.
– It is a unit of electrical charge equal to the amount of charge
transferred by a current of 1 ampere (amp) in 1 second
An alternate way of saying this is that
Ampere (amp) = 6.242 x 1018 electrons = 1 Coulomb
sec
sec
Charge of 1 mole of electrons = 96,485 C =
In moving 1 mole of electrons through a circuit, the numerical value of the work
done by a voltaic cell is the product of the Faraday constant (F) times the
potential difference between the electrodes.
19-12