Redox and Electrochemistry
Oxidation:
Historically means the combination of a
substance with oxygen
2Mg + O2 à 2MgO
Oxygen is called the oxidizing agent, and
magnesium is the substance oxidized.
Reduction:
Historically, reduction is associated with the
removal of oxygen from a substance.
CuO + H2 à Cu + H2O
Hydrogen is called the reducing agent, and
copper (II) oxide is the substance that is
reduced.
Oxidation and reduction arises out of the
competition for electrons in a chemical reaction.
To know what really happens during these
processes, we need a method of assigning
electrically charges to atoms and ions.
Oxidation Numbers (Oxidation States)
An oxidation number is the charge that an atom
or ion has- or appears to have- when certain
1 rules are applied. (appears to have because
covalent compounds do not have charges; we
just treat them as if they do)
Rules:
1. The oxidation number of a free element is
zero. Free element = any element that is
not combined with another element.
Therefore, Na, He, O2, O3 all have the
oxidation number zero.
2. The oxidation number of a simple ion is its
charge. For example, the oxidation number
of Cl- is –1, the oxidation number of Al+3 is
+3.
3. The metals in Groups 1 and 2 have
oxidation numbers of +1 and +2
respectively
4. Hydrogen in combination usually has an
oxidation number of +1. An exception to
this rule are the metal hydrides (such as
NaH), in which hydrogen has the oxidation
number –1.
5. Oxygen in combination usually has an
oxidation number of –2. Exceptions to this
rule include peroxides (such as H2O2), in
which the oxidation number of oxygen is –1,
and oxygen-fluorine compounds, in which
the oxygen number is positive.
2 6. In a molecular or ionic compound, the sum
of the oxidation number totals must add to
zero since these compounds are electrically
neutral.
7. In a polyatomic ion, the sum of the
oxidation number totals must add to the
charge on the ion.
Questions
Find the oxidation number in each of the
following:
a.
S in H2SO3
b.
Cr in Na2CrO7
c.
Fe and Cl in FeCl3
d.
P in PO43e.
O in OF2
Formal Definitions of Oxidation and Reduction
2Mg0 + O02 à 2Mg2+O2Magnesium increased oxidation number from
zero to +2
Magnesium is oxidized
Oxidation = a loss of electrons
2Mg0 à 2Mg +2 + 4 e- (oxidation halfreaction)
3 Oxygen decreases oxidation number from zero
to –2
Oxygen is reduced
Reduction = a gain of electrons
O02 + 4e- à 2O2Redox reactions = reactions involving oxidation
and reduction
LEO the lion says GER
Oxidizing agent = the substance that is reduced
(causes oxidation)
Reducing agent = the substance that is oxidized
(causes reduction)
Redox Equations
Is this (unbalanced) equation:
NH3 + O2 à NO + H2O
a redox equation? To answer this question, we
rewrite the equation and include the oxidation
number for each element. Then, look for
changes in the oxidation numbers: an increase
means oxidation and a decrease means
reduction.
N3-H+3 + O02 à N2+O2- + H+2O2Nitrogen changes from –3 to +2; its oxidation
number increases, therefore it has been
oxidized.
4 Oxygen changes its oxidation number from zero
to –2; its oxidation number decreases, and
therefore it has been reduced.
Half- Reactions
Redox equations can be split into half reactions
by identifying the elements that are oxidized
and reduced.
NH3 + O2 à NO + H2O
N3-H+3 + O02 à N2+O2- + H+2O2N3- à N2+ + 5e- (oxidation half reaction)
O0 + 2 e- à O2- (reduction half reaction)*
unbalanced*
Question:
Indicate whether each of the following equations
is a redox equation. If it is a redox equation,
write the half-reactions representing oxidation
and reduction.
a. Fe + 2HCl à FeCl2 + H2
b. HNO3 + I2 à HIO3 + NO2 + H2O
(unbalanced)
c. HCl + NaOH à H2O + NaCl
Balancing Redox Equations by the Half-Reaction
Method
Redox equations may be very difficult to
balance simply by inspection.
5 Cu + HNO3 à Cu(NO3)2 + NO + H2O
Redox equations can be balanced, however, by
identifying the elements that are oxidized and
reduced and then writing their half reactions.
Steps:
1. Rewrite the equation with its oxidation
numbers
Cu0 + H+N5+O32- à Cu2+(N5+O32-)2 + N2+O2+ H2+O22. Identify the elements that are oxidized
and reduced.
Cu is oxidized
N is reduced
3. Write the oxidation and reduction half
reactions
Cu0 à Cu2+ + 2eN5+ + 3e- à N2+
4. Balance the half reactions. (This step is
necessary because electric charge must be
conserved.) Multiply each half reaction so
that the number of electron lost by the
oxidized element is equal to the number of
electrons gained by the reduced element.
3 (Cu0 à Cu2+ + 2e-) = 3Cu0 à 3Cu2+ +
6e2 (N5+ + 3e- à N2+) = 2N5+ + 6e- à
2N2+
6 5. Add the two balanced half reactions
together, eliminating the electrons. The
result is the balanced skeleton redox
equation.
3Cu0 + 2N5+ à 3Cu2+ 2N2+
6. Insert the coefficients from the skeleton
back into the original equation by matching
each element and its oxidation number.
There is one exception: Do not insert the
coefficient of any item that appears in more
than one place in the equation.
3Cu0 + _H+N5+O2-3 à 3Cu2+(N5+O2-3)2 +
2N2+O2- + _H+2O2since N5+ appears twice, do not insert the
coefficient for it.
3Cu + HNO3 à 3Cu(NO3)2 + 2NO +H2O
7. Balance the rest of the equation (the
nonredox part) by inspection.
3Cu + 8HNO3 à 3Cu(NO3)2 + 2NO
+4H2O
Question:
Balance the following redox equations by the
half reaction method:
a. Cu + HNO3 à Cu(NO3)2 + NO2 + H2O
b. HNO3 + I2 à HIO3 + NO2 + H2O
7 Steps for Balancing Redox Equations in Acidic
Solution
Since the solution is acidic, H+ and H2O can
be added either to the reactants or the
products to balance hydrogen or oxygen
1.
Divide the equation into two complete
half reactions, one for oxidation and the
other for reduction
2.
Balance each half reaction.
a.First, balance the elements other than
H and O.
b.
Next, balance the O atoms by
adding H2O
c.Then, balance the H atoms by adding H+
d.
Finally, balance the charge by
adding e- to the side with the greater
overall positive charge.
3.
Multiply each half reaction by an
integer so that the number of electrons lost
in one half reaction equals the number
gained in the other.
4.
Add the two half reactions and simplify
where possible by canceling species
appearing on both sides of the equation
5.
Check the equation to make sure that
there are the same number of atoms of
8 each kind and the same total charge on both
sides.
Questions:
Complete and balance the following oxidationreduction equations using the method of half
reactions. Both reactions occur in acidic
solution
a. Cu(s) + NO3-(aq) à Cu2+(aq) + NO2(g)
b. Mn2+(aq) + NaBiO3(s) à Bi3+(aq) +
MnO4-(aq)
Balancing Equations for Reactions Occurring in
Basic Solution
OH- and H2O are used instead of H+ and H2O
The half reactions can be balanced initially as if
they occurred in acidic solution. The H+ ions
can then be “neutralized” by adding an equal
number of OH- ions to both sides of the
equation.
Question:
Complete and balance the following equations for
oxidation reduction reactions that occur in basic
solution:
a. NO2-(aq) + Al(s) à NH3(aq) +
Al(OH)4(aq)
9 b. Cr(OH)3(s) + ClO-(aq) à CrO42-(aq) +
Cl2(g)
Reduction Potentials
Every half reaction has a potential, or voltage,
associated with it. A table of half reactions
can be used to determine the tendency for a
substance to be oxidized or reduced.
Insert table of Reduction Potentials AP review
The table provides reduction potentials. To
determine oxidation potentials, reverse the
reaction and change the sign on the voltage
given.
Reduction potential for Li+
Li+ + e- à Li
Eo = -3.05 V not a
favorable reaction
10 Oxidation potential for Li
Lià Li+ + e- Eo = 3.05 V favorable reaction
Li has a large oxidation potential, making it
likely to lose electrons. Li is a very strong
reducing agent.
Reduction potentials can be used to determine
the potential of a redox reaction. If the
potential for the redox reaction is positive, the
reaction is favored and will be spontaneous.
· Add the potential for the oxidation half
reaction to the potential for the reduction
half reaction
· Never multiply the potential for a half
reaction by a coefficient
E0cell = Eoxidation + Ereduction
Is the following reaction a spontaneous
reaction?
Sn(s) + 2Ag+ à Sn2+ + 2Ag(s)
Oxidation: Sn à Sn2+ + 2e- Eo = 0.14 V
Reduction: Ag+ + e- à Ag E0 = 0.80 V
E0cell = Eoxidation + Ereduction
Eocell = 0.14 + 0.80 = 0.94 V
Notice, the coefficient for silver does not enter
into the equation for the voltage for the cell.
11 Question:
Calculate E0 for the redox reaction:
Fe(s) + Pb2+ (aq) à Fe2+(aq) + Pb(s)
Is the reaction spontaneous?
Galvanic Cells (Voltaic Cells)
In every example discussed thus far, there
has been a transfer of electrons between the
substance that was oxidized and the substance
that was reduced. If we could direct these
electrons through an external wire, we would
have the source of an electric current.
An electrochemical cell uses a spontaneous
redox reaction to provide a source of electrical
energy. It is designed so that the oxidation
and reduction half reactions occur in separate
half-cells that are connected to one another.
Insert cell diagram in Let‛s Review
Chemistry
12 Oxidation occurs in the left half-cell. The
electrons that are released travel through the
external wire and enter the Cu(s). Reduction
occurs in the right half-cell. The metal strips
are called electrodes. The electrode that
participates in oxidation is called the anode; the
electrode that participates in reduction, the
cathode. The direction of the electron flow in
an electrochemical cell is always from the anode
(where oxidation occurs) to the cathode (where
reduction occurs).
AN OX
RED CAT
As the reaction progresses, there would be
an electrical imbalance in the half-cells that
would stop the electrochemical cell from
functioning. However, the porous barrier (salt
bridge*) allows the migration of ions between
the half-cells, keeping them electrically neutral.
If we connect the external wire to an
external circuit, we will have a usable source of
electrical energy; we have created a battery!
The anode of the cell is the negative terminal
because electrons are flowing out of it; the
cathode is the positive terminal because
electrons are flowing into it.
13 After a period of time, the redox reaction
of the cell reaches equilibrium and the cell no
longer operates. The battery is dead!
*Salt Bridges
We can build an electrochemical cell whose
half cells are completely separated, but we
must provide a device known as a salt bridge to
allow for the flow of ions between the halfcells. As shown in the accompanying diagram,
the salt bridge contains an electrolyte, such as
KCl, which is dispersed throughout a gel, such
as agar. The gel provides firmness but allows
ions to flow through it.
Insert salt bridge diagram Let‛s Review
Chemistry
14 Since Zn2+(aq) is produced in the left halfcell, negative ions are required to keep the
solution electrically neutral. Therefore, Cl- ions
flow into this half-cell. At the same time
Cu2+(aq) is being used up in the right half-cell,
which requires positive ions to keep the solution
electrically neutral. Therefore, K+ ions flow
into this half-cell.
Nerst Equation
Under standard conditions (STP and all
concentrations are 1 M) the voltage of the cell
is the same as the total voltage of the redox
reaction. Under nonstandard conditions, the
cell voltage can be computed using the Nerst
equation.
Ecell = E0cell – - RT ln Q nF Ecell = cell potential under nonstandard
conditions (V)
E0cell = cell potential under standard
conditions (V)
R = the gas constant, 8.31 (volt-coulomb) /
(mol-K)
T = absolute temperature (K)
n = the number of moles of electrons
exchanged in the reaction (mol)
15 F = Faraday‛s constant, 96,500 coulombs /
mole
Q = the reaction quotient (same as the
equilibrium expression, but with initial
concentrations instead of equilibrium
concentrations)
At 25oC, the Nerst equation reduces to:
Ecell = E0cell – 0.0592 log Q
n
As the concentration of the products of a
redox reaction increases, the voltage decreases;
and as the concentration of the reactants in a
redox reaction increases, the voltage increases.
Electrolytic Cells and Electrolysis
Calculate E0 for the redox reaction involving
molten NaCl (l)
2NaCl à 2Na(s) + Cl2(g)
Half reactions:
2Na+ + 2e- à 2Na(s) (Na+ is reduced)
2Cl- à Cl2(g) + 2e- (Cl- is oxidized)
E0 = ECl + ENa
Then,
= -1.36V + -2.71V =
-4.07V
16 The negative sign indicates the reaction is not
spontaneous
It would be useful if we could force this
reaction to occur: we could generate two
elements that do not occur freely in nature. To
accomplish this purpose, we construct a special
type of cell.
Electrolytic cells are used to force
nonspontaneous redox reactions to occur. Unlike
galvanic or voltaic cells, they do not generate
electrical energy—they use it. In practice, an
electric current is passed through the substance
inside the cell. The process is called
electrolysis.
Insert operation of an electrolytic cell diagram
Let‛s Review Chemistry
Electroplating
In this process an electric current is used to
deposit a layer of metal, such as silver, on the
17 object to be plated. The accompanying diagram
illustrates how electroplating is accomplished.
Insert Electroplating diagram Let‛s Review
Chemistry
The plating solution is a salt of the metal. A
bar of metal is made the anode (positive
electrode) and provides positive ions needed for
plating. The material to be plated is made the
cathode (negative electrode) and receives the
positive ions which it reduces to the metallic
layer of electroplate.
18 Additional Application of Redox and
Electrochemistry
Reduction of Metals
Most metals do not occur freely in nature,
but occur in oxidized state (i.e. as positive
ions). A metal that belongs to Group 1 or 2 is
recovered by electrolysis of its fused salt.
Other metals are recovered by reduction of
their ores; the method used depends on the
activity of the metal and the nature of the ore.
In the production of chromium metal,
chromium (III) oxide is fairly stable and is
treated with aluminum metal, which is a
relatively strong reducing agent:
2Al + Cr2O3 à Al2O3 + 2Cr
Metals such as zinc and iron are extracted
by the reduction of their oxides. Carbon, in
the form of coke, or carbon monoxide is used as
the reducing agent.
ZnO + C + Heat energy à Zn + CO
Fe2O3 + 3CO + Heat energy à 2Fe +
3CO2
19 Preventing the Corrosion of Metals
Corrosion occurs when a metal is attacked
slowly by elements in its environment. In many
cases, the metal ceases to be useful. Corrosion
is a redox reaction, and agents such as moisture
may contribute to the process.
Metals such as aluminum and zinc oxidize
readily, but the oxide adheres to the metal‛s
surface tightly producing a protective coating.
Some metals, such as iron, form oxides that
flake off the metal face allowing oxidation of
the remaining metal surface to occur. Iron can
be protected by plating it with a corrosionresistant metal such as chromium or nickel or
coating it with paint, oil, or porcelain.
Another protection technique is to coat the
iron with zinc-a process known as galvanizing.
If the coating is broken, exposing fresh iron
and zinc, the more active zinc will be oxidized
first and will produce a protective oxide coating.
Commercial Batteries
Lead Acid Battery
Car and boat batteries
The positive electrode (cathode) is PbO2, the
negative electrode (anode), Pb. A solution of
sulfuric acid is the electrolyte for the redox
20 reaction- a reaction that lead changes its
oxidation state.
Pb + PbO2 + 2H2SO4 CHARGEàDISCHARGEß 2PbSO4
+ 2H2O
Since the redox reaction is reversible, the
battery may be recharged for further use.
Nickel Oxide-Cadmium Battery
NiO2 is the positive electrode, Cd is the
negative electrode. A solution of KOH is the
electrolyte, and its concentration does not
change.
NiO2 + Cd + 2H2O àchargeßdischarge Ni(OH)2 +
Cd(OH)2
21