Unit 14
Oxidation States
Oxidation versus
Reduction
 Redox reactions


• Identifying
• Half Reactions
• Balancing

Voltaic (Galvanic)
Cells
• Cell potential
• Standard reduction
potentials
• Standard hydrogen
electrode


Strength of Agents
Nernst Equation
• Calculations
• Equilibrium
• Gibbs Free Energy

Electrolytic cells
• Electroplating
• Stoichiometry
 Actual
charge of a monatomic ion
 Rules for assigning oxidation numbers
1.

2.



For an atom in its elemental form (uncombined
with other elements) the oxidation number is
always 0.
Example: H2 = 0
For any monatomic ion, the oxidation number
equals the charge of the ion.
Alkali metals always = +1
Alkaline earth metals always = +2
Aluminum always = +3
3. Nonmetals usually have negative oxidation
numbers (although they can be positive)
Oxygen is usually -2 (exception: peroxide)
Hydrogen is usually +1 (-1 when bonded to metals)
Fluorine is -1.





Other halogens have oxidation number of -1 in binary
compounds.
In oxyanions, halogens usually have positive oxidation states.
4. The sum of oxidation numbers in neutral
compounds is zero. The sum of oxidation
numbers in polyatomic ions equals the
charge of the ion.
 What
is the oxidation state of S?
• H2S = -2
• SCl2 = +2
• S8 = 0
• Na2SO3 = +4
• SO4-2 = +6
 Oxidation
– A process in which an element
attains a more positive oxidation state (loses
electrons)
Na(s)  Na+ + e-
 Reduction
– A process in which an element
attains a more negative oxidation state
(gains electrons)
Cl2 + 2e-  2Cl-
LEO says GER
Lose Electrons = Oxidation
Gain Electrons = Reduction
 OR…
 Redox
Reaction: reaction in which
oxidation and reduction occur
• Oxidation and reduction occur spontaneously
Mg
+
S→
Mg2+
+
S2-
(MgS)
The magnesium atom (which has zero charge) changes to a magnesium
ion by losing 2 electrons, and is oxidized to Mg2+
The sulfur atom (which has no charge) is changed to a sulfide ion by
gaining 2 electrons, and is reduced to S2-
0
1
0
1
2 Na  Cl 2  2 Na Cl
Each sodium atom loses one electron (oxidation):
1
0
Na  Na  e

Each chlorine atom gains one electron (reduction):
0

1
Cl  e  Cl
 Active
metals:
Lose electrons easily
Are easily oxidized
Are strong reducing agents

 Active
nonmetals:
Gain electrons easily
Are easily reduced
Are strong oxidizing agents
A
common approach for listing species that
undergo REDOX is as half-reactions.
For 2Fe3+ + Zno(s) = 2Fe2+ + Zn2+
Fe3+ + e-  Fe2+
(reduction)
Zno(s)
 Zn2+ + 2e- (oxidation)
 They
are individual equations showing just
the oxidation or just the reduction.
 The
amount of each element must be the
same on both sides (like normal)
 BUT
the gain/loss of electrons must also
be balanced
• If an element loses a certain amount of electrons,
another element must gain the same amount
#1
Divide the equation into two half
reactions
#2
Balance each half reaction
 Balance all elements other than H and O
 Balance the O atoms by adding H2O
 Balance the H atoms by adding H+
 Balance the charge by adding e-
#3
Multiply the half reactions by integers
so that the electrons lost in one halfreaction equals electrons gained in
the other reaction
#4
Add the two half-reactions together
 Simplify by cancelling out species
appearing on both sides of the equation
Balancing Redox Reactions
Balance the following reaction in acidic solution.
Fe2+ + Cr2O72-
Fe3+ + Cr3+
1. Separate the equation into two half-reactions.
+2
Oxidation:
Reduction:
Fe2+
+6
Cr2O7
2-
+3
Fe3+
+3
Cr3+
2. Balance the atoms other than O and H in each halfreaction.
Cr2O722Cr3+
19.1
Balancing Redox Reactions
3. For reactions in acid, add H2O to balance O atoms and H+
to balance H atoms.
Cr2O7214H+ + Cr2O72-
2Cr3+ + 7H2O
2Cr3+ + 7H2O
4. Add electrons to one side of each half-reaction to balance
the charges on the half-reaction.
Fe2+
6e- + 14H+ + Cr2O72-
Fe3+ + 1e2Cr3+ + 7H2O
5. If necessary, equalize the number of electrons in the two
half-reactions by multiplying the half-reactions by
appropriate coefficients.
6Fe2+
6Fe3+ + 6e6e- + 14H+ + Cr2O72-
2Cr3+ + 7H2O
19.1
7. Add the two half-reactions together and balance the
final equation by inspection. The number of electrons
on both sides must cancel.
Oxidation:
6Fe2+
Reduction: 6e- + 14H+ + Cr2O7214H+ + Cr2O72- + 6Fe2+
6Fe3+ + 6e2Cr3+ + 7H2O
6Fe3+ + 2Cr3+ + 7H2O
19.1
 Cu
+ NO3-  Cu+2 + NO2
Answer:
Cu + 4H+ + 2NO3-  Cu+2 + 2NO2 + 2H2O
 Cr2O7-2
+ Cl-  Cr+3 + Cl2
Answer:
14H+ + Cr2O7-2 + 6Cl-  2Cr+3 + 7H2O + 3Cl2
 If
redox reaction occurs in “basic
solution”…
• Follow all steps as usual
• Before half-reactions are combined, add 1 OH- to
BOTH sides of the equation for every 1 H+ in the
equation
• Combine H+ and OH+ to form H2O when they
appear on the same side of the equation
 The
following reactions occur in basic solutions
 CN-
+ MnO4-  CNO- + MnO2
Answer:
3CN- + H2O + 2MnO4-  3CNO- + 2MnO2 + 2OH NO2-
+ Al  NH3 + Al(OH)4-
Answer:
NO2- + 2Al + 5H2O + OH-  NH3 + 2Al(OH)4-
 There
are two general ways to conduct an
oxidation-reduction reaction
#1 Mixing oxidant and reductant together
Cu2+ + Zn(s)
This approach does
not allow for
control of the reaction.
Cu(s) + Zn2+
 Electrochemical cells
• Each half reaction is put in a separate ‘half cell.’ They
can then be connected electrically.
• Called a “voltaic” or “galvanic” cell
• This permits better control over the system.
Cu2+ + Zn(s)
eZn
Zn2+
Cu(s) + Zn2+
eCu
Cu2+
Electrons are
transferred from
one half-cell to
the other using
an external metal
conductor.
 Electrodes
• Two solid metals that are connected by the
external circuit
Anode
• The electrode at which oxidation occurs
Cathode
• The electrode at which reduction occurs
Just remember the…
Red Cat
“Reduction occurs
at the cathode”
 Oxidation
occurs at the anode
 Reduction occurs at the cathode
 This
means that
electrons flow from
the anode to the
cathode through the
external circuit
e-
e-
To complete the
circuit, a salt
bridge is used.
salt bridge
 Salt bridge
• Allows ions to migrate without mixing
electrolytes (keeps solutions neutral).
• It can be a simple porous disk or a gel saturated
with a non-interfering, strong electrolyte like
KCl.
Cl-
Cl- is released
to Zn side as Zn
is converted to Zn+2
KCl
K+
K+ is released
as Cu+2 is
converted to Cu
 Salt
bridge
• Consists of a U-shaped tube that contains the
electrolyte solution (ex. NaNO3)
 Usually solution is a paste or gel
• Ions of electrolyte solution will not react with other
ions in the cell or with the electrode materials
 Ions migrate to neutralize the cells
 Anions migrate toward the anode
 Cations migrate toward the cathode

Rather than drawing an entire cell, a type of
shorthand can be used.
• For our copper - zinc cell, it would be:
Zn | Zn2+ (1M) || Cu2+ (1M) | Cu

The anode is always on the left.
| = boundaries between phases
|| = salt bridge

Other conditions like concentration are listed just
after each species.
 Electromotive
Force (EMF): Driving force
that pushes electrons through the external
circuit
• Measures how willing a species is to gain or lose
electrons
• Also referred to as “cell potential” or “cell voltage”
• Denoted
Ecell
• Spontaneous cell reaction = positive (+) Ecell
• Measured in volts (1 V = 1Joule/Coulomb)
 EMF
under standard state conditions (Eocell)
• All soluble species are at 1 M
• Slightly soluble species must be at saturation.
• Any gas is constantly introduced at 1 atm
• Any metal must be in electrical contact
• Other solids must also be present and in contact.
 Eocell
is determined from the reduction
potentials of the two half reactions
Eocell = Eored(cathode) – Eored(anode)
 Reduction
potentials can be looked up for
all half reactions (table given on AP test)
Half reaction
Eo, V
F2 (g) + 2H+ + e-
2HF (aq)
3.053
Ce4+ + e-
Ce3+ (in 1M HCl)
1.28
O2 (g) + 4H+ + 4e-
2H2O (l)
1.229
Ag+ + e-
Ag (s)
0.7991
2H+ + 2e-
H2 (g)
0.000
Fe2+ + 2e-
Fe (s)
-0.44
Zn2+ + 2e-
Zn (s)
-0.763
Al3+ + 3e-
Al (s)
-1.676
Li+ + e-
Li (s)
-3.040
 The
table of standard reduction
potentials relates to the activity series of
metals
• Flip the table upside down and only look at the
metals (solids), and it is the activity series list
• The AP test does not give you an activity series
list so you have to use the standard reduction
potentials to predict products of reactions

E0 is for the reaction as written

The more positive E0 the greater the
tendency for the substance to be reduced

The half-cell reactions are reversible
•

The sign of E0 changes when the reaction is
reversed (do this if you need the “oxidation
potential” of an element)
Changing the stoichiometric coefficients of
a half-cell reaction does not change the
value of E0
 Reversing
the sign
 For the reduction of Zn+2 + 2e- 
• The reduction potential of Zn is -0.76
Zn Ered = -0.76
 If
we are looking at the reverse reaction and
want the oxidizing potential, the sign is reversed
• Zn  of
 We
Zn+2 + 2e-
Eox = +0.76
will do redox calculations using only
“reduction potentials”
 Standard Hydrogen Electrode (SHE)
• We can’t measure the standard reduction
potential of a half reaction directly
• SHE used as a reference point
 Reduction potential = 0 V
 Consists of…
• Platinum wire connected to piece of platinum foil
• Electrode encased in a glass tube so that
hydrogen at standard conditions can bubble
over the platinum

To solve for the Ecell, subtract the reduction potential for
the half-reaction at the cathode from the reduction
potential for the half-reaction at the anode.
Eocell = Eocathode – Eoanode

What is the standard emf of an electrochemical cell
made of a Cd electrode in a 1.0 M Cd(NO3)2 solution
and a Cr electrode in a 1.0 M Cr(NO3)3 solution?
2Cr (s) + 3Cd2+ (1 M)
Cd2+ (aq) + 2eCr3+ (aq) + 3e-
3Cd (s) + 2Cr3+ (1 M)
Cd (s) E0 = -0.40 V
Cr (s) E0 = -0.74 V
E0cel l = E0 cathode - E0 anode
E0cell = -0.40 – (-0.74)
E0cell = 0.34 V

Given that the standard reduction potential of Zn+2 to
Zn(s) is -0.76 V, calculate the E0red for the reduction of
Cu+2 to Cu(s).
Zn(s) + Cu2+ (aq,1 M)
Cu(s) + Zn+2(aq,1 M)
E0cel l = E0 cathode - E0 anode
1.10 = E0 cathode - (-0.76)
E0cathode = 0.34 V
E0cell = 1.10 V
 The
more positive the value of E0red, the
greater the driving force for reduction
under standard conditions
• Meaning…the reaction at the cathode has a more
positive value of E0red than the reaction at the
anode

A voltaic cell is based on the following two standard
half reactions. Determine which half-reaction occurs at
the cathode and which occurs at the anode as well as
the standard cell potential.
Cd2+ (aq) + 2eSn2+ (aq) + 2e-
Cd (s)
Sn (s)
E0red of (Cd+2/Cd) is -0.403 and E0red of (Sn+2/Sn) is -0.136
Reaction of Sn+2/Sn is more positive so it occurs at the cathode

Cd2+ (aq) + 2e-
Cd (s)
Sn2+ (aq) + 2e-
Sn (s)
Cathode reaction is reduction/Anode reaction is oxidation
Cathode: Sn2+ (aq) + 2eAnode:
Cd (s)
Sn (s)
Cd2+ (aq) + 2e-
E0red = -0.136
E0red = -0.403
E0cel l = E0 cathode - E0 anode
E0cell = -0.136 – (-0.403)
E0cell = 0.267 V


The more readily an ion is reduced (the more
positive its E0red value), the stronger it is as an
oxidizing agent.
The half-reaction with the smallest reduction
potential (most negative E0red) is most easily
reversed as an oxidation
• Smallest reduction potential is strongest reducing agent

The stronger the oxidizing agent, the weaker it
acts as a reducing agent (and vice versa)
E
versus E0
• E means not at standard conditions
E
versus Ecell
• E doesn’t have to occur in a voltaic cell so there
is no cathode and anode
• E = Ereduction - Eoxidation

Standard potentials assume a concentration of 1 M. The
Nernst equation allows us to calculate potential when
the two cells are not 1.0 M.
RT
EE 
ln(Q)
nF
0
R = 8.31 J/(molK)
T = Temperature in K
n = moles of electrons transferred in balanced redox equation
F = Faraday constant = 96,500 coulombs/mol eQ
(Charge of one mole of electrons)
= reaction quotient
room temperature (25 C or 298 K) the
Nernst equation can be simplified
 At
0.0591
EE 
log(Q )
n
0


As the concentration of products increases in a
redox reaction, voltage decreases
As the concentration of reactants in a redox
equation increases, voltage increases
 Determine
the potential of a Pt indicator
electrode if dipped in a solution
containing 0.1M Sn4+ and 0.01M Sn2+.
Sn4+ + 2e-
Sn2+
Eo = 0.15V
0.0592
0.01 M
E = 0.15V - 2
log 0.1M
= 0.18 V
 At
equilibrium, the forward and
reverse reactions are equal
• The cell potential is 0 V
• The battery is “dead”
 Modifying
the Nernst equation for
equilibrium (at 25°C)gives us…
0.0591
0 volts  E 
log( K )
n
0
 Calculate
the equilibrium constant, K, for the
following reaction.
Zn + Cu2+  Zn2+ + Cu
E0 = + 1.10 V
0.0591
0 volts 1.10 
log( K )
2
(1.10)(2)
 log( K )
0.0591
37.2  log( K )
1037.2  K 1.58 x1037
 Any
reaction that can occur in a voltaic
cell must be spontaneous
• Not all redox reactions are spontaneous
A
positive value of E indicates a
spontaneous process
• A negative value means it is not spontaneous
 Both
∆G and E can be used to tell
whether or not a reaction is spontaneous
• Negative ∆G or positive E means spontaneous
 The
relationship between the two is
0
G
=
0
-nFE
 Calculate
∆G for the following reaction:
Zn + Cu2+  Zn2+ + Cu
E0 = + 1.10 V
coulombs
Joules
G   (2 mol e )(96 485
)(1.10
)

mol e
Coulomb
0

G0   212267 Joules   212 kJ
 As
reactants are converted to products, Q
increases and E decreases until E = 0
 Using
the Nernst equation, when E = 0, G = 0
• When G = 0, the system is at equilibrium
• No net reaction is occurring

Earlier, we explained that G and the equilibrium
constant can be related. Since Ecell is also related
to K, we know the following.
Q
E
Forward change, spontaneous
<K
G
-
At equilibrium
=K
0
0
Reverse change, spontaneous
>K
+
-
+
 Portable, self-contained
power source that
contains one or more voltaic cells
• Greater voltages are
achieved by connecting
multiple voltaic cells
• Cathode is labeled with a
(+) sign
• Anode is labeled with a (-)
sign
• Life of battery depends on
quantities of substances
packed in the battery to
oxidize and reduce at the
anode and cathode

An outside source is used to force a nonspontaneous
redox reaction to take place
• Two electrodes are in a molten salt or a solution
• Outside source example is a battery

E0cell is negative and ∆G is positive (not spontaneous)
Example: Electrolysis of Water
In acidic solution
Anode rxn:
Cathode rxn:
2H2O O2  4H   4e -1.23 V
4H2O  4e  2H2  4OH 
2H2O  2H2  O2
-0.83 V
-2.06 V
 Uses
electrolysis to deposit a thin layer of
one metal onto another metal to improve
beauty or to resist corrosion
 We
will sometimes be asked find how
much metal “plates out” when
electroplating occurs.
#1
Given the current used (amps) and time, calculate
the charge (coulombs)
q
I=
t
#2
I = current (amperes, A)
q = charge (coulombs, C)
t = time (sec)
Solve for moles of electrons involved using
Faraday’s constant of 96,500 C/mol
___ Coulombs
___ Mole of e-
=
96,500 C
1 mole e-
#3
Identify the balanced half reaction that you are
working with (with electrons transferred)
Example:
#4
Au+3 + 3e-  Au(s)
Use stoichiometry to find moles of metal, then
grams of metal
e- transferred  moles of metal  grams of metal
How much Ca will be produced in an electrolytic cell of
molten CaCl2 if a current of 0.452 A is passed through
the cell for 1.5 hours?
Ca2+ (l) + 2Cl- (l)
Ca (s) + Cl2 (g)
Step 1: determine the charge, C
q
I=
t
q
0.452 =
5400 sec
q = 2440.8 C
How much Ca will be produced in an electrolytic cell of
molten CaCl2 if a current of 0.452 A is passed through
the cell for 1.5 hours?
Ca2+ (l) + 2Cl- (l)
Ca (s) + Cl2 (g)
Step 2: determine the moles of e- involved
2440.8 Coulombs
___ Mole of e-
=
96,500 C
1 mole e-
Mole of e- = 0.026
How much Ca will be produced in an electrolytic cell of
molten CaCl2 if a current of 0.452 A is passed through
the cell for 1.5 hours?
Ca2+ (l) + 2Cl- (l)
Ca (s) + Cl2 (g)
Step 3: Determine the balanced half reaction
Cathode: Ca2+ (l) + 2e-
Ca (s)
How much Ca will be produced in an electrolytic cell of
molten CaCl2 if a current of 0.452 A is passed through
the cell for 1.5 hours?
Ca2+ (l) + 2Cl- (l)
Ca (s) + Cl2 (g)
Cathode: Ca2+ (l) + 2eCa (s)
Step 4: Stoichiometry time
1 mol Ca(s)
40.08 g Ca(s)
0.026 mol e- =
x
= 0.52 g Ca(s)
2 mol e1 mol Ca
0.52 grams of Ca will “plate out” or be produced