326
Electrochemistry
327
Test Content from AP Chemistry Course Description
III. Reactions (35-40%)
A. Reaction types
3. Oxidation-reduction reactions
a. Oxidation number
b. The role of the electron in oxidation-reduction
c. Electrochemistry: electrolytic and galvanic cells; Faraday's laws; standard
half-cell potentials; Nernst equation; prediction of the direction of redox
reactions
E. Thermodynamics
4. Relationship of change in free energy to equilibrium constants and electrode
potentials
Types of Calculations found on Exam adapted from AP Chemistry Course
Description
Summary of types of problems either explicitly or implicitly included in the preceding
material.
7. Faraday's laws of electrolysis
9. Standard electrode potentials and their use; Nernst equation
10. Thermodynamic and thermochemical calculations
OXIDATION NUMBERS
Oxidation number is the charge of an atom in a molecule if all the bonding is
considered ionic.
Oxidation number is different from formal charge.
Using oxidation numbers gives us another way to account for electrons in chemical
changes.
Rules of assigning oxidation numbers
1. The oxidation number of an atom as an element is always zero.
2. The oxidation number of an atom as an ion is the same as the charge of the ion.
3. In polyatomic compounds, consider the oxidation number of the most
electronegative element to be its “normal” ionic charge.
4. The sum of all the oxidation numbers of the atoms equals the charge of the
compound.
328
DEFINITIONS
Oxidation – process of losing electrons
Reduction – process of gaining electrons
Oxidizing agent – substance that causes oxidation in another substance
Reducing agent – substance that causes reduction in another substance
**Source of confusion**
Oxidizing agent is reduced.
Reducing agent is oxidized.
EXCLUSION STATEMENT Essential Knowledge 3.B.3
Language of reducing agent and oxidizing agent is beyond the scope of this
course and the AP Exam.
0
+1 -1
+2 -1
0
Consider 2 Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
Zn metal is losing e-. (oxidation)
Zn (s)  Zn2+ (aq) + 2 eH+ ions are gaining e-. (reduction)
2 H+ (aq) + 2 e-  H2 (g)
Zn is the reducing agent. (It causes H+ to reduce.)
H+ is the oxidizing agent. (It causes Zn to oxidize)
HALF-REACTIONS
Reduction-Oxidation (Redox) reactions can be broken into two parts: an oxidation
reaction and a reduction reaction.
These two parts are called half-reactions.
0
+2
+2
0
Consider Zn (s) + CuSO4 (aq)  ZnSO4 (aq) + Cu (s)
Oxidation: Zn (s)  Zn2+ (aq) + 2 eReduction: Cu2+ (aq) + 2 e-  Cu (s)
Half-reactions must occur in pairs
- We can never have only oxidation or only reduction.
- e- lost in oxidation must be gained through reduction.
# of e- lost = # of e- gained
329
Learning objective 3.8 The student is able to identify redox reactions and justify
the identification in terms of electron transfer. (See SP 6.1; Essential
Knowledge 3.B.3)
Learning objective 3.9 The student is able to design and/or interpret the result
of an experiment involving a redox titration (See 4.2, 5.1; Essential Knowledge
3.B.3)
BALANCING REDOX EQUATIONS
1. Divide equation into half-reactions
2. Balance each half-reaction
a.) Balance atoms other than H and O
b.) Balance oxygen atoms by adding water to the appropriate side
c.) Balance hydrogen atoms by adding H+ to appropriate side
i.) If reaction occurs in basic conditions, “neutralize” H+ by adding OH- to both
sides and canceling extra water molecules.
d.) Balance charge by adding e- to appropriate side
3. Multiply each equation by integer so that
# of e- lost = # of e- gained
4. Add equations together; cancel spectator species
Example: IO3- (aq) + Cr3+ (aq)  I2 (s) + Cr2O72- (aq) in acidic conditions
Balance: IO3- (aq)  I2 (s)
a) 2 IO3- (aq)  I2 (s)
b) 2 IO3- (aq)  I2 (s) + 6 H2O (l)
c) 2 IO3- (aq) + 12 H+ (aq)  I2 (s) + 6 H2O (l)
d) 2 IO3- (aq) + 12 H+ (aq) + 10 e-  I2 (s) + 6 H2O (l)
Balance: Cr3+ (aq)  Cr2O72- (aq)
a) 2 Cr3+ (aq)  Cr2O72- (aq)
b) 2 Cr3+ (aq) + 7 H2O (l)  Cr2O72- (aq)
c) 2 Cr3+ (aq) + 7 H2O (l)  Cr2O72- (aq) + 14 H+ (aq)
d) 2 Cr3+ (aq) + 7 H2O (l)  Cr2O72- (aq) + 14 H+ (aq) + 6 e3{2 IO3- (aq) + 12 H+ (aq) + 10 e-  I2 (s) + 6 H2O (l)}
5{2 Cr3+ (aq) + 7 H2O (l)  Cr2O72- (aq) + 14 H+ (aq) + 6 e-}
6 IO3- + 36 H+ + 30 e- + 10 Cr3+ + 35 H2O  3 I2 + 18 H2O + 5 Cr2O72- + 70 H+ + 30 e6 IO3- + 10 Cr3+ + 17 H2O  3 I2 + 5 Cr2O72- + 34 H+
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BALANCING REDOX EQUATIONS (alternative)
1. Assign oxidation numbers
2. Divide equation into half-reactions
3. Balance each half-reaction
a.) Balance atoms other than H and O
b.) Add electrons to balance electrons based oxidation numbers
4. In acidic solution
a.) Balance H atoms by adding H+
b.) Balance O atoms by H2O
5. In basic solution
a.) Balance O atoms by adding H2O
b.) Balance charge by adding OH6. Multiply each equation by integer so that
# of e- lost = # of e- gained
7. Add equations together; cancel spectator species
Learning objective 3.12 The student can make qualitative or quantitative
predictions about galvanic or electrolytic reactions based on half-cell reactions
and potentials and/or Faraday’s laws. (See SP 2.2, 2.3, 6.4; Essential
Knowledge 3.C.3)
Learning objective 3.13 The student can analyze data regarding galvanic or
electrolytic cells to identify properties of the underlying redox reactions. (See
SP 5.1; Essential Knowledge 3.C.3)
VOLTAIC CELLS
Voltaic cells are batteries.
In 1800 Alessandro Volta discovered that stacking plates of metal between layers of
wet blotter paper created electricity. (Voltaic ‘pile’)
Electrons spontaneously flow from reducing agent to oxidizing agent.
- Batteries exploit spontaneous redox reaction!
Consider reaction in a nickel/cadmium battery
Cd (s) + Ni2+ (aq)  Cd2+ (aq) + Ni (s)
Cd is being oxidized as Ni2+ is being reduced.
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Schematic of a Voltaic Cell
flow of e-
anode
cathode
-
+
K+
Cl-
K+
Cl-
Ni
Cd
Cl-
K+
Cd
Cd
2+
salt bridge
Cd2+
Cd2+
Cd (s)  Cd2+ (aq)
Ni2+
Ni2+
Ni
Ni2+
Ni2+ (aq)  Ni (s)
Definitions
Electrode where oxidation occurs is anode.
Electrode where reduction occurs is cathode.
Note: Oxidation  Anode
(both vowels)
Reduction  Cathode (both consonants)
At the anode, Cd (s) loses e- to form Cd2+ (aq)
- The cadmium metal slowly dissolves.
At the cathode, Ni2+ (aq) gains e- to form Ni (s)
- The nickel ions plate on electrode. (It gets bigger.)
EXCLUSION STATEMENT Essential Knowledge 3.B.3
Labeling an electrode as positive or negative is beyond the scope of this course
and the AP Exam.
Salt Bridges
The electrons flowing from anode to cathode can be used to do electrical work such as
lighting a light bulb or heating a toaster.
Charges must flow in a circuit so that electrical charge is conserved.
Thus a salt bridge is needed to allow charges to flow.
A salt bridge is a glass tube filled with brine (concentrated salt) solution with
semipermeable plugs at the ends of the tubes.
For a battery to work over any substantial period of time, the two metal solutions must
be separated.
The semipermeable plugs allow only ions (such as K+ and Cl-) from brine solution to leave.
The flow of oppositely charged ions out of the salt bridge completes the electrical
circuit in a battery.
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Voltaic Cell Notation
An abbreviated way of illustrated has been developed by electrochemists where the
essential features of a voltaic can be written on one line.
1. All phase boundaries are indicated with a vertical line.
Cd (s) | Cd2+ (aq) (The solid is a different phase than the aqueous solution.)
2. The salt bridge is indicated by a double vertical line.
3. The anode is written first (on the left), then the cathode (on the right)
Cd (s) | Cd2+ (aq) || Ni2+ (aq) | Ni (s)
Note the two ions in the cell come in “contact” with each other via the salt
bridge.
ELECTROMOTIVE FORCE (EMF)
EMF is the strength of the reducing agent’s e- pushing combined with the strength of
the oxidizing agent’s e- pulling.
The SI unit to measure EMF is the volt.
1V 
1J
1C
1 volt of EMF gives 1 Coulomb of charge 1 Joule of energy.
The charges in a system (e. g., a copper wire) move randomly until a voltage is applied.
e-
ee
-
e
e-
e-
-
e-
e-
Voltage causes charges to move a single direction. Voltage induces electrical current.
e-
+
ee-
e-
ee-
e-
e-
-
Voltage comes from a difference in potential energy.
- Electrons flow from high potential energy to low potential energy.
- Flow of electrons is like flow of water in a waterfall from a height (high potential
energy) to a depth (low potential energy).
Cell potentials and standard reduction potentials
For the Ni/Cd cell, the EMF at std. conc. is 0.12 V.
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The total voltage is the sum of the voltages due to each half-reaction.
Cd(s)  Cd2+ (aq) + 2 eNi2+ (aq) + 2 e-  Ni (s)
Cd (s) + Ni2+ (aq)  Cd2+ (aq) + Ni (s)
E = 0.40 V
E = - 0.28 V
E = 0.12 V
The voltages for many half-reactions are tabulated as standard reduction potentials.
Ni2+ (aq) + 2 e-  Ni (s)
E = - 0.28 V
2+
Cd (aq) + 2 e  Cd (s)
E = - 0.40 V
*Note when redox reaction is reversed, potential has opposite sign.*
**Source of confusion**
When calculating standard cell potentials, one does not multiply standard reduction
potential by stoichiometric coefficients.
This method is different than the procedure to calculate Hrxn, Srxn and
Grxn.
Example: Calculate the standard EMF for the galvanic cell reaction:
Cr (s) + Cu2+ (aq)  Cr3+ (aq) + Cu (s)
Balanced Equation: 2 Cr (s) + 3 Cu2+ (aq)  2 Cr3+ (aq) + 3 Cu (s)
Find standard reduction potentials in table
Cr3+ (aq) + 3 e-  Cr (s)
Cu2+ (aq) + 2 e-  Cu (s)
E = - 0.74 V
E = 0.34 V
Cr (s)  Cr3+ (aq) + 3 eCu2+ (aq) + 2 e-  Cu (s)
Total cell potential
M(s) + 3 Ag+(aq)  3 Ag(s) + M3+(aq)
Ag+(aq) + e-  Ag(s)
E = 0.74 V
E = 0.34 V
E = 1.08 V
E = +2.46 V
E = +0.80 V
MC Question: According to the information above, what is the standard reduction
potential for the half-reaction M3+(aq) + 3 e-  M(s)?
(A)
(B)
(C)
(D)
(E)
-1.66 V
-0.06 V
0.06 V
1.66 V
3.26 V
334
FR Question: Answer the following questions about electrochemical cells.
It is observed that when silver metal is placed in aqueous thallium(I) fluoride, TlF, no
reaction occurs. When the switch is closed in the cell represented above, the voltage
reading is +1.14 V.
a) Write the reduction half-reaction that occurs in the cell.
Ag+ + e− → Ag
b) Write the equation for the overall reaction that occurs in the cell.
Tl + Ag+ → Tl+ + Ag
c) Identify the anode in the cell. Justify your answer.
The anode is where oxidation occurs. In the overall reaction Tl is oxidized to Tl+,
so the anode is the Tl electrode in the left cell.
d) On the diagram above, use an arrow to clearly indicate the direction of electron
flow as the cell operates.
The arrow should show electron flow in the direction from the Tl electrode
through the wire to the Ag electrode.
e) Calculate the value of the standard reduction potential for the Tl+/Tl half-reaction.
E 0cell  E 0red  E 0ox
0
0
0
 Eox
= Ered
- Ecell
 0.80 V  1.14 V  -0.34 V
The standard reduction potential, E°, of the reaction Pt2+ + 2 e− → Pt is 1.20 V.
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f) Assume that electrodes of pure Pt, Ag, and Ni are available as well as 1.00 M
solutions of their salts. Three different electrochemical cells can be constructed
using these materials. Identify the two metals that when used to make an
electrochemical cell would produce the cell with the largest voltage. Explain how
you arrived at your answer.
Ni2+ + 2 e− → Ni
Ag+ + e− → Ag
Pt2+ + 2 e− → Pt
E°(V)
−0.25
0.80
1.20
The two metals that yield the largest E 0cell are those with the biggest difference
in E°, namely, Pt and Ni (see E 0cell calculation below).
E 0cell  E 0red  E 0ox  1.20 V   0.25 V   1.45 V
g) Predict whether Pt metal will react when it is placed in 1.00 M AgNO3(aq). Justify
your answer.
When Pt metal is added to 1.00 M AgNO3, the only redox reaction that could occur
would be for Pt to become oxidized as Ag+ is reduced.
0
0
0
Ecell
= Ered
- Eox
= 0.80 V -  +1.20 V  = -0.40 V
Because E 0cell for that reaction is negative, no reaction will occur.
STANDARD REDUCTION POTENTIAL
We are not able to measure the potential of a reduction half-reaction by itself.
- Cell potential is due to both ox. and red.
Ecell = Ered - Eox
- To find reduction potential, we must set a standard, i.e. reference point.
The standard that is chosen is the reduction of H+ ion.
2 H+ (aq, 1M) + 2 e-  H2 (g, 1atm) at 25 C
E = 0.00 V by definition.
Thus the standard reduction potentials are measured relative to the reduction of
hydrogen ion.
Standard concentration for all standard reduction potentials is 1M.
Standard reduction potentials are the source of the activity series.
*Source of Confusion*
Electrical potentials are intensive quantities.
- They do not depend on stoichiometric coeff.
- To find potential of cell, add potentials of half-reactions.
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Strength of Oxidizing and Reducing Agents
The more positive the reduction potential, the stronger the reduction.
- Therefore, the better the oxidizing agent.
F2 (g) + 2 e-  2 F- (aq)
E = 2.87 V
+
H2O2 (aq) + 2 H (aq) + 2 e  2 H2O (l)
E = 1.776 V
- Fluorine gas is one of the best oxidizing agents.
- Hydrogen peroxide is also a good oxidizing agent.
The more negative the reduction potential, the more positive the oxidation potential.
Thus the stronger the oxidation.
- Therefore, the better the reducing agent.
Li+ (aq) + e-  Li (s)
Mg2+ (aq) + 2 e-  Mg (s)
E = - 3.07 V
E = - 2.37 V
- Lithium is one of the best reducing agents.
- Magnesium is also a good reducing agent.
SPONTANEOUS ELECTROCHEMICAL RXNS
Using chemical intuition, which reaction is spontaneous?
Cl2(g) + 2 e-  2 Cl-(aq)
E = 1.36 V
+
Na (aq) + e  Na(s)
E = - 2.71 V
- Chlorine taking e is spontaneous.
In general, reactions with E positive are spontaneous.
Reactions with E negative are not spontaneous.
Note: To decide spontaneity, one needs to consider total cell potential, not only halfreaction.
Free Energy and Electric Potential
For a general electrochemical reaction,
G = -n F E
n – number of moles of electrons transferred in reaction, i. e., stoichiometric coefficient
of e- in balanced redox equation.
F – Faraday’s constant = charge of 1 mole of electrons
F = 96,485 C/mol = 96,485 J/Vmol
E – cell potential
Equation relates intensive property (EMF) to extensive property (G) via moles of etransferred.
Since equation is general
G = -n F E

G = -n F E
- Thus we can calculate G of a reaction by measuring E.
- Also note that equation shows us that cells with positive EMF spontaneously react.
- Spontaneous electrochemical reactions create electrical current.
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Example: Calculate G for the reaction Zn (s) + Cl2 (g)  ZnCl2 (aq) from the
standard reduction potentials of the half-reactions.
Zn (s) + Cl2 (g)  Zn2+ (aq) + 2 Cl- (aq)
Zn (s)  Zn2+ (aq) + 2 eCl2 (g) + 2 e-  2 Cl- (aq)
E = 0.76 V
E = 1.36 V
E = 2.12 V
G = -n F E = - 2  (96,485 J/Vmol)  2.12 V = - 409000 J/mol = - 409 kJ/mol
Example: For the reaction, Ni(NO3)2 (aq) + Cd (s)  Ni (s) + Cd(NO3)2 (aq), what is
the expected voltage in a voltaic cell where [Ni2+] = [Cd2+] = 1 M?
Grxn = -23.2 kJ/mol
  23,200 J / mol
 G 0
0
E 

 016
. V
nF
2  96,485 J / V  mol
Nernst Equation
Equation relates EMF with concentration of electrolyte
Recall:
G = G + RT ln Q 
E  E0 
-n F E = -n F E + RT ln Q
RT
ln Q
nF
0.0257 V
ln Q
n
Example: For the reaction, Cr (s) + Ni2+ (aq)  Cr3+ (aq) + Ni (s), calculate the EMF
when
a) [Ni2+] = [Cr3+] = 1 M
At 25 C, E  E 0 
Cr (s)  Cr3+ (aq) + 3 eNi2+ (aq) + 2 e-  Ni (s)
Total cell potential
E = 0.74 V
E = - 0.23 V
E = 0.51 V
b) [Ni2+] = 2.0 M; [Cr3+] = 0.010 M
Balanced equation is 2 Cr(s) + 3 Ni2+(aq)  2 Cr3+(aq) + 3 Ni(s)
 Cr 
Q
 Ni 
3 2

2 3
 0.010 2

 125
.  105
3
 2.0
0.0257 V
RT
ln Q  0.51 V 
ln1.25  10 5 
nF
6
 0.51 V  0.048 V  0.56 V
E  E0 
High amount of reactant and low amount of product creates greater driving force in a
reaction that is already spontaneous.
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EXCLUSION STATEMENT Essential Knowledge 3.B.3
The Nernst equation is beyond the scope of this course and the AP Exam.
Review relationships between G, E, and Q(or K)
G  G 0  RT ln Q
G = -n F E
RT
E  E0 
ln Q
nF
at eq.
at eq.
G 0   RT ln K
G = 0, E = 0
at eq.
E0 
RT
ln K
nF
ELECTROLYSIS
The process of adding electrical energy to make a nonspontaneous process,
spontaneous.
Applications of electrochemical cells
Galvanic cells
- source of electrical energy (battery)
Electrolytic cells
- refining of ores
- produce alkali, alkaline earth metals
- produce H2 and O2 from water (fuel cells)
- metal plating (gold, chromium, etc…)
Consider the formation of KBr
Br2(l) + 2 e-  2 Br-(aq)
K(s)  K+(aq) + e-
E = 1.07 V
E = 2.92 V
E = 3.99 V
Balanced reaction is 2 K (s) + Br2 (l)  2 KBr (aq)
Calculate K (equilibrium constant)
3.992
RT
0.0257
E 
ln K  K  e
 e 311  10135
nF
0
This reaction is definitely spontaneous!!
Is there any way to make K and Br2 from KBr?
Yes!, by reversing the voltage, i.e., forcing e- on K+ and ripping e- from Br-.
When an external voltage is applied that is greater than 3.99 V, the reverse reaction
occurs.
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Schematic of an Electrolytic Cell
EMF source
(battery)
-
flow of e-
+
cathode
anode
-
+
Pt
Pt
Br-
K+
K(l)
Br-
Br-
Br-
K+
Br+
K
K+
K+
K+
K+
K+ (l)  K (l)
Br2(g)
Br-
Br-
2 Br- (l)  Br2 (g)
Liquid potassium is depositing on the cathode.
Gaseous bromine is forming on the anode.
Note: cell reaction is 2 KBr (l)  2 K (l) + Br2 (g).
We must use the molten salt rather than the aqueous solution because if we attempt the
reaction 2 KBr (aq)  2 K (l) + Br2 (g) other complicated reactions occur, especially
hydrolysis of the water.
*Source of Confusion*
The following table gives the signs of the electric potential at each of the electrodes in
each type of cell.
Galvanic
Electrolytic
Cathode
+
-
Anode
+
However current always flows from anode to cathode.
The amount of K(l) or Br2 (g) can be controlled by controlling the number of moles of
electrons in an oxidation/reduction process.
- The number of moles of electrons used (i.e., total charge used) is controlled by
adjusting the time that the electrical current is allowed to flow.
amount of ch arg e
I
- current 
time
1C
- SI unit of current is Ampere (A) 1 A 
1s
-Q=It
Q – charge I – current t – time
340
Example: An iron spoon is placed in an electrochemical cell with AuCl3 solution to be
gold plated. How long must the spoon be in the electrolytic cell if the spoon
is to be plated with 0.293 g of Au and the current of the cell is 1.03 A.
1st Question: How many moles of e- are needed for the reaction Au3+(aq) + 3 e- 
Au(s)
0.293 g 
1 mol Au 3 mol e 

 0.00446 mol e 
196.97 g 1 mol Au
2nd Question: How many coulombs is 0.00446 mole-?
0.00446 mole-  96,485 C/mol = 431 C
3rd Question: If current is 1.03 A, how long does it take for 431 C to flow?
Q It  t 
Q
431 C

 418 s
I 103
. C/s
Example: How many grams of barium metal can be produced by the passage of 173 C
of charge in an electrolytic cell of molten BaCl2?
Ba2+ (l) + 2 e-  Ba (s)
173 C 
1 mol e 
1 mol Ba 137.33 g

 0123
.
g Ba
96,485 C 2 mol e  mol Ba
