Electrochemistry
Name: ____________________
AP Chemistry Lecture Outline
During oxidation-reduction (redox) reactions, the oxidation states of two substances change.
oxidation =
reduction =
Oxidation-Reduction Reactions
oxidizing agent (oxidant): is reduced (or has a component that is reduced)
reducing agent (reductant): is oxidized (or has a component that is oxidized)
e.g.,
Zn(s) + 2 H+(aq) 
EX. Identify the oxidant and the reductant.
2 H2O(l)
+
Al(s)
+
MnO4–(aq)

Al(OH)4–(aq)
+
MnO2(s)
Balancing Oxidation-Reduction Reactions
-- conserve mass AND conserve charge
half-reaction: oxidation by itself, or reduction by itself
EX. Write half-reactions for…
Sn2+(aq) + 2 Fe3+(aq)  Sn4+(aq) + 2 Fe2+(aq)
In line notation, this reaction would be written:
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Steps in Balancing Equations by the Method of Half-Reactions
1. Break overall equation into two half-reactions.
2. a. Balance everything but H and O.
b. Balance O by adding H2O as needed.
c. Balance H by adding H+ as needed (assuming acidic solution).
d. Add e– as needed.
e. Multiply each half-reaction by integers to cancel e–.
3. Add the two half-reactions and simplify.
*4. BASIC SOLN ONLY: Add enough OH– to cancel any H+. Simplify again.
EX. Balance this reaction, which takes place in acidic solution.
Cr2O72– + Cl–  Cr3+ + Cl2
EX. Balance this reaction, which takes place in basic solution.
CN– + MnO4–  CNO– + MnO2
Voltaic (or Galvanic) Cells
In a voltaic (or galvanic) cell, e– transfer occurs via an external pathway that links the
reactants.
e.g.,
electrodes: the two solid metals in a voltaic/galvanic cell
anode
cathode
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Consider a solution of Zn(NO3)2(aq) and Cu(NO3)2(aq) with electrodes as shown…
Zn anode
Cu cathode
salt bridge containing
electrolyte (e.g., NaNO3)
in a porous gel
WHY do the e– go the way they do?
Cell EMF
PE of anode’s e–
PE of cathode’s e–
Thus…
Difference in
PE
is important;
charge
Voltage (or “potential”) difference is also called the electromotive force (emf).
For a particular cell, (i.e., a particular anode and cathode), the cell’s emf is written Ecell and is
called the cell potential.
---- standard emfs occur at 25oC
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-- To calculate Ecell, look up tabulated standard reduction potentials for each half-cell…
e.g.,
Ag+(aq) + e–  Ag(s)
…and then use the equation:
The reference point for reduction potentials is the standard hydrogen electrode (SHE):
2 H+(aq, 1 M) + 2 e–  H2(g, 1 atm)
Multiples of coefficients don’t affect Eored.
e.g.,
EX.
For Cr(s) + Cu2+(aq)  Cr2+(aq) + Cu(s), Eocell is measured to be 1.25 V. Given
that Eored for Cr2+ to Cr is –0.91 V, find Eored for the reduction of Cu2+ to Cu.
EX.
A galvanic cell has half-rxns:
(a)
Al3+(aq) + 3 e–  Al(s)
Eored = –1.66 V
(b)
Ba2+(aq) + 2 e–  Ba(s)
Eored = –2.90 V
Calculate Eocell and write the balanced equation.
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For a half-reaction, the more (+) the Eored value, the greater the tendency for that reaction to
“go” in that direction (i.e., reduction). Strongest oxidizer is…
Other strong oxidizers are…
and oxyanions in which the central atom has a large ___ charge.
e.g.,
Poorest oxidizer is…
Eored = –3.05 V
(–) sign indicates poor
tendency to “go” in this
direction, but large
magnitude (i.e., 3.05 V)
shows strong tendency to
“go” in other direction
(i.e., oxidation).
A–
In comparing the reduction potentials
(+) V
A–
of two half-reactions, consider the scale
shown. The “higher-up” reaction is the
C–
0V
B–
reduction half-cell; the “lower-down”
reaction is the oxidation half-cell.
(–) V
B–
C–
Spontaneity of Redox Reactions
Eo = Eored, reduction – Eored, oxidation
(same equation as before)
-- The Activity Series is based on standard reduction potentials.
Relationship between E and G…
n = # of mol of transferred e–
F = Faraday’s constant
In standard states…
=
5
EX.
For…
5 Fe2+ + MnO4– + 8 H+  5 Fe3+ + Mn2+ + 4 H2O
(a) What is n?
(b) Find Go.
Effect of Concentration on Cell EMF
Cell emf drops gradually due to changing concentrations of reactants and products.
When emf = 0 V, cell is “dead.”
At 25oC (298 K):
Nernst equation:
EX.
Fe(s) + Cd2+(aq)  Cd(s) + Fe2+(aq)
Find emf at 25oC when [Cd2+] = 0.030 M and [Fe2+] = 2.0 M.
If [Cd2+] = 2.0 M and [Fe2+] = 0.030 M…
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concentration cell:
Ni bar A
Ni bar B
salt bridge
[Ni2+] = 1.00 M
[Ni2+] = 0.001 M
-- Eo for a [ ] cell =
-- In the above example, cell will act to equalize [Ni2+]s, so…
Bar A:
Bar B:
Add…
At equilibrium…
Thus, in general…
EX. A [ ] cell has Cell A with [Cd2+] = 2.35 M and Cell B with [Cd2+] = 2.25 x 10–3 M. Identify
anode and cathode, and calculate emf. Assume 298 K.
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At equilibrium, G = ___, E = ___ and Q = ___. The Nernst equation
can be rearranged to give the relationship between K and E o.
Corrosion
-- spontaneous redox reactions in which a metal reacts with some substance in its
environment to form an unwanted compound
-- For some metals (e.g., Al and Mg)…
-- Galvanized iron is coated with a protective layer of ________.
-- cathodic protection:
 oxidized metal is called the ____________________________
EX.
Electrolysis:
-- Electrolysis occurs in electrolytic cells, which consist of two electrodes in a molten salt
or a solution.
 reduction at cathode; oxidation at anode
Fe2+/Fe  Eored = –0.44 V
Consider plating chromium onto an iron pipe:
Cr3+/Cr  Eored = –0.74 V
For Cr to plate out on the Fe pipe, the equation is:
and Eo (for the galvanic cell) would be:
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electroplating:
Ni anode
Vext
Fe cathode
to be plated
with Ni
For electrolysis, current in amperes
(1 A = 1 C/s) is passed through the
liquid. One mole of transferred e–
carries with it 96,500 C of charge.
Ni2+(aq)
EX. For how long must a 50.0 A current be passed through molten BaBr2 in order to produce
500. g of barium?
For a voltaic cell, the maximum work the system
can do on the surroundings is given by:
where E is the reduction potential of the system at the conditions specified
For an electrolytic cell, the work that the surroundings do
(i.e., that the external “oomph” does) on the system is given by:
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The unit for electric power is the watt (1 W = 1 J/s). Electric companies often measure
electrical energy in the kilowatt-hour (kWh).
How many J is 1 kWh?
EX. Calculate the ideal number of kWh required to produce 10.0 kg of calcium from the
electrolysis of molten calcium chloride if the applied emf is 75 V. Assume 100%
efficiency.
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