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)
Steps in Balancing Equations by the Method of Half-Reactions
1. Break overall equation into two half-reactions.
2. Balance each half-reaction, as follows…
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. BASIC SOLN ONLY: Add enough OH– to cancel H+.
3. Multiply each half-reaction by integers to cancel e–; simplify, where possible.
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EX. Balance this reaction, which takes place in acidic solution.
Cr2O72–(aq) + Cl–(aq)  Cr3+(aq) + Cl2(g)
EX. Balance this reaction, which takes place in basic solution.
CN–(aq) + MnO4–(aq)  CNO–(aq) + MnO2(s)
Voltaic (or Galvanic) Cells
In a voltaic (or galvanic) cell, e– transfer occurs via an external pathway rather than directly
between reactants.
e.g.,
electrodes: two solid metals connected via an external circuit
anode
cathode
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Consider a solution of Zn(NO3)2(aq) and Cu(NO3)2(aq) with electrodes as shown…
e–
e–
Zn anode
Cu cathode
NO3–
Zn2+
Zn2+
NO3–
Na1+
salt bridge containing
electrolyte (e.g., NaNO3)
in a porous gel
Cu2+
Cu2+
NO3–
Why do the e– go the way they do?
Cell EMF
We can think of the PE of each of the anode’s e– as being > PE of each of the cathode’s e–
Thus…
Difference in
PE
is important;
ch arg e
Voltage (or “potential”) difference is called the electromotive force (emf)
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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 emf’s occur at 25oC
-- To calculate Ecell, look up tabulated standard reduction potentials for each half-cell…
e.g.,
Ag+(aq) + e–  Ag(s)
Eored = 0.799 V
…and then use the equation:
The reference point for reduction potentials is the standard hydrogen electrode (SHE):
SHE:
2 H+(aq, 1 M) + 2 e–  H2(g, 1 atm)
Eored = 0 V
Multiples of coefficients don’t affect Eored
e.g.,
EX. For Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s), Eored is measured to be 1.10 V. Given
that Eored for Zn2+ to Zn is –0.76 V, find Eored for the reduction of Cu2+ to Cu.
EX.
A galvanic cell based on
Al3+(aq) + Mg(s)  Al(s) + Mg2+(aq)
(a)
Al3+(aq) + 3 e–  Al(s)
Eored = –1.66 V
(b)
Mg2+(aq) + 2 e–  Mg(s)
Eored = –2.37 V
Write the balanced equation and calculate Eocell.
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has half-rxns:
EX. For a voltaic cell with the half-reactions…
(a)
Cd2+(aq) + 2 e–  Cd(s)
Eored = –0.403 V
(b)
Sn2+(aq) + 2 e–  Sn(s)
Eored = –0.136 V
…find the standard cell potential and write the balanced equation.
For a half-reaction, the more (+) the Eored value, the greater the tendency for that reaction to
“go” in that direction (i.e., reduction).
Other strong oxidizers are the halogens…
and oxyanions in which the central atom has a large (+) charge.
e.g.,
Poorest oxidizer (i.e., hardest to reduce / easiest to oxidize) is…
Li+(aq) + e–  Li(s)
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–
C–
of two half-reactions, consider the scale
0V
at right. The “higher-up” reaction is the
reduction half-cell; the “lower-down”
reaction is the oxidation half-cell.
B–
(–) V
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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…
G = – nFE
n = # of mol of transferred e–
F = Faraday’s constant
In standard states…
Go = – nFEo
= 96,500 J/V-mol e–
EX. For 3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH–(aq)  3 Ni(s) + 2 CrO42–(aq) + 8 H2O(l)…
(a) What is n?
(b) Find Go.
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Effect of Concentration on Cell EMF
Cell emf gradually drops due to changes in concentration of reactants and products.
When emf = 0 V, cell is “dead.”
Nernst equation:
At 25oC (298 K):
EX.
2 Al(s) + 3 I2(s)  2 Al3+(aq) + 6 I–(aq)
Find emf at 25oC when [ Al3+ ] = 0.0040 M and [ I– ] = 0.010 M.
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concentration cell:
Ni bar A
Ni bar B
salt bridge
[ Ni2+ ] = 0.001 M
[ Ni2+ ] = 1.00 M
-- Eo for a [ ] cell =
-- In the above example, cell will act to equalize [ Ni2+ ]s, so…
Bar A:
Bar B:
Add…
Thus, Q =
In general,
At equilibrium…
EX. A concentration cell with two Zn(s)–Zn2+(aq) half-cells has Cell A with [ Zn2+ ] = 1.35 M
and Cell B with [ Zn2+ ] = 3.75 x 10–4 M. Identify anode and cathode, and calculate emf.
Assume 298 K.
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At equilibrium, G = ___ and Q = ___. The Nernst equation
can be rearranged to give the relationship between K and E o.
Batteries
-- consist of one or more voltaic cells
For batteries in series…
Primary cells can’t be recharged; secondary cells can.
Types of Batteries:
-- lead-acid battery
-- nickel-metal-hydride battery
-- alkaline battery
-- lithium-ion battery
-- nickel-cadmium battery
-- fuel cells
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.
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Electrolysis:
-- electrolysis occurs in electrolytic cells, which consist of two electrodes in a molten salt
or a solution
 reduction at cathode; oxidation at anode
EX. Electrolysis of AgF(aq) in acidic solution forms Ag(s) and O2(g). Find the two halfreactions and the minimum external emf.
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 the electrolysis of molten MgCl2, calculate the mass of Mg formed when 60.0 A of
current are passed through the liquid for 4.00 x 103 s.
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For spontaneous processes, the maximum useful work that can be extracted is equal to G.
For a voltaic cell…
Voltaic cells have +E, so G and wmax are (–). Recall that…
For an electrolytic cell -- for which E is (–) and G is (+) --…
The unit for electric power is the watt (1 W = 1 J/s). Electric companies often measure
electrical energy in the kilowatt-hour (kWh).
1 kWh =
EX. Calculate the number of kWh required to produce 1.00 kg of magnesium from electrolysis
of molten MgCl2 if the applied emf is 5.00 V. Assume 100% efficiency.
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