Download General Electrochemistry Notes

Chemistry 122 [Tyvoll]
Fall 2008
Electrochemistry: Electrochemical reactions are of two basic types:
(1) galvanic or voltaic -- an oxidation-reduction reaction is used to produce a flow of electrons
through an external circuit or wire
(2) electrolytic ("electrolysis") -- electrons from an external source, e.g. a battery, are used to
cause an oxidation- reduction reaction to occur in a molten electrolyte or electrolyte
solution
Summary: an electrolytic cell uses electrical current to drive a chemical reaction, and a galvanic
cell uses a chemical reaction to produce an electrical current
Galvanic Cells and Electrode Potentials:
A galvanic (voltaic) cell is one in which a chemical reaction is used to produce an electrical
current with some associated voltage or potential.
Example.
Zn ' Zn2+ + 2e1- (reversible reaction)
When the concentration of Zn2+ is specified, the potential for the reaction is constant for a given
concentration.
Factors influencing voltage for a given reaction:
*(1) concentration of ions involved in the reaction
*(2) the value of KC for the reaction
(3) temperature [remember that KC = f(T)]
Definition: Standard hydrogen electrode (SHE) (250C, 1 atm, 1.0 M H+ )
where 2 H+ + 2 e1- ' H2(g) (remember that "'" indicates a reversible process)
We arbitrarily assign a voltage of 0.000 volts to this reaction, i.e. for both H2(g) ' 2 H+ + 2 e1(oxidation) and 2H+ + 2 e1- ' H2(g) (reduction), E0 = 0.000 volts
Note that reversing the direction of the reaction changes only the sign of the voltage, not its
magnitude.
We use the hydrogen electrode (standard hydrogen electrode) and its assigned potential to
measure quantitatively the ease with which other half-reactions proceed as reduction reactions.
Example. Zn2+ + 2 e1- ' Zn0
E0
= ??
Let concentration (actually “activity”) of the Zn2+ be 1.0 M to get the standard potential for this
reaction, and construct an electrochemical cell as shown below:
Zn0 / Zn2+(aq) // H+(aq) / H2(g), 1.00 atm
anode (oxidation)
cathode (reduction)
// = “salt bridge”
Reactions: Each oxidation or reduction reaction by itself is called a half-reaction, where
Oxidation Zn2+ ' Zn0 + 2 e1Reduction 2H+ + 2 e1- ' H2(g)
E0 = ??
E0 = 0.000 V (assigned value)
Overall Rxn. Zn0 + 2 H+ = Zn2+ + H2(g)
E0cell = +0.76 V (experimentally measured E0cell)
This reaction proceeds spontaneously, so, in the presence of H+, Zn0 is oxidized to Zn2+ and H+ is
reduced to molecular hydrogen, where
E0cell = E0 (H2/H+) - E0 (Zn/Zn+2) = +0.76 V = 0.00 - E0 (Zn/Zn+2)
or, E0 (Zn/Zn+2) = 0.00 - 0.76 = - 0 .76 V
This means that H+ is more easily reduced than Zn2+.
Electromotive Series: a tabulation of E0 values for reduction half-reactions.
1. Negative E0 indicates a tendency toward spontaneous oxidation.
2. Positive E0 indicates a tendency toward spontaneous reduction.
The magnitude of E0 can be used as a measure of a reaction’s relative tendency towards
oxidation or reduction., e.g.
K+ + e1- ' K0
E0 = -2.925 V
Mg2+ + 2 e1- ' Mg0
E0 = -2.37 V
Here, Mg2+ shows a greater tendency toward reduction than K+, although both reactions have a
relative tendency toward oxidation.
Half-reactions with more negative E0 values (the upper part of the table) tend to proceed as
oxidation reactions when coupled with a half-reaction below them in the table of reduction
potentials, i.e. the lower half-reaction will proceed as the reduction in the overall reaction.
Electrolytic Cells: Electrolytic cells are used to drive processes (electrochemical) which are not
normally spontaneous. For an electrolytic cell, the following definitions hold:
anode – site where oxidation occurs; the positive electrode
cathode – site where reduction occurs; the negative electrode
Faraday's Laws: Recall that electrons are transferred from one species (an oxidizing agent) to a
second species (a reducing agent) during redox reactions.
Anode Rx. Red0 Redn+ + n e1Cathode Rx. Oxn+ + n e1- Ox0
Example.
Na+ + 1 e1- Na0
where Na+ is reduced to Na0
1 mol electrons = 96,487 coulombs (C) = 1 Faraday (F) (a coulomb is a quantity of charge, Q)
so,
1 mol of electrons = 1 faraday (1F) = 6.02 x 1023 electrons = 96,487 coulombs (C)
The most common unit of electrical current is the ampere (amp), where 1 amp = 1 C/second
Thus, we can measure amps and time (seconds) and then calculate coulombs, i.e.
Q = I x t (amps x seconds) amp·seconds
Example. A current of 0.100 amp flowed for 150. s. How many electrons were transferred
during this time?
C = 0.100 amp x 150 sec. = 15.0 C
# e1- = C x (6.02 x 1023 e1- / 96,487 C)
# e1- = C x 6.24 x 1018
# e1- = 15.0 C x 6.24 x 1018 = 9.36 x 1019 e1How many faradays is this? F = 15 C/ x (96,487 C/F) = 1.55 x 10-4 F
Faraday found experimentally that:
(1) the mass (m) of an element deposited during electrolysis is directly proportional to the
quantity of charge (C) that passes so m ∝ Q or m ∝ I x t since Q = I·t
where m = mass in grams, Q = charge (C), I = current (amps) and t = time (s)
(2) the mass of an element oxidized or reduced also depends upon the "electrochemical
equivalent" (z) for that element, i.e. m ∝ z, where z = equivalent weight (EW) of an
element/96,485 C/eq and EW = atomic mass/n (n= # electrons transferred per atom or ion)
Thus, consider for example, Ag+ + e1- Ag0
where n = 1
-3
g/coulomb (g/C)
z = (107.868/1) / 96,487 = 1.12 x 10
but, for
Cu0 Cu2++ 2 e1- , EW Cu = 63.546/2 = 31.773 g/eq
and z = 31.773/96,487 = 3.29 x 10-4 g/C
Other examples: Al3++ 3e1- Al0 z = (26.9815/3)/96,487 = 9.32 x 10-4 g/C
but, Ni4+ + 2e1- Ni2+ where z = (58.71/2)/96,487 = 3.04 x 10-4 g/C
Final relationship:
m = zIt
Practical Application:
Assume that aluminum is produced from its ore by the reaction shown above. An aluminum
beer can has a mass of 16.0 grams. How long will it take to produce enough aluminum to make
the cans needed for 10 cases of beer if the current used for the electrolysis is 100 amps ?
First, calculate the time needed for 1 can: (substitute C s for amps)
m = zIt
so, 16.0 g = ( 9.32 x 10-4 g/C ) x 100 amps x t
t = 16.0 g / ( 9.32 x 10-4 g/C ) x 100 C s) = 1.72 x 102 s ≅ 2.9 minutes
Then, t = (2.9 min/can)(10 cases)(24 cans/case) = 7.0 x 102 minutes = 11.6 hours