Unit 6
Electrochemistry
Chemistry 020, R. R. Martin
Electrochemistry
Electrochemistry is the study of the interconversion
of electrical and chemical energy.
We can use chemistry to generate electricity.... this is
termed a Voltaic (or sometimes) Galvanic Cell
Or we can use electricity to bring about chemical
changes.
This would be termed an Electrolytic cell.
We will consider both as separate topics, though the
underlying theory is the same in each case.
Consider the following reaction that you can do in a
beaker on the bench:
Add zinc metal to a solution of copper sulphate. We
will observe that the blue colour associated with the
copper sulphate fades as a reddish brown precipitate
is formed. Simultaneously, the zinc metal dissolves.
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The equation for the reaction is:
Zn(s) +
CuSO4(aq) → Zn SO4(aq) + Cu(s)
The brown precipitate is copper metal.
In terms of the ions involved ( SO42- is a
spectator) :
1) Zn → Zn2+ + 2e2) Cu2+ + 2e- → Cu
These are so-called half-reactions (remember
balancing redox equations)
in 1) zinc is oxidised (the oxidation # goes
from 0 to 2+) .
Zinc is acting as the reducing agent.
in 2) copper ions are reduced to copper (gain
electrons) metal.
Copper ions are acting as the oxidising agent.
This is a redox reaction that occurs spontaneously
2
If the reaction is allowed to proceed in the
beaker, the liberated energy will appear as heat. If we
make each of the half reactions happen in separate,
but electrically connected containers, the two e- can
be made to travel along a ‘wire’ and do electrical
work.
Some definitions:
Reduction reactions occur at the cathode (red
cat).
Oxidation reactions occur at the anode (an ox)
Thus: Cations are the +ve species that
move to the cathode,
Anions are the -ve species that
move to the anode
The movement of electrons through a wire requires
an unbalanced electrical force. This force called the
electromotive force (emf) is the electrical driving
force that pushes the electrons generated in the
oxidation half-reaction towards the reaction where
reduction takes place.
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The difference in potential energy between the
reactants and products (the emf) of an
electrochemical reaction is measured in volts (V).
The greater the potential energy difference between
the two electrodes, the larger the emf.
Charge is measured in coulombs, the charge on one
electron = 1.602 x 10-19 coulombs.
A difference of one volt causes a charge of one
coulomb to acquire kinetic energy equal to one joule
1V=1J/C
We measure the emf of a cell with a voltmeter . The
potential energy difference, commonly referred to as
the potential of the cell is designated Ecell.
In our setup above, as this reaction occurs, there is a
build-up in cation concentration at the Zn anode, and a
decrease in cation concentration at the Cu cathode. So
anions migrate through the salt bridge from Cu side to Zn
side and cations pass in the opposite direction to preserve
the charge balance.
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We write the Daniel Cell as:
Zn
/
Zn2+
Boundary
between Zn &
solution
//
salt bridge
or partition
Cu2+ / Cu
boundary between
Cu & solution
By convention, the oxidation side is shown on the left
and the reduction side is shown on the right.
When we connect a voltmeter to measure the cell
emf, we connect the red or +ve connection to the
right or reduction electrode:
Remember:
RED, RIGHT, REDUCTION:
this is the side of the cell where electrons are
RECEIVED
If the cell is connected in this manner, a positive
indicated voltage shows that the reaction is
proceeding in the expected spontaneous direction.
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Experimentally it is found that the voltage and
direction of current flow generated by a galvanic
cell depend upon the electrode materials, ion
concentration and temperature, T.
We must define standard conditions:
Pure metals for solids, 1 M concentration for
solutions, 25 0C and 1 atmosphere pressure.
Under any other conditions, the cell voltage will
change (We do not deal with the effect of concentration in
this course. If you are interested read about the Nernst
equation)
For the Daniell cell:
Zn /
Zn2+
//
Cu2+ / Cu Eocell = 1.10v
Eo refers to standard conditions
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Eo refers to standard conditions
But we should ask, how much of that voltage is
‘generated’ at the oxidation side and how much at the
reduction side?
Suppose we consider cells with the same anode
reaction, but different cathode reactions, what
might we expect? The LHS side reaction
remains the same and must be associated with
the same potential.
Cells with the same anode reaction, but different cathode
reaction exhibit different Eo.
Zn / Zn2+ // Cu2+ / Cu
Voltage = 1.10 V
Zn / Zn2+ // Cd2+ /Cd
Voltage = 0.36 V
Zn / Zn2+ // Ag+ / Ag
Voltage = 1.56 V
Since the tendency of the Zn to oxidize remains
the same, the tendency of the other half-reaction
to reduction must be different. Since the RHS
involves reduction, we can say that the easier
the reduction is, the larger the voltage
difference.
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Thus the tendency towards reduction
increases in the order:
Ag+ > Cu2+ > Cd2+
Conversely, the tendency to oxidation
increases in the order:
Cd > Cu > Ag
Thus we might say that Cd2+ wants to stay as
Cd2+ more than Ag+ wants to stay as Ag+
Anthropomorphic: Attribution of human motivation,
characteristics, or behavior to inanimate objects,
animals, or natural phenomena.
For large EMF’s we want:
Negative electrode (LH electrode) with a
strong tendency to oxidize
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Positive electrode (RH electrode) with strong
tendency to reduce
The cell voltage is determined by the
difference in the tendencies of the two
electrode metals to oxidize, or difference in
the tendencies of the two ions to be
reduced
Standard Electrode Potentials.
The voltmeter in our picture of the Daniell cell
measures the difference between the potentials of the
two electrodes. We cannot measure the potential of
each electrode individually because the electrons
have nowhere to go or to come from.
Thus, we need a reference system. We arbitrarily
assign a potential value to a particular half reaction:
2 H+( 1 Molar aq) + 2e- → H2 (g, 1 atm )
This is the Standard Hydrogen Electrode (SHE), it
has :
Eo = 0.0 V
By definition.
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We can now construct a cell that combines the
standard hydrogen electrode (SHE) with any other
standard ‘half-cell’ and the observed voltage is then
all attributed to the reaction in the second half-cell. In
this way we can construct a table of standard
electrode potentials.
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Some things to note in the table:
1) In all cases, standard conditions are used.
2) All of the reactions are shown as proceeding in the
direction of reduction no matter how likely or
unlikely that is....
3) The reduction potentials E0RED in the table have a
negative sign if the SHE in the cell is found to be the
positive electrode. (reduction is taking place at the
SHE, oxidation at the other electrode)
4) To construct a cell from any two of these
reactions, one of the half reactions must be reversed
and written as an oxidation and so the sign of the
potential is reversed. But which one????
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Consider the following reaction:
Fe(s) + 2Ag+ (aq) → Fe2+(aq) + 2Ag(s)
We can write this in ‘cell’ form...
Fe/Fe2+ // Ag+/Ag
We can ask: Is this reaction spontaneous in the
direction as indicated, and what would be the cell
emf under standard conditions?
From the table:
1)
Ag+(aq) + e → Ag(s)
E0 = +0.8 V
2)
Fe2+(aq) + 2e → Fe(s)
E0 = -0.41V
But in the reaction, we are looking for the reverse of
2) ie Fe(s) → Fe2+(aq) + 2e
E0 = +0.41V
Since the oxidation half-reaction produces two moles
of electrons per mole of Fe we need to reduce two
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moles of Ag+ so multiply equation 1 by 2. Note E0
remains the same since it reflects only a difference in
energy between the two states.
2Ag+(aq) +2 e → 2Ag(s)
Now we have:
2Ag+(aq) +2 e → 2Ag(s)
Fe(s) → Fe2+(aq) + 2e
E0 = +0.8 V
E0 = +0.8 V
E0 = +0.41V
Adding these two yields:
Fe(s) + 2Ag+ (aq) → Fe2+(aq) + 2Ag(s) E0cell =
+1.21V
The sign is positive and therefore the reaction is
spontaneous in the direction shown, and the cell emf
would be + 1.21 volts
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Q. Is the following reaction spontaneous?
Fe/Fe2+//Zn2+/Zn
From the tables
1. Zn2+ + 2e → Zn
2. Fe2+ + 2e → Fe
E0 = -0.762v
E0 = -0.409v
But we are looking for the reverse of 2.
Fe → Fe2+ + 2e
Zn2+ + 2e → Zn
Fe → Fe2+ + 2e
Sum
2+
Zn + Fe → Fe2+ + Zn
E0 = +0.409v
E0 = -0.762v
E0 = +0.409v
E0 = -0.300v
The negative sign of the cell tells us that the reaction
is not spontaneous in the indicated direction, and in
fact that the reverse reaction will take place:
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Zn + Fe2+ → Zn2+ + Fe and that the emf of this
cell under standard conditions would be +0.300v.
Q. Calculate the emf of a cell which has the
following set up. Assume all reactants in standard
state. Write a balance equation for the reaction.
Al / Al3+ // Ag+ / Ag
Al(s) + 3Ag+(aq) → Al3+(aq) + 3Ag(s)
From the table:
Al3+ + 3e- → Al
Ag+ + e- → Ag
E0 = - 1.68 V
E0 = 0.8 V
But we want the reverse of the first equation:
Al → Al3+ + 3e
E0 = +1.68v
We now have:
Al → Al3+ + 3e
E0 = +1.68v
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3Ag+ + 3e- → 3Ag
E0 = 0.8 V
Adding yields:
Al + 3Ag+ → Al3+ + 3Ag
E0cell = +2.48v
Lets take another look our table of reduction
potentials. What sorts of things might we predict from
looking at the Eo values?
Li+ + e- → Li
Mg2+ + 2e- → Mg
Eo = - 3.040 V
Eo = - 2.357 V
2H+ + 2e- → H2 Eo = 0.000 V
Ag+ + e- → Ag
Eo = 0.799 V
F2 + 2e- → 2F- Eo = 2.889 V
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e- → Li the large
For the reaction Li+ +
negative value of the reduction potential tells us that
it is unlikely for this reaction proceed in the indicated
direction. We are aware that Li metal is extremely
reactive and wants to be in the form of positive (cat-)
ions.
That is to say, Li+ is a VERY weak oxidising agent,
but conversely, Li (metal) is a POWERFUL reducing
agent
In fact we can conclude that Li metal must be the
MOST powerful reducing agent and Li can reduce
any of the species in the left hand column of the
table.
We can predict that:
2Li + Mg2+ → 2Li+ + Mg
will proceed as written.
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At the other end of the table, the large positive value
of reduction potential for F2 + 2e → 2Findicates that this reaction is VERY likely to proceed
as written. F2 is an extremely reactive gas that wants
to be in the form of negative (an-) ions.
That is to say, F2 is a very powerful oxidising agent
and conversely F- is a very weak reducing agent.
In fact we can conclude that F2 must be the most
powerful oxidising agent and F2 is capable of
oxidising any of the species in the right hand column
of the table.
We can predict that:
F2 + 2Ag → 2Ag+ + 2Fwill proceed as written.
As an aside, notice that H2O is above F- in the right hand column. Yes, Fluorine can
oxidise water.
Possible add-on for me to use as an in-class exercise:
Look at the full table of reduction potentials and find:
any four species that can reduce Al3+ to Al
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any metal which can reduce Sn4+ to Sn2+ but not Cr3+
to Cr2+
all the metals which will dissolve in aqueous acids
a metal ion which can reduce Pb2+ to Pb but not Al3+
to Al
ELECTROLYTIC CELLS
In an electrolytic cell, a non-spontaneous redox
reaction is made to go by applying a large
enough voltage (larger than the spontaneous
voltage and opposite in sign) to the cell and
pumping electrical energy into the system.
Thus: for the reaction:
Zn + Cu2+ → Zn2+ + Cu Eo = 1.10 V
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If we apply a reverse voltage > 1.10 V to the Cu
electrode, the reverse reaction should occur:
Zn2+ + Cu → Zn + Cu2+
If we pass a current through a molten mixture of
KHF2 and anhydrous HF we achieve the following
process:
2HF(l) → H2(g) + F2(g)
There is no chemical means of achieving this result
(WHY?)
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Quantitative relationships
As always the stoichiometry of a balanced
reaction as written implies the molar ratios
involved
Thus the reaction:
Cu2+ + 2 e- → Cu(s)
Shows that one mol of Cu2+ reacts with 2 mol of
electrons.
We now need to know the total charge on a
mol of electrons.
Unit of charge is the Coulomb (C).
One electron has a charge of 1.602 x 10-19 C
therefore 1 mol e- carries a charge of 1.602 x 10-19 x
6.022 x 1023 = 96490 C mol-1
The charge on a mole of electrons ( approx
96,500 coulombs) is called Faraday’s Constant
or one Faraday.
Electrical current is defined in terms of the rate
of flow of charge per unit time thus:
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1 Amp = 1 coulomb per second
and
Total charge = current(amps) x time (seconds)
eg, in our copper purification experiment,
suppose a current of 0.2A passed for 30 min.
How many grams of copper are deposited at the
cathode?
We know that the number of moles of Cu
deposited will be one half the number of moles
of electrons supplied, (Cu2+ + 2 e- → Cu(s) ).
Now the number of moles of electrons supplied
will be the total charge delivered divided by the
charge on a mole of electrons and:
Total Charge (in C) = Current (in amps) x Time
(seconds)
Total charge = [0.2 C/sec] x [ 30 min x 60
sec/min]
= 360 C
Moles e- = (360 C) / (96,490 Cmol-1)
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= 0.00373
Finally:
Moles Cu deposited = (moles e- supplied)/2
Moles Cu deposited = 0.00373/2
Mass Cu = moles x MM
Mass Cu = (0.00373/2) x 63.54 = 0.1185 g
Chromium metal can be electroplated from an
acidic solution of CrO3
1. How many g of Cr will be plated by 1 x104 C?
2. How long will it take to plate one g of Cr using
a current of 6 A?
The reaction is:
CrO3(aq) + 6 H+ + 6 e- → Cr = H2O
1 mol CrO3 requires 6 mol e1 mol CrO3 contains 1 mol of Cr or 52.0 g Cr
1. If 1 x 104 C are supplied;
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moles e- = Charge/Faraday = 1 x 104 C/ (96,490
C/mol) = .1036 moles eMoles Cr = (Moles e-)/6 = 0.0172
Mass Cr = moles x MM = 0.0172 x 52 = 0.894g
2. Time to deposit 1 g Cr:
moles Cr = mass/MM = 1/52 = 0.01923 moles
moles e- required = moles Cr x 6 = 0.01923 x 6
= 0.1154 moles
Charge on a mole of e- = Faraday = 96, 490 C
Total charge required = moles e- x Faraday
= 0.1154 x 96,490 = 1.113 x 104 C
Finally:
Total Charge = Current x time (sec)
1.113 x 104 C = 6 C s-1 x time
time = 1.85 x 103 sec = 30.8 minutes
C. Reversibility: Electrolytic Cells
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In Voltaic cells, we converted the chemical
energy of a spontaneous reaction into electricity (e.g.
using a battery)
However, we can often do the reverse: use
electrical energy to force a non-spontaneous reaction
(e.g. recharge a battery). In this context, where we
apply electrical energy, we have an electrolytic cell.
Consider the lead-acid battery, which can be
used in both voltaic- and electrolytic-cell modes
(discharge and recharge).
There are separate Pb and PbO2 lead plates
•
As a voltaic cell, the we have the following
reactions
0
o
Anode E ox = +0.36 V
2−
Pb(s) + SO4 ¬ PbSO4(s) + 2 e
0
−
+
−
o
Cathode E red = +1.69 V PbO2(s) + 4 H + 2 e
2−
+ SO4 ¬ PbSO4(s) + 2 H2O
o Cell notation Pb │PbSO4 ║ PbO2 │PbSO4
0
o E cell = approx +2 V, and six of these are
connected in series to provide for a car battery of 12
V
o Some interesting notes:
. As the car battery gets discharged, H2SO4
decreases, and so does the density of the electrolyte.
Some batteries have a green “eye” that lets you
estimate its condition… when the electrolyte is dense,
the eye floats up and appears green.
PbSO4(s) coats the cathode as the battery
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discharges. Thus, the more discharged the battery,
the thicker the coating, and the less contact with the
electrolyte. If you have a dead battery and wait a few
minutes, you can crank your engine again… the
electrolyte needs time to diffuse through the coating.
. Remember that cell potential does not tell you
the rate of the chemical reaction. In the automotive
industry, battery makers aim for the highest number of
Cold Cranking Amps, a measure of how many
Coloumbs per second can be produced.
•
To recharge the battery, the products of the
discharge needs to be converted back to starting
materials (electrolytic cell).
0
o
Cathode E red = −0.36 V
−
PbSO4(s) + 2 e ¬ Pb(s) + SO4
2−
0
Anode E ox = −1.69 V PbSO4(s) + 2 H2O ¬
+
−
2−
PbO2(s) + 4 H + 2 e + SO4
o Cell notation PbSO4 │PbO2 ║ PbSO4 │Pb
o Note that in recharge mode (electrolytic cell),
the anode and cathode have switched places!
Nonetheless, the anode is always the electrode where
oxidation occurs.
0
o E cell = approx −2 V, which indicates the
reaction is not spontaneous. This is expected!
o To recharge, we must apply an external
voltage of at least 2 V to force this undesirable
reaction to occur. When a redox reaction is driven by
electricity, it is called electrolysis.
The recharging of a battery is just one example of
o
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electrolysis, but they all rely on the same principles.
•
Example: Industrial electrolysis of molten
(not aqueous) NaCl for the generation of Na(s) and
Cl2
o A battery or power source drives the
reaction. A small amount of CaCl2 is also added to
help lower melting point
o
−
Anode 2 Cl (l) ¬ Cl2(g) + 2 e
+
−
−
o
Cathode 2 Na (l) + 2 e ¬ 2 Na(l)
o Try and see if you could calculate the
minimum external voltage needed for this electrolysis
o Industrial problem: how do you keep Cl2(g)
from reacting with the Na(s) product?
o Solution: A brilliant piece of engineering that
incorporates an inverted funnel to isolate the chlorine
gas product (known as a Downs electrolysis cell)
o Another question: Why do we need to use
molten NaCl? Why not simply use NaCl dissolved in
water?
•
Electrolysis of NaCl in water results in the
−
same oxidation of Cl to Cl2, but H2O is reduced much
+
more easily than Na
o
o
−
Anode 2 Cl (aq) ¬ Cl2(g) + 2 e
Possible cathode reactions:
−
+
−
0
2 Na (aq) + 2 e ¬ 2 Na(s) E red = −2.71 V
−
−
0
2 H2O + 2 e ¬ H2(g) + 2 OH E red = −0.83 V
o Reduction of water is much more favourable
−
2 Cl (aq) + 2 H2O ¬ Cl2(g) + H2(g) + 2 OH
−
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o So, we don’t have any Na(s) being formed.
Is this bad? No! All of Cl2, H2, and NaOH are
industrially useful, so the electrolysis of salt water is
very important!
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D. Quantitative Analysis (now some
physics!)
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