Unless otherwise stated, all images in this file have been reproduced from:
Blackman, Bottle, Schmid, Mocerino and Wille,
Chemistry, 2007 (John Wiley)
ISBN: 9 78047081 0866
Slide 34-1
Chem 1101
A/Prof Sébastien Perrier
Room: 351
Phone: 9351-3366
Email: [email protected]
Prof Scott Kable
Room: 311
Phone: 9351-2756
Email: [email protected]
A/Prof Adam Bridgeman
Room: 222
Phone: 9351-2731
Email: [email protected] 34-2
Highlights of last lecture
Voltaic cells...
CONCEPTS
Half-reaction
Table of standard reduction potential
Effect of concentration
Link between E , Q and K
CALCULATIONS
Work out cell potential from reduction potentials;
Work out cell potential for any concentration
(Nernst equation)
“Nernst equation”
Ecell = E 0 − 2.303×
RT
log(Q)
nF
Slide 34-3
Concentration cells
If the electrochemical potential
is a function of concentration (as
in the Nernst eq’n), then can we
build a voltaic cell using the same
half-reaction, but with different
concentrations in each half-cell?
0.75V
How does this work?
Afterall:
Cu2+
Cu → Cu2+ + 2e+ 2e- → Cu
????
E 0=-0.34
E 0=+0.34
E = 0.0V
Slide 34-4
Concentration cells
It works because the standard half-cell potential is
based on 1.0 M concentrations. Even though in this
experiment E 0 = 0, the potential for Q ≠ 1 is non-zero.
The measured cell potential in our experiment was
0.75V. Let’s work out what the [Cu2+] was in the beaker:
Cu2+
Cu → Cu2+ (y M) + 2e(0.1M) + 2e- → Cu
Cu2+ (0.1M) → Cu2+ (y M)
Ecell = ____V
0.75
0.0592
 y  0.75
0
.
0
0
.
0296
log
=
−
Ecell = E −
log(Q )

 = ____ V
 0.1 
n
0
1E-26 M
Solve for y: _______________________
Slide 34-5
Applications of concentration cells
Consider the following concentration cell:
pH meter
H2(g) → 2H+(aq, unknown) + 2e2H+ (aq, 1M) + 2e- → H2(g)
2H+ (1M) → 2H+ (unknown)
Ecell = E
0
−
0.0592
n
 [H + ]2 

log(Q ) = −0.0296 × log

 1.0 
+
Ecell = ?
In case you
missed it!
= −0.0592 log[H ]
pH ≡ -log[H+]
E cell = 0.0592 x pH
i.e. measurement of the cell potential provides pH directly!
Slide 34-6
Applications of concentration cells
Concentration cells are used all around us, e.g.
nerve signalling
ion pumps across cell membranes
Na+ / K+ pump, Ca2+ pump
energy production and storage in cells
ATP
pH meters
concentration gradients produce electrical current
these usually use a silver/silver chloride reference rather
than SHE
ion selective electrodes
different choice of reference can then be specific to
specific ions.
Slide 34-7
Other applications of voltaic cells
Batteries (later lecture)
Refs: Most General Chemistry texts have good sections on
batteries, e.g.
Silberberg, pp.923-5
McMurray and Fay, pp.781-6
Petrucci, Harwood and Herring, pp.844-848
However, Housecroft and Constable does not(!)
Corrosion (later lecture)
Ref: Most General Chemistry texts also have a section on
corrosion:
Silberberg, pp.926-8
McMurray and Fay, pp.786-9
Petrucci, Harwood and Herring, pp.849-50
Again, Housecroft and Constable does not(!)
Slide 34-8
Electrolysis
So far we have considered electrochemical cells where:
the reaction is spontaneous, or
the cell potential is positive, or
the cell produces electricity
Many of these reactions are equilibrium reactions. Can we choose
conditions to force the reverse reaction?
YES! Just force electrons the other way around the circuit…
This is called “electrolysis”.
A cell that performs this is called an “electrolytic cell”.
This process is responsible for several of the industrial processes
that we have glossed over, including #9 and #10 on “top ten” list
(NaOH and Cl2), refining of metals, and production of aluminium.
Slide 34-9
Comparison – voltaic vs electrolytic
Electrolytic Cell
Voltaic Cell
-
Zn→
→Zn2++2e-
+
Cu2++2e-→Cu
-
Zn2++2e-→Zn
+
Cu→
→Cu2++2e-
Anode = oxidation = negative
Anode = oxidation = positive
Cathode = reduction = positive
Cathode = reduction = negative
Ecell > 0
Ecell < 0
Slide 34-10
Why does the sign change?
Assigning the term “anode” and “cathode” is not based
on sign, but on whether the chemical species is oxidised
or reduced.
Oxidation ALWAYS occurs at the anode.
In a voltaic cell, the anode liberates the electrons from solution and
the electrode becomes negatively charge.
In an electrolytic cell, the electrons are withdrawn from the
electrode, which therefore becomes positive.
Reduction ALWAYS occurs at the cathode.
In a voltaic cell, the cathode loses electrons to the solution and the
electrode becomes positive.
In an electrolytic cell, the cathode has electrons pumped into it
from the battery, and the electrode becomes negative.
Slide 34-11
Electrolysis of water
At standard state (all @ 1 mol/L):
Oxidation:
2H2O → O2 + 4H+ + 4eReduction:
Overall:
4H2O + 4e- → 2H2 + 4OH-
E 0 = -1.23 V
E 0 = -0.83 V
6H2O(l) → 2H2(g) + O2(g) + 4H+(aq) + 4OH-(aq)
E 0 = -2.06 V @ [H+] = [OH-] = 1 M
But at [H+] and [OH-] = 10-7 M (use Nernst eq):
Oxidation:
2H2O → O2 + 4H+ + 4eReduction:
4H2O + 4e- → 2H2 + 4OHOverall:
E = -0.82 V
E = -0.41 V
6H2O(l) → 2H2(g) + O2(g) + 4H+(aq) + 4OH-(aq)
E = -1.23 V @ pH 7
Slide 34-12
Electrolysis of water
At standard state (all @ 1 mol/L): E 0 = -2.06 V
But at [H+] and [OH-] = 10-7 M : E = -1.23 V
So it should take between 1.23 and 2.06 V
to drive the electrolysis of water
Turns blue
Turns red
Oxidation:
2H2O → O2 + 4H+ + 4e-
Reduction:
4H2O + 4e- → 2H2 + 4OH-
Vmin = 1.23-2.06 V
but Vexp ≥ 3 V??
Slide 34-13
Over-potential
It is frequently observed that the required voltage for
electrolysis is higher than predicted using the reduction
potential tables. Several reasons:
Resistance in the electrical circuit;
Rate of electron transfer limited by electron transfer at the
electrode interface;
If half-reaction has a high barrier for electron transfer then
the rate is slowed.
Empirically:
Deposition and dissolution of metals usually involves only a very
small over-potential;
Production of gases at the electrode can require quite a high
over-potential. For example, H2 and O2 require an over-potential of
typically 0.4-0.6 V
Present theory is still unable to predict the size of the over-potential
Slide 34-14
Predicting electrolysis reactions
Given any combination of anode and cathode materials
and electrolyte, can you predict what reaction will
result?
Things to think about:
- Is the battery providing enough voltage?
- Which species are oxidised / reduced most easily?
- Is an over-voltage needed?
Slide 34-15
What can be electrolysed?
Cations in solution that have a more positive
reduction potential than water (E 0 > -0.41 to -0.82 V),
can be reduced.
Anions in solution that have a more positive oxidation
potential than water (E 0 > -0.82 to -1.23 V), can be
oxidised.
If the redox potential is in the “water range” then
probably both will happen.
Slide 34-16
Reduction potential table
Au3+(aq) + 3e− Au(s)
Half-cell potential
(V)
+1.50
+1.36
O2(g) + 4H+(aq) + 4e− 2H2O(l)
+1.23
O2(g) + 4H+(aq) + 4e− 2H2O(l)
+0.82 @ pH7
Ag+(aq) + e− Ag(s)
+0.80
I2(s) + 2e− 2I− (aq)
+0.51
Cu2+(aq) + 2e− Cu(s)
+0.34
2H+(aq) + 2e− H2(g)
0.00
Sn2+(aq) + 2e− Sn(s)
-0.14
2H2O(l) + 2e− H2(g) + 2OH−(aq)
-0.41 @ pH7
Fe2+(aq) + 2e− Fe(s)
-0.44
Zn2+(aq) + 2e− Zn(s)
-0.76
2H2O(l) + 2e− H2(g) + 2OH−(aq)
-0.83
Mg2+(aq) + 2e− Mg(s)
-2.37
Oxidised by electrolysis
Cl2(g) + 2e− 2Cl− (aq)
Reduced by electrolysis
Half-reaction
Slide 34-17
What can be electrolysed?
Cations in solution that have a more positive
reduction potential than water (E 0 > -0.41 to -0.82 V),
can be reduced.
Anions in solution that have a more positive oxidation
potential than water (E 0 > -0.82 to -1.23 V), can be
oxidised.
If the redox potential is in the water range then
probably both will happen.
So many metal ions can be reduced to metal,
e.g. Cu, Pb, Sn, Ni, Co, Cd, …
But many cannot,
e.g. Na, Mg, Al, Ti, Mn, …
Slide 34-18
Electrolysis of aqueous solutions
What products do you expect when the following
aqueous solutions are electrolysed (all at 1 M)?
a) KI;
b) AgNO3
a) Two possible reduction reactions:
c) MgSO4
pure water (pH7)
4H2O + 4e- → 2H2 + 4OH- E 0 = -0.41 ⇒ -0.81 to -1.01 V with over-pot’l
E 0 = -2.92 V
K+ + e- → K
KI electrolysis
Two possible oxidation reactions:
2I− → I2(s) + 2e-
2H2O → O2 + 4H+ + 4e-
E 0 = -0.51 V
E 0 = -0.82
⇒ -1.2 to -1.4 V
Therefore we expect H2 at the cathode
and I2 at the anode.
Slide 34-19
Electrolysis of aqueous solutions
b) 1 M AgNO3 ?
What will be reduced?
Ag+(aq) or H2O(l)
Check out their E 0 values…
What will be oxidised?
H2O(l) or NO3-.
NO3- cannot be oxidised
c) 1 M MgSO4 ?
What will be reduced?
Mg+(aq) or H2O(l)
What will be oxidised?
SO42-(aq) or H2O(l)
Slide 34-20
Faraday’s Laws of Electrolysis
Faraday’s first law:
The mass of substance liberated at an electrode
during electrolysis is proportional to the quantity of
charge (in Coulombs) passing through the electrolyte.
Faraday’s second law:
The number of Coulombs needed to liberate one mole
of different products are in whole number ratios.
charge (Q) = current (I) x time (t)
Coulombs = Amperes x seconds
Faraday 1:
mass ∝ charge
Faraday 2: Experimentally, the charge on 1 mol of electrons is 96,485
Coulombs. This value is now known as the “Faraday
constant” and given the symbol “F”
Slide 34-21
Example question:
In the electro-refining of Cu, what mass of Cu is
deposited in 1.00 hr, by a current of 1.62 Amp?
Approach:
1.
How many moles of electrons pass
through the circuit?
2. How many moles of copper can be
produced from this many electrons?
3. What is the mass of this many moles
of copper?
Michael Faraday
(1791-1867)
Slide 34-22
Example question:
1.
Q=Ixt
[1 hr = 3600 s]
= 1.62 x 3600
= 5830 C
96485 C per mol electrons, therefore
5830/96485 = 0.060 mol electrons
2. The copper half-reaction requires 2 electrons for each Cu:
Cu2+ + 2e- → Cu
Therefore 0.030 mol Cu produced
3. Atomic weight of Cu = 63.55 g/mol
therefore 0.030 x 63.55 = 1.92 g Cu.
Slide 34-23
Summary
CONCEPTS
Nernst equation
E 0 and K
Concentration cells
How pH meters work
CALCULATIONS
Work out cell potential for any concentration
(Nernst equation)
Work out K from E 0
Work out pH from concentration cell
Slide 34-24