Chem 1101
A/Prof Sébastien Perrier
Room: 351
Phone: 9351-3366
Email: [email protected]
Prof Scott Kable
Unless otherwise stated, all images in this file have been reproduced from:
Room: 311
Blackman, Bottle, Schmid, Mocerino and Wille,
Chemistry, 2007 (John Wiley)
ISBN: 9 78047081 0866
Phone: 9351-2756
Email: [email protected]
A/Prof Adam Bridgeman
Room: 222
Phone: 9351-2731
Email: [email protected] 34-2
Slide 34-1
Highlights of last lecture
Concentration cells
Voltaic cells...
CONCEPTS
Half-reaction
Table of standard reduction potential
Effect of concentration
Link between E , Q and K
CALCULATIONS
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?
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 34-4
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 −
Cu → Cu2+ (y M) + 2e(0.1M) + 2e- → Cu
Cu2+ (0.1M) → Cu2+ (y M)
E 0=-0.34
E 0=+0.34
E = 0.0V
????
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+ + 2eCu2+ + 2e- → Cu
Slide 34-3
Concentration cells
Afterall:
0.0592
n
 [H + ]2 

log(Q ) = −0.0296 × log

 1.0 
+
Ecell = ?
In case you
missed it!
= −0.0592 log[H ]
Ecell = ____V
0.75
pH ≡ -log[H+]
0.0592
 y  0.75
Ecell = E 0 −
log(Q ) = 0.0 − 0.0296 log
 = ____ V
 0.1 
n
1E-26 M
Solve for y: _______________________
Slide 34-5
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
Other applications of voltaic cells
concentration gradients produce electrical current
ion pumps across cell membranes
Na+ / K+ pump, Ca2+ pump
energy production and storage in cells
ATP
pH meters
these usually use a silver/silver chloride reference rather
than SHE
ion selective electrodes
Batteries (later lecture)
Refs: Most General Chemistry texts have good sections on
batteries, e.g.
Corrosion (later lecture)
Ref: Most General Chemistry texts also have a section on
corrosion:
different choice of reference can then be specific to
specific ions.
Silberberg, pp.923-5
McMurray and Fay, pp.781-6
Petrucci, Harwood and Herring, pp.844-848
However, Housecroft and Constable does not(!)
Silberberg, pp.926-8
McMurray and Fay, pp.786-9
Petrucci, Harwood and Herring, pp.849-50
Again, Housecroft and Constable does not(!)
Slide 34-7
Electrolysis
Slide 34-8
Comparison – voltaic vs electrolytic
Electrolytic Cell
Voltaic Cell
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…
Zn→
→Zn2++2e-
Cu2++2e-→Cu
Zn2++2e-→Zn
Cu→
→Cu2++2e-
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
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.
Anode = oxidation = negative
Anode = oxidation = positive
Cathode = reduction = positive
Cathode = reduction = negative
Ecell > 0
Ecell < 0
Electrolysis of water
At standard state (all @ 1 mol/L):
Oxidation:
2H2O → O2 + 4H+ + 4eReduction:
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.
4H2O + 4e- → 2H2 + 4OH-
E 0 = -1.23 V
E 0 = -0.83 V
6H2O(l) → 2H2(g) + O2(g) + 4H+(aq) + 4OH-(aq)
Overall:
Oxidation ALWAYS occurs at the anode.
Slide 34-10
E = -2.06 V @ [H+] = [OH-] = 1 M
0
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
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
Over-potential
At standard state (all @ 1 mol/L): E 0 = -2.06 V
But at [H+] and [OH-] = 10-7 M : E = -1.23 V
It is frequently observed that the required voltage for
electrolysis is higher than predicted using the reduction
potential tables. Several reasons:
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-
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
Vmin = 1.23-2.06 V
but Vexp ≥ 3 V??
Slide 34-13
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?
What can be electrolysed?
Things to think about:
- Is the battery providing enough voltage?
- Which species are oxidised / reduced most easily?
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.
- Is an over-voltage needed?
Slide 34-15
Slide 34-16
Reduction potential table
Half-cell potential
(V)
+1.50
+1.36
+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)
O2(g) + 4H+(aq) + 4e− 2H2O(l)
Reduced by electrolysis
Half-reaction
Au3+(aq) + 3e− Au(s)
What can be electrolysed?
Slide 34-17
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
Electrolysis of aqueous solutions
b) 1 M AgNO3 ?
c) MgSO4
What will be reduced?
pure water (pH7)
a) Two possible reduction reactions:
Ag+(aq) or H2O(l)
E 0 = -0.41 ⇒ -0.81 to -1.01 V with over-pot’l
4H2O + 4e- → 2H2 + 4OH-
E 0 = -2.92 V
K+ + e- → K
What will be oxidised?
KI electrolysis
H2O(l) or NO3-.
Two possible oxidation reactions:
2I−
→ I2(s) +
2e-
2H2O → O2 + 4H+ + 4e-
E = -0.51 V
0
⇒ -1.2 to -1.4 V
Faraday’s first 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)
What will be oxidised?
SO42-(aq) or H2O(l)
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?
Coulombs = Amperes x seconds
Faraday 1:
Mg+(aq) or H2O(l)
Slide 34-20
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:
What will be reduced?
Slide 34-19
Faraday’s Laws of Electrolysis
NO3- cannot be oxidised
c) 1 M MgSO4 ?
E 0 = -0.82
Therefore we expect H2 at the cathode
and I2 at the anode.
Check out their E 0 values…
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:
1.
Q=Ixt
Michael Faraday
(1791-1867)
Slide 34-22
Summary
[1 hr = 3600 s]
CONCEPTS
= 1.62 x 3600
= 5830 C
96485 C per mol electrons, therefore
5830/96485 = 0.060 mol electrons
Nernst equation
E 0 and K
Concentration cells
How pH meters work
2. The copper half-reaction requires 2 electrons for each Cu:
Cu2+ + 2e- → Cu
CALCULATIONS
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
Work out cell potential for any concentration
(Nernst equation)
Work out K from E 0
Work out pH from concentration cell
Slide 34-24