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
Slide 25-2
Email: [email protected]
Slide 25-1
Highlights of last lecture
2 cases from last 2 lectures…
Equilibrium and thermochemistry in production
of the world’s Top 10 chemicals
Aluminium metal cannot be refined
using smelting;
NaOH and Cl2 production;
CONCEPTS
Control of equilibrium in industrial processes.
Energy recycling
Endo- and exothermic processes in industry
Methods of manufacture of H2SO4 and NH3
Uses of these chemicals
An understanding of both these processes
requires knowledge of electrochemistry.
CALCULATIONS
Industrial-type equilibrium and
thermochemistry problems
Slide 25-3
Slide 25-4
Electrochemistry
Themes:
Electrochemistry
generating electricity from
chemical reactions
Key chemical concepts:
•
Redox and half reactions;
driving unfavorable chemical
reactions using electricity
•
Cell potential;
References: Any general chemistry text will
have a satisfactory chapter on
redox and electrochemistry, e.g.
Blackman, et al.: Chap 12
Slide 25-5
•
Voltaic and electrolytic cells;
•
Concentration cells.
Key Calculations:
•
Calculating cell potential;
•
Calculating amount of product for given current;
•
Using the Nernst equation for concentration cells.
Slide 25-6
Activity series of metals
Half-Reactions
LEO, the lion…
… goes GER
Loss of Electrons = Oxidation
Gain of Electrons = Reduction
OXIDATION
REDUCTION
Gain of Electrons
Oxidizing agent is
reduced
Oxidation number
decreases
An example:
2H+(aq) + 2e− → H2(g)
Loss of Electrons
Reducing agent is
oxidised
Oxidation number
increases
An example:
Zn(s) → Zn2+(aq) + 2e−
Consider the same reaction for
5 metals:
Cu(s) +
2H+(aq)
→
Cu2+(aq)
Observations
+ H2(g)
Sn(s) + 2H+(aq) → Sn2+(aq) + H2(g)
Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g)
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)
Slide 25-7
Activity series of metals
Consider the same reaction for
5 metals:
Cu(s) +
2H+(aq)
Sn(s) +
2H+(aq)
→
Cu2+(aq)
+ H2(g)
→
Sn2+(aq)
+ H2(g)
Slide 25-8
Activity series of metals
Observations
Activity of metals, compared to H+:
Mg > Zn > Fe > Sn > Cu
Consider:
(1)
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
(2)
(1)-(2)
Fe(s) + 2H+(aq) → Fe2+(aq) + H2(g)
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)
Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Which direction? Zn is a stronger reducing agent than Cu, so reaction will
proceed to right.
Movie 32_1
Activity: Mg > Zn > Fe > Sn > Cu as reducing agents
Movie 32_1 cont’d
Slide 25-9
Activity series of metals
Slide 25-10
Half reactions
In a redox reaction we consider the electrons to be transferred from
one element to another (obvious in this case).
The molecular interpretation…
Can these
electrons be
harnessed?
Consider the two half-reactions:
Zn
Cu2+ + 2e-
Zn2+
2e-
→
+
→ Cu
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Note: state
labels are
left off halfreactions
If we can separate the half-reactions then we can
harness the electrons.
Movie 32_2
Figure 12.1 and 12.2 Blackman
Slide 25-11
Slide 25-12
Galvanic cells
Galvanic cells
~1.1 V
Movie 32_3
Slide 25-13
Slide 25-14
Galvanic cells
Short-hand nomenclature
Anode:
Cathode:
Oxidation is occuring
Reduction is occuring
Negative terminal
Positive terminal
Rather than writing out the full chemical
equation, we can use a short-hand notation:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Salt bridge
Anode half-cell
Cathode half-cell
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
Phase boundary
Electrons flow this way
Slide 25-15
Slide 25-16
Cell potentials
Galvanic cells
all reagents at standard
concentration of 1.0 M
2H+(aq)
Potential(E0)
Exp’t 1:
Zn(s) +
Exp’t 2a:
Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)
Exp’t 2b:
Cu2+(aq) + H2(g) → Cu(s) + 2H+(aq)
Exp’t 3:
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
→
Zn2+(aq)
Phase boundary
+ H2(g)
+0.76V
-0.34V
+0.34V
+1.10V
= Exp’t 1 + Exp’t 2b
Conclusion 1: Reverse the reaction – reverse the sign
of the potential
Every different galvanic cell has a different cell
potential.
We have seen that you can add cell potentials, so…
We could tabulate the cell potential against H2 and
then calculate any other combination…
or, we could assign a “half-cell” potential to each
“half-reaction”, and then add up two half-reactions to
get the full reaction and full potential.
Let’s see how this is done…
Conclusion 2: You can add cell potentials when you
add chemical reactions.
Slide 25-17
Slide 25-18
Half-cell potentials
Half-cell potentials
Consider the half-reactions for each exp’t:
ε0 (V)
Zn → Zn2+ + 2e2H+ + 2e- → H2
Calculated: +0.76V
Definition: +0.00V
Observed: +0.76V
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
Cu2+ + 2e- → Cu
H2 → 2H+ + 2e-
Calculated: +0.34V
Definition: +0.00V
Cu2+(aq) + H2(g) → Cu(s) + 2H+(aq)
Observed: +0.34V
Zn → Zn2+ + 2eCu2+ + 2e- → Cu
Calculated: +0.76V
Calculated: +0.34V
Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq)
Calculated: +1.10V
Slide 25-19
Slide 25-20
Reduction potential table
Strong
oxidising
agent
Half-reaction
Au3+(aq) + 3e− Au(s)
Cl2(g) +
Weak
oxidising
agent
Half-cell
potential (V)
2e−
Using the reduction tables
Weak
reducing
agent
+1.50
2Cl− (aq)
+1.36
O2(g) + 4H+(aq) + 4e− 2H2O(l)
+1.23
Ag+(aq) + e− Ag(s)
+0.80
Cu2+(aq) + 2e− Cu(s)
+0.34
2H+(aq) + 2e− H2(g)
0.00*
Sn2+(aq) + 2e− Sn(s)
-0.14
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
* by definition
Write down the two half-reactions
Work out which is the oxidation and which
is the reduction half-reaction.
Balance the electrons
Add up the half-reactions to get full
reaction
Add up half-cell potentials to get E 0
A spontaneous (working) voltaic cell,
ALWAYS has a positive cell potential, E 0
Strong
reducing
agent
Slide 25-21
Slide 25-22
Example calculation (1)
Reduction potential table
What will the cell potential be for a galvanic cell with Mg2+(aq) | Mg(s)
in one half-cell and
Sn2+(aq)
| Sn(s) in the other?
Sn2+ + 2e- → Sn
E 0 = -0.14V
+
→ Mg
E
Which half reaction is turned around?
Mg2+
2e-
0
Strong
oxidising
agent
= -2.37V
We saw that Mg is a stronger reducing agent than Sn, i.e. it likes to
be oxidised. Turn the Mg reaction around to an ox. reaction.
Sn2+
+
2e-
→ Sn
Mg → Mg2+ + 2eMg(s) + Sn2+(aq) → Mg2+(aq) + Sn(s)
Half-reaction
Au3+(aq) + 3e− Au(s)
Half-cell
potential (V)
+1.50
Cl2(g) + 2e− 2Cl− (aq)
+1.36
O2(g) + 4H+(aq) + 4e− 2H2O(l)
+1.23
Ag+(aq) + e− Ag(s)
+0.80
Cu2+(aq) + 2e− Cu(s)
+0.34
2H+(aq) + 2e− H2(g)
0.00*
E = -0.14V
Sn2+(aq) + 2e− Sn(s)
-0.14
E 0 = +2.37V
Fe2+(aq) + 2e− Fe(s)
-0.44
0
E 0 = +2. 23V
In general, you turn around the reaction
lowest in the table of reduction potentials
Slide 25-23
Weak
oxidising
agent
Zn2+(aq) + 2e− Zn(s)
-0.76
2H2O(l) + 2e− H2(g) + 2OH−(aq)
-0.83
Mg2+(aq) + 2e− Mg(s)
-2.37
* by definition
Weak
reducing
agent
Strong
reducing
agent
Slide 25-24
Example calculation (2)
Summary
What will the cell potential be for a galvanic cell with Ag+(aq) | Ag(s) in
one half-cell and Cr3+(aq) | Cr(s) in the other?
Data: Ag+(aq)|Ag(s) = +0.80V;
Cr3+(aq)|Cr(s) = -0.74V
Cr half reaction is lowest (most negative), turn it around…
Ag+ + e- → Ag
E 0 = +0.80V
Cr → Cr3+ + 3eBalance the electrons…
3Ag+ + 3e- → 3Ag
Cr → Cr3+ + 3eCr(s) +
3Ag+(aq)
Note: E
→
0
Cr3+(aq)
+ 3Ag(s)
E 0 = +0.74V
0
Note!
E
CONCEPTS
Oxidation and reduction
Half-reactions
Galvanic cell
Cell notation
= +0.80V
CALCULATIONS
E 0 = +0.74V
E = +1.54V
0
does not depend on stoichiometry!
Slide 25-25
Work out cell potential from reduction
potentials;
Slide 25-26