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
Reduction potential table
Strong
oxidising
agent
Introduction to electrochemistry..
Half-reaction
Au3+(aq) + 3e− Au(s)
CONCEPTS
Revision of oxidation No., and redox;
Half-reactions
Galvanic cell
Cell potential (voltage)
CALCULATIONS
none from last lecture
Slide 25-3
Active and inactive electrodes
We have seen voltaic (galvanic) cells where the electrodes themselves
are part of the chemical reaction (remember the Zn electrode in the
Cu2+ solution?)
These are called “active electrodes”.
There are many species for which there is no appropriate electrode
material (remember the H2 electrode?). In these cases we use other
inert materials, such as C(graphite), or Pt as the electrode.
These are called “inactive electrodes”.
The inactive electrodes conduct electrons in and/or out of the halfcell, but do not take part in the reaction.
Electrodes (both active and inactive), are placed at the outside of the
shorthand notation for a voltaic cell.
Slide 25-5
Half-cell
potential (V)
Weak
oxidising
agent
Weak
reducing
agent
+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*
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
Strong
reducing
agent
Slide 25-4
Active and inactive electrodes
For example:
Half-reactions:
2I- I2 + 2eMnO4- + 8H+ + 5e- Mn2+ + 4H2O(l)
Overall reaction*:
10I-(aq) + 2MnO4-(aq) + 16H+(aq)
5I2(s) + 2Mn2+(aq) + 8H2O(l)
Cell notation (note phase
boundaries and electrodes):
graphite |I2(s)| I-(aq) || H+, MnO4-, Mn2+| graphite
* Check that you can balance redox reactions (see homework at end of lecture)
Slide 25-6
Standard Electrodes
Standard Electrodes
For routine measurements, it is nice to have one half-reaction that
we use all the time, and just change the other half-cell.
Several choices:
Normal calomel electrode, E 0 = 0.28 V
0
Saturated calomel electrode (SCE), E = 0.242 V
Silver/Silver chloride, E 0 = 0.22 V
Calomel: Hg2Cl2(s) + 2e− → 2Hg(l) + 2Cl−(aq);
E0=0.24V
SCE:
- Based on the reaction between Hg and
Hg2Cl2 (‘calomel’). The aqueous phase in
contact with Hg and Hg2Cl2 is a
saturated solution of KCl in water.
Q: The standard reduction potential of Zn2+/Zn is -0.76 V.
What would be the observed cell potential for the Zn/Zn2+
couple when measured using the SCE as a reference?
Calomel: Hg2Cl2(s) + 2e− → 2Hg(l) + 2Cl−(aq);
Zn → Zn2+ + 2e;
E0=0.24V
E0=0.76V
Ans: The Zn will be oxidised (lower reduction potential), so
E0(cell) = 0.76 + 0.24 = 1.00 V
So to get E 0 using the SCE as cathode, subtract 0.24 V from the reading.
- The electrode is linked via a porous
frit (salt bridge) to the solution in which
the other electrode is immersed.
Slide 25-7
Slide 25-8
The effect of concentration
Let’s explore what happens when the
concentration of the cell is changed…
calomel || I3-(aq), I-(aq) | graphite
[I3-] (M)
Eobs (V)
E0 (I3-,I-) (V)
0.2
0.05
0.33V
0.57V
0.2
0.005
0.29V
0.53V
0.05
0.005
0.35V
0.58V
[I-] (M)
Standard concentration
What happened in this experiment
when the solutions are diluted?
Dilution
Change in
potential (V)
Implication: We need to
stipulate concentration if
referring to cell potential.
constant I−,
I3− diluted
0.05M → 0.005 M
Calomel: Hg2Cl2(s) + 2e− → 2Hg(l) + 2Cl−(aq); E0 = 0.24 V
I3-,I-:
I3-(aq) + 2e- → 3 I-(aq);
E 0 = +0.53 V
2Hg(l) + 2Cl−(aq) + I3-(aq) → Hg2Cl2(s) + 3I-(aq);
E0 = +0.29V
Conclusion: The cell
potential depends on
concentration.
Corollary: It is useful to
define a standard
concentration, which is 1 M
constant I3−,
I− diluted
0.2M → 0.05 M
To get E 0 using the SCE as anode, add 0.24V to the readingSlide 25-9
The effect of concentration
Slide 25-10
The effect of concentration
I3-(aq) + 2e- 3 I-(aq); E 0 = +0.53 V
Do these results make sense in terms of
what you have learned about equilibrium?
I3-(aq)+ 2e- 3 I-(aq) ; E 0 = +0.53 V
Eobs = E 0 − A × log
A = 2.303 ×
1. Reduced [I-], reaction shifts to right, cell
potential increases.
RT
nF
The factor of 2.303
comes from using log10
rather than loge or ln.
where R = 8.314 J
T = temperature (K)
n = number of electrons transferred per mole of reagent
F = charge on 1 mol of electrons = 96485 C = “Faraday constant”
K-1
2. Reduced [I3-], reaction shifts to left, cell
potential decreases.
mol-1
Ecell = E 0 − 2.303 ×
Can we quantify this?…
 [I − ]3 

− 
 [I3 ] 
The equation that fits the
experimental measurements is:
Slide 25-11
 [I − ]3 
RT

log
− 
nF
 [I3 ] 
Slide 25-12
Nernst Equation
Nernst Equation
I3-(aq) + 2e- 3 I-(aq); E 0 = +0.53 V
Ecell = E 0 − 2.303 ×
“Nernst equation”
 [I − ]3 
RT

log
− 
nF
 [I3 ] 
Q =
The ratio looks like something we have seen before:
“Nernst equation”
Ecell = E 0 − 2.303×
[I − ]3
[I3 − ]
Ecell = E 0 − 2.303×
RT
log(Q)
nF
Walther Nernst derived this equation in 1889 at the age of 25. He also
derived the 3rd Law of Thermodynamics and explained the principle of the
solubility product. He was the awarded the Nobel Prize in Chemistry in
1920.
RT
log(Q)
nF
At 25ºC,
2.303× RT
Slide 25-13
F
= 0.0592V
Ecell = E 0 −
0.0592
n
log(Q )
Slide 25-14
Example calculation (1)
Use the Nernst Equation to calculate the expected cell
potential for two similar experiments to the ones in the
last lecture, i.e.
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
E0 = 1.1 V
i) [Cu2+] = 1.0 M; [Zn2+] = 10-5M
ii) [Cu2+] = 10-5M; [Zn2+] = 1.0 M
0.0592
Ecell = E 0 −
log(Q )
n
Firstly, work out the value of n :
Cu2+ + 2e− Cu
n=2
Slide 25-15
Example calculation (1)
0.0592
n
log(Q )
Ecell = E 0 −
(n=2)
Q =
log(Q)
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Slide 25-16
0.0592
n
log(Q )
where n is the number of moles of
electrons transferred in the reaction
(2 in our case above for Zn/Cu).
[Zn 2 + ]
[Cu 2 + ]
Ecell (V)
[Zn2+]
(M)
1.0
[Cu2+]
(M)
1.0
1.0
0.0
1.10
10-5
1.0
10-5
-5.0
1.25
1.0
10-5
105
5.0
0.95
Q
2 mol e- transferred
per mole of reaction
Zn2+ + 2e−
Nernst Equation
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Ecell = E 0 −
Zn
Slide 25-17
Slide 25-18
E 0 and K
E 0 and K
Q: What do you think happens as the reaction
proceeds until equilibrium is reached?
A: The reaction stops, therefore the voltage, or
electrical potential, is zero (the battery is flat!)
In mathematical terms:
Eobs = E 0 −
0.0592
n
Let’s examine the relationship between E 0
and K…
log(Q ) = 0
At eq’m Q=K
E0 =
0.0592
n
log(K )
So the equilibrium constant determines the cell potential!
Large K products favoured large cell potential, E
Slide 25-19
Slide 25-20
Example question (2)
Example question (2)
Q: A voltaic cell consisting of a Ni/Ni2+ half-cell and
Co/Co2+ half-cell is constructed with the following
initial concentrations: [Ni2+] = 0.80M; [Co2+]=0.2M.
a) What is the initial Ecell?
b) What is the [Ni2+] when the voltage reaches 0.025V?
c) What are the equilibrium concentrations of the ions?
Given: E 0 (Ni2+|Ni) = -0.25 V; E 0 (Co2+|Co) = -0.28 V
a) What is the initial Ecell?
Q=
[Co 2+ ]
[Ni 2+ ]
=
0.2
= 0.25
0.8
Ecell = E 0 −
0.0592
= 0.03 −
Ni2+ + 2e- → Ni
E 0 = -0.25V
Co → Co2+ + 2e-
E 0 = +0.28V
Co(s) + Ni2+(aq) Co2+(aq) + Ni(s)
0.0592
n
log(Q )
Co(s) + Ni2+(aq) Co2+(aq) + Ni(s)
0.80-x
Q = 1.47 =
Slide 25-22
Example question (2)
0.005
= 0.169
0.0296
Q = 1.47
1.176 − 1.47x = 0.2 + x
2.47 x = 0.976
x = 0.40
What are the equilibrium concentrations of the ions?
E0 =
0.0592
log(K )
n
0.80-x
0.20+x
1.47 (0.8 − x ) = (0.2 + x )
c)
Co(s) + Ni2+(aq) Co2+(aq) + Ni(s)
[Co 2+ ]
( 0.2 + x )
=
[Ni 2+ ] (0.8 − x )
0.0592
log( 0.25)
2
Slide 25-21
0.025 = 0.03 − 0.0296× log(Q )
log(Q ) =
log(Q )
= 0.048V
E 0 = 0.03 V
What is the [Ni2+] when the voltage reaches 0.025V?
Ecell = E 0 −
n
= 0.03 − 0.0296 × ( −0.602)
Example question (2)
b)
E 0 = 0.03 V
Co(s) + Ni2+(aq) Co2+(aq) + Ni(s)
So when Ecell = 0.025 V
[Co2+] = 0.60 M
[Ni2+] = 0.40 M
Slide 25-23
0.03
= 1.014
0.0296
K = 10.24
0.20+x
[Co 2+ ]
(0.2 + x )
=
[Ni 2+ ] ( 0.8 − x )
10.24(0.8 − x ) = (0.2 + x )
K = 10.24 =
0.03 = 0.0296 × log(K )
log(K ) =
8.192 − 10.24x = 0.2 + x
11.24x = 7.986
x = 0.71
So at equilibrium,
[Co2+] = 0.91 M
[Ni2+] = 0.09 M
Slide 25-24
Summary
CONCEPTS
Half-reaction
Table of standard reduction potential
Effect of concentration
Link between E and Q
CALCULATIONS
The lecture should finish about here.
After this are some slides on
balancing redox equations for home
revision.
Work out cell potential from reduction potentials;
Work out cell potential for any concentration
(Nernst equation)
Slide 25-25
Slide 25-26
Balancing Redox Reactions
Balancing Redox Reactions
Cr2O72−(aq) + I−(aq) → Cr3+(aq) + I2(s)
Step 2: Balance the atoms and charges in
each half reaction:
Cr2O72− → Cr3+
Step 1: Determine the half-reactions.
Cr2O72− → Cr3+
I− → I2
Cr2O72− → 2Cr3+
Step 2: Balance the atoms and charges in
each half reaction:
Balance atoms other than O and H
Balance O by adding H2O
Balance H by adding H+
Balance charge by adding e−
Balance O by adding H2O
Cr2O72− → 2Cr3+ + 7H2O
Balance H by adding H+
Balance charge by adding e−
14H+ + Cr2O72− → 2Cr3+ + 7H2O
Reduction
Slide 25-27
Step 2: Balance the atoms and charges in
each half reaction:
Balancing Redox Reactions
I− → I2
Step 3: Multiply each half reaction by an integer
so that the number of e− in each reaction are the
same.
6e− + 14H+ + Cr2O72− → 2Cr3+ + 7H2O
(2I− → I2 + 2e−) ×3
Balance atoms other than O and H
2I− → I2
6e− + 14H+ + Cr2O72− → 2Cr3+ + 7H2O
Slide 25-28
Balancing Redox Reactions
Balance atoms other than O and H
Step 4: Add the balanced half reactions and
include the states of matter.
6e− + 14H+ + Cr2O72− → 2Cr3+ + 7H2O
6I− → 3I2 + 6e−
Balance charge by adding e−
Oxidation
2I− → I + 2e−
____________________________________________________________________________________________________
2
6I−(aq) + 14H+(aq) + Cr2O72−(aq) → 3I2(s) + 7H2O(l) + 2Cr3+(aq)
• Step 5: Check the atoms and charges are balanced.
Slide 25-29
Reactants (6I, 14H, 2Cr, 7O; 6+) → Products (6I, 14H, 2Cr, 7O; 6+)
Slide 25-30
Balancing Redox Reactions in
Basic Solutions
Balance the following reaction under basic conditions:
−
2−
2−
MnO4 (aq) C2O4 (aq) → MnO2(s) + CO3 (aq)
Initially balance the reaction as an acidic reaction
C2O42− → CO32−
Identify ½ Eqn
C2O42− → 2CO32−
Balance C
2−
2−
C2O4 + 2H2O → 2CO3
Balance O
2−
2−
+
C2O4 + 2H2O → 2CO3 + 4H
Balance H
2−
2−
+
−
C2O4 + 2H2O → 2CO3 + 4H + 2e
Balance e−
STEPS
Loss of electrons = OXIDATION
1 & 2
MnO4− → MnO2
Identify ½ Eqn
−
MnO4 → MnO2 + 2H2O
Balance O
MnO4− + 4H+ → MnO2 + 2H2O
Balance H
−
+
−
MnO4 + 4H + 3e → MnO2 + 2H2O
Balance e−
Gain of electrons = REDUCTION
Slide 25-31
Balancing Redox Reactions in
Basic Solutions
One additional step is necessary at Step 4.
The acidic result would be:
2MnO4− + 2H2O + 3C2O42− → 2MnO2 + 6CO32− + 4H+
Add OH− to each side to cancel the H+, therefore add 4OH−.
2MnO4− + 2H2O + 3C2O42− + 4OH → 2MnO2 + 6CO32− + (4H+ + 4OH−)
2MnO4− + 2H2O + 3C2O42− + 4OH− → 2MnO2 + 6CO32− + 4H2O 2H2O
2MnO4−(aq) + 3C2O42−(aq) + 4OH−(aq) → 2MnO2(s) + 6CO32−(aq) + 2H2O(l)
Slide 25-33
Balancing Redox Reactions in
Basic Solutions
STEP 3
C2O42− + 2H2O → 2CO32− + 4H+ + 2e−
MnO4− + 4H+ + 3e− → MnO2 + 2H2O
×3
×2
3C2O42− + 6H2O → 6CO32− + 12H+ + 6e−
2MnO4− + 8H+ + 6e− → 2MnO2 + 4H2O
STEP 4
2H2O
4H+
3C2O42− + 6H2O → 6CO32− + 12H+ + 6e−
2MnO4− + 8H+ + 6e− → 2MnO2 + 4H2O
2MnO4− + 2H2O + 3C2O42− → 2MnO2 + 6CO32− + 4H+
Slide 25-32