The Nernst Equation
Galvanic and Electrolytic Cells
1.
Galvanic cells and Electrolysis Cells
: in an electrolysis cell, the cell
reaction runs in the non--spontaneous direction electrolysis of water
2.
Thermodynamics of the relationship of DE and DG, electrical work
DG = - n F DE
3.
Concentration dependence of the cell potential, the Nernst Equation, the
pH meter, the glass electrode, the calomel electrode
4.
Adding Half-Cell potentials (problem 12.23a)
5.
Disproportionation reactions (problem 12.23b)
Galvanic Cells and Electrolysis Cells
Galvanic cell: cell reaction runs in the spontaneous direction
Electroytic cell: cell reaction runs in the non-spontaneous direction, driven by an applied
electrical potential (a voltage) -that is greater than the spontaneous potential
negative
anode if galvanic
cathode if electrolytic
positive
Ag(s)
Cu(s)
Cu2+(aq)
salt bridge
Ag+(aq)
galvanic: Eo= 0.+4554 V
Cu(s) + 2 Ag+(aq)
Cu2+(aq) + 2 Ag(s)
electrolytic: E (applied) > Eo
cathode if galvanic
anode if electrolytic
Cathode and Anode in Galvanic Cells and Electrolysis Cells
By definition, the cathode is where reduction occurs and the anode is where oxidation
occurs.
In the galvanic cell, the copper electrode is the anode, silver is the cathode. In the
electrolytic cell, the silver electrode is the anode, copper is the cathode.
Ag(s)
(positive)
Cu(s)
(negative)
Cu2+(aq)
salt bridge
Ag+(aq)
Note that the sign of the potential doesn't change - copper is always negative, silver is always
positive. But the designations "cathode" and "anode" change between galvanic and
electrolytic cells.
The Galvanic Cell
This galvanic cell “burns” H2(g) and O2(g) to form water:
voltmeter
cathode
Pt(s)
-
+
anode
Pt(s)
H2(g)
O2(g)
salt bridge
H3O+(aq)
O+(aq)
H3
2H3O3+(aq) + 2e-
Overall cell reaction:
H2(g) + 2H2O
2 H2(g) + O2(g)
O2(g) + 4 H3O+(aq) + 4 e-
2 H2O (aq)
6 H2O (l)
DEo=+1.229 V
The Electrolysis Cell
The electrolysis cell runs in the non-spontaneous direction, driven by an
external voltage source:
power supply: E > DEocell
-
anode Pt(s)
+
Pt(s)
H2(g)
cathode
O2(g)
salt bridge
H3O+(aq)
H3O+(aq)
2H3O3+(aq) + 2e-
H2(g) + 2H2O
Reaction in the electrolysis cell:
O2(g) + 4 H3O+(aq) + 4 e-
2 H2(g) + O2(g)
2 H2O (aq)
6 H2O (l)
Faraday's Laws of Electrolysis
1.
Mass is proportional to electric charge passed through the cell.
2.
Equivalent masses of different substances require equal amounts of electric
charge passed through the cell.
An electrical current of 1 ampere equals one coulomb per second Q (coulombs) = I (amperes) * t (time)
1 mole of electrons has a charge of -96,485 C (the Faraday)
moles of e- = (coulombs passed through cell) / 96485
= (I*t ) / 96,485
3 Cu(s) + 2 Au3+(aq)
3 Cu2+(aq) + 2 Au(s)
How much gold is deposited if 100 A is passed through this
electrolysis cell for an hour?
(Ans. 1.24 moles or 245 g)
Thermodynamics of Electrochemical Cells
DG = DH - TDS
(const T)
= DE + pDV - TDS
(const P)
= (qrev + w) + pDV - qrev
(reversible, const T)
w
=
- pDV
+
expansion
work
wel = - Q DE
= - n F DE
wel,rev
electrical
work
work = charge * (potential difference)
Joules
Coulombs
Volts
DG = wel,rev
= - n F DE
(const T, P, reversible)
Concentration Effects and The Nernst Equation
DGo is the standard DG - all solute concentrations must be 1 M
and partial pressures must be 1 atm.
DG = DGo + RT ln Q - this allows you to calculate DG at
non-standard concentrations and
partial pressures.
DG = - n F DEcell
- this is always true; the cell potential is a very
fundamental quantity, just like DG
DEcell = DEocell - (RT/nF) lne Q - the Nernst equation. It allows you to
calculate DG at non-standard conditions
Measuring Equilibrium Constants
DGo = - RT lne K
DG = - n F DEcell
- K can be calculated from DGo, which
for many compounds can be obtained
from thermodynamic tables.
- this is always true; the cell potential is a
very fundamental quantity, just like DG
DEo = (RT/nF) lne K
- For reactions in solution, K and
can be measured from a standard cell potential.
DGo,
DEcell = DEocell - (RT/nF) lne Q - The Nernst equation relates DEcell to DEocell
The Nernst Equation
DEcell = DEocell - (0.0592/n) log10 Q
at T = 25oC
PbO2(s) + SO42-(aq) + 4 H3O+(aq) + 2 e-
PbSO4(s) + 6 H2O(l)
Eo=+1.685 V
E=
Eo
- (.0592/2) log10
1
O+]4
[H3
Pt(s)
2-]
[SO4
E changes by 0.118 V for each order of magnitude
change in [H3O+] or by 0.0296 V for each order of
magnitude change in [SO42-]:
SO42-(aq)
pH = 0
E = +1.685 V
pH = 4
E = +1.211 V
H3O+(aq)
slurry of
PbO2(s) +
PbSO4(s)
pH measurement
Standard Hydrogen Electrode:
2 H3O+(aq, 1M) + 2 e-
H2(g) + 2 H2O(l)
E = +0.241 V - (.0592/2) log10 [H3O+]2
Calomel Electrode:
Hg2Cl2(s) + 2 e-
= +0.241 + .0592 * pH
2Hg(l) + 2 Cl-(sat’d)
Pt(s)
Ecell  pH
Pt(s)
H2(g)

salt bridge
sat’d
KCl(aq)
H3O+(aq)
standard hydrogen electrode
E° = +0.000 V
paste of Hg(l)
and Hg2Cl2(s)
KCl(s)
calomel reference electrode
E° = +0.241 V
The Glass Electrode
and the pH Meter
The glass electrode is non-metallic
electrode composed of a special glass
that is porous to H3O+.
A diffusion potential develops across
the membrane in response to a pH
gradient. The potential varies by 59.6
mV for each unit change in pH.
Ecell = Eocell -
0.0592 V
[H+]out2
log10
2
[H+]in2
Ecell = constant + (0.0592 v) * pH
Notice that Ecell is uniformly sensitive
to pH over 14 orders of magnitude!!
Adding Half-Cell Potentials
12.23 a. Calculate the half-cell potential Eo for the half-reaction
Cu2+ + e- = Cu+
E3o = ??
using these tabulated potentials:
Cu2+ + 2 e- = Cu(s)
E1o = +0.340
Cu+ + e- = Cu(s)
E2o = +0.522
You can’t simply add the potentials because the numbers of electrons differ.
However, you can always add DGo’s:
Cu2+ + 2 e- = Cu(s)
DG1o = -n F E1o
Cu+ + e- = Cu(s)
DG2o = -n F E2o
Cu2+ + e- = Cu+
DG3o = DG1o - DG2o
Adding Half-Cell Potentials (Problem 12.23 a)
12.23 a. Calculate the half-cell potential Eo for the half-reaction
Cu2+ + e- = Cu+
E3o = ??
using these tabulated potentials:
Cu2+ + 2 e- = Cu(s)
E1o = +0.340
Cu+ + e- = Cu(s)
E2o = +0.522
DG3o = DG1o - DG2o
n3 F E3o = n1 F E1o - n2 F E2o
E3o =
n1 E1o - n2 E2o
n3
Ans: + 0.158 v
Reduction Potential Diagrams and Disproportionation
same species is both
oxidized and reduced
12.23 b. Will Cu+(aq) disproportionate in aqueous solution?
2 Cu+(aq) = Cu(s) + Cu2+(aq)
The oxidation-reduction reactions of Cu2+/Cu+/Cu(s) are described on a
reduction potential diagram of the form:
Cu2+
0.158 v
Cu+
0.522 v
Cu(s)
0.342 v
Disproportionation occurs only if the potential to the right of Cu+ is greater
than that to the left of Cu+
Cu2+ + e- = Cu+
Cu+ + e- = Cu(s)
Reduction Potential Diagrams and Disproportionation (ii)
Cu2+
0.158 v
Cu+
0.522 v
Cu(s)
0.342 v
Disproportionation occurs only if the potential to the
right of Cu+ is greater than that to the left of Cu+
Why?
Disproportionation requires that you run
this backward
Cu2+ + e- = Cu+
and
this forward
Cu+ + e- = Cu(s)
which occurs only if this Eo is more positive than this Eo
Ans: yes