Title:
The Thermodynamics of a Galvanic Cell
Aim:
(a) To Verify the Nernst Equation
(b) To Measure S, H, and G for a cell reaction
Intro:
The electrochemical cell used on this experiment is comprised of a standard zinc
electrode, which provides a reference throughout, and a redox electrode involving
platinum metal immersed in the redox system to be investigated. The overall cell reaction
is given by:
Zn + 2 [ Fe(CN)6 ] 3Zn 2+ + 2 [Fe(CN)6]4Reaction for the Cathode:
Zn2+ + 2e-
Zn
The electrode potential of this half-cell, Ec, is expressed by the Nernst Equation:
E c = Ec 0 +
.
RT
ln
nF
[ Zn2+ ]
[Zn]
F is the Faraday constant and n is the number of electrons involved in the cell reaction.
Since the activity of the cathode remains at 1.0 during the experiment, the electrode
potential of this half-cell Ec is always equal to E0c.
Reaction for the Anode:
[Fe(CN)6]3- + e-
[Fe(CN)6]4-
ferricyanide
Ea = Ea
.
0
+ R T ln
nF
ferrocyanide
3+
[Fe ]
[Fe2+]
Complete Cell
The overall cell equation depends on the redox processes occurring in both half-cells.
Hence, the EMF of the cell is the difference between the electrode potentials of the anode
and cathode half-cells:
Ecell = E0cell + R T ln [ Fe3+ ]
.
.
nF
[ Fe2+ ]
The standard Gibbs free energy change for a cell with reversible electrodes is
G = - n F E0cell
Since G is also given by:
G = H - TS (  )
then:
and so
S = n F
-n F E0cell = H - TS
d E0cell
dT
()
()
P
assuming that H changes only slightly with temperature. Hence, following E0cell as a
function of temperature, S may be determined from the slop of the E0cell vs. T plot.
Since G and S are known, H may then be determined.
Pre-Practical Questions
1. The salt bridge completes the electric circuit within the cell. To keep the solutions
from mixing, a salt bridge is used to join the two electrode compartments and
complete the electrical circuit.
2.
Zn (s) | Zn 2+ (aq) || [ Fe(CN)6 ] 3-(aq) | [ Fe(CN)64- (aq)
Procedure:
Constant Temperature Measurements
Prepare 100 cm3 of 0.2 M potassium ferricyanide solution K3Fe(CN)6. Rinse the Zn
electrode in dilute nitric acid solution and then rinse with deionised water. Assemble the
cell. Place the cell in a 2-litre beaker filled with water at room temperature and add
10cm3 of 1M ZnSO4 to the zinc half-cell. Using the potassium ferrocyanide solution
provided, make up 10cm3 of a 50/50 mixture of ferro and ferricyanide solution and add it
to the other half-cell. Connect the electrodes to the potentiometer. Ensure the cell is in
thermal equilibrium and record the temperature.
Mix 1cm3 of the same potassium ferricyanide solution with 9cm3 of distilled water and
add 5cm3 of the resulting solution to 5cm3 of ferrocyanide solution to produce a ferro :
ferro ratio of 10:1. Measure the EMF, noting carefully the polarity of the electrodes;
Invert the dilutions above to make a 10:1 ferro-ferri ratio and again, measure the EMF;
Measure the EMF of the cell in which the ferro:ferri ratio is 1:1 – i.e. ln (Fe 3+ / Fe 2+) = 0
Temperature Variation
Using a mixture of 1:1 ferro-/ ferricyanide mixture in the half-cell containing the Pt
electrode, place the assembly in a two-litre beaker filled with ice-water mixture, with the
thermometer in the ferro-/ferricyanide solution electrolyte. At thermal equilibrium, record
the temperature and EMF. Replace the freezing mixture in the beaker with water and
assemble the temperature control so that it can take eight to ten readings of the EMF at 5
intervals over a temperature range up to 323K.
1.14
y = -0.0195x + 1.0657
R2 = 0.9993
1.12
1.1
E cell
1.08
Series1
1.06
Linear (Series1)
1.04
1.02
1
0.98
-4
-2
0
2
4
Results:
When the cell was in thermal equilibrium, its temperature was 292K.
Ecell - Volts
1.110
1.067
1.020
aoxid
1
1
10
ared
10
1
1
The slope of this line = -0.0195.
From the intercept, Ecell = 1.0657 V.
Part 2:
Determine G.
G = -n F E0cell
Ln( aoxid / ared )
-2.30259
0
2.302589
G = - ( 2 ) ( 9.64853 * 104 ) ( 1.0657 )
= - 2.0565 * 10^5 kJ . mol-1 .
EMF – Volts
0.729
0.735
0.740
0.743
0.750
0.757
0.763
0.765
0.773
0.775
Temperature K
273.9
283
288
293
298
303
308
313
318
323
Plot of EMF as a function of Temperature
y = 0.001x + 0.4508
2
R = 0.9883
0.78
0.77
EMF - Volts
0.76
Series1
0.75
Linear (Series1)
0.74
0.73
0.72
270
280
290
300
310
Temperature - K
320
330
From eqn () above:
S = n F * (slope of graph)
= (2) (9.64853 * 10 4) (0.001)
= 1.9297 * 10^2 kJ . mol-1 K-1
From eqn () above:
At TK, H = G + T  S
G = H - TS
Let T = 298K,
 H  = - 1.4814 * 105 kJ . mol-1
Let T = 323K,
 H  = - 1.4332 * 105 kJ . mol-1
Post Practical Questions
2. The value of the slope from the plot compares quite accurately with that predicted by
the Nernst equation.
3. Sr =  n Sm (products) -  n Sm (reactants)
Sr = { -106.5 + 33.4 } - { 41.6 – 98.7 }
= { -73.1 } - { -57.1 }
= - 16 J K-1 mol –1
This is quite small because there is no consumption or release of gas which would usually
correspond to a larger entropy in one direction or the other.
1. The entropy change of the cell reaction of quite large because any species with 3
gained electrons is reluctant to accept a fourth. It would be “more natural” for the species
to rid itself of some of the extra electrons it has at the moment rather than accept another.
Also, because there is this large increase in entropy to facilitate this extra electron the
entropy is going to be positive.
--------------------------Paul Walsh – 2000
pwalsht@maths.tcd.ie