Group Cell Experimental cell potential 1 Fe (s) Fe2+ (aq,1M) Cu2+ (aq,1M) Cu (s)
ID: 1058335 • Letter: G
Question
Group
Cell
Experimental cell potential
1
Fe (s) Fe2+ (aq,1M) Cu2+ (aq,1M) Cu (s) 5°C
.664
2
Fe (s) Fe2+ (aq,1M) Cu2+ (aq,1M) Cu (s) 50°C
.885
3
Fe (s) Fe2+ (aq,1M) Cu2+ (aq,1M) Cu (s) 75°C
.926
4
Zn(s) Zn2+ (aq,1M) Cu2+ (aq,1M) Cu (s) 5°C
.793
5
Zn(s) Zn2+ (aq,1M) Cu2+ (aq,1M) Cu (s) 50°C
.833
6
Zn(s) Zn2+ (aq,1M) Cu2+ (aq,1M) Cu (s) 50°C/75°C
.724
Group
Cell
Experimental cell potential
1
Fe (s) Fe2+ (aq,1M) Cu2+ (aq,1M) Cu (s) 5°C
.664
2
Fe (s) Fe2+ (aq,1M) Cu2+ (aq,1M) Cu (s) 50°C
.885
3
Fe (s) Fe2+ (aq,1M) Cu2+ (aq,1M) Cu (s) 75°C
.926
4
Zn(s) Zn2+ (aq,1M) Cu2+ (aq,1M) Cu (s) 5°C
.793
5
Zn(s) Zn2+ (aq,1M) Cu2+ (aq,1M) Cu (s) 50°C
.833
6
Zn(s) Zn2+ (aq,1M) Cu2+ (aq,1M) Cu (s) 50°C/75°C
.724
cell bath with plate. Plauu ulu voltage. You need to record the water off the hot the the them to the voltmeter and measure temperatures of both half cells. Use the average of both for your calculations 3. Share your data with your class. an VS. Determine AG at 25 oC and compare them to your results C 5. palculate the theoretical values for AH°, ASo, AGo,Explanation / Answer
Ans.
We know that
Delta G = -RTlnK = -nFE0
Delta G0 = Delta H0 - T Delta S0
-RTlnK = Delta H0 - T Delta S0
lnK = -Delta H0 / RT + DeltaS0 / R
-nFE0 = Delta H0 - T Delta S0
Delta S0 = nF[dE0 /dT]
so if we plot a graph between E0 and temperature we can obtain delta S as slope of the graph
DS° = n F * (slope of graph)
Emf of the cell Temperature T K
0.62 5 278
0.589 50 323
0.71 75 348
y = 0.001x + 0.307
Slope = 0.001 = Delta S° = n F * (slope of graph)
Delta S0 = 2 X 96485 X 0.001
Delta S0 = 1.9297 * 10^2 kJ . mol-1K-1
E0 (298K) = Ecathode - Eanode = 0.34 - (-0.41) = 0.75 V
Delta G° = - ( 2 ) ( 96485 ) ( 0.75 ) = 144727.5 = 1.45 10^5 KJ / mol
delta G° = Delta H° - TDeltaS°
Delat H° = DG° + T D S°
Delta H° = -1.45 X 10^5 + 298 X 1.9297 * 10^2 kJ . mol-1K-1
Delta H° = 0.8722 X 10^5 KJ / Mole
2) for second cell
Emf of the cell Temperature T K
0.781 5 278
0.713 50 323
0.741 75 348
Slope = 0
So DeltaS = 0
E0 (298K) = Ecathode - Eanode = 0.34 - (-0.76 = 1.10 V
Delta G° = - ( 2 ) ( 96485 ) ( 1.10 ) = 212267 = -2.12 10^5 KJ / mol
delta G° = Delta H° - TDeltaS°
Delat H° = DG° + T D S°
Delta H° = -2.12 10^5+ 298 X 0
Delta H° =-2.12 10^5 KJ / Mole.
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