[12%] A galvanic cell is set up as follows and the cell potential measured as a function of temperat — Electrochemistry Chemistry Question
Problem Context
[12%] A galvanic cell is set up as follows and the cell potential measured as a function of temperature to give the graph shown.
salt bridge
V
Pb Zn
1.00 M Pb2+ 1.00 M Zn2+
Which electrode is the anode and which is the cathode?
Model Answer
The Zn electrode is the anode and the Pb electrode is the cathode.
The standard reduction potential for Zn2+(aq) at 298 K is –0.762 V. What is the standard reduction potential for Pb2+(aq) at 298 K?
Model Answer
At 298 K the (standard) cell potential is (0.7887 V) – (0.000515 V K-1)(298 K) = 0.635 V.
E°(Pb2+/Pb) – E°(Zn2+/Zn) = E°cell
E°(Pb2+/Pb) – (–0.762 V) = 0.635 V
E°(Pb2+/Pb) = –0.127 V
What are ∆G° (at 298 K), ∆H°, and ∆S° for the reaction shown below?
Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)
Model Answer
∆G° = –nFE° = –2(96500 J V-1 mol-1)(0.635 V) = –123 kJ mol-1
∆G° = ∆H° – T∆S° = –nFE°
E° = –∆H°/nF + (∆S°/nF)(T)
From the equation of the line,
∆H° = –nF(intercept) = –2(96500 J V-1 mol-1)(0.7887 V) = –152 kJ mol-1
∆S° = nF(slope) = 2(96500 J V-1 mol-1)(–0.000515 V K-1) = –99.4 J mol-1 K-1
Explain the sign of ∆S° for the reaction given in part c.
Model Answer
The Pb2+ ion is much larger and hence less tightly solvated than the Zn2+ ion, so it has a much more positive S° value (10.5 J mol-1 K-1 compared to –112.1 J mol-1 K-1 for Zn2+(aq)). Thus replacing Pb2+ in solution with Zn2+ results in a net ordering of the solvent molecules and a net decrease in entropy.
The temperature-dependence of the cell is remeasured with the same cell, except that the concentration of Pb2+(aq) in the left-hand compartment is changed to 0.100 M. Plot the results on the graph shown below (with the results shown from the standard cell given again for reference), and briefly justify your plot.
Model Answer
Under nonstandard conditions, the Nernst equation indicates that
E = E° – (RT/nF)ln([Zn2+]/[Pb2+])
E = E° – {(8.314 J mol-1 K-1)/(2 • 96500 J mol-1 V-1)}•ln(10)•T
E = E° – (9.92 × 10-5 V K-1)T
So the graph will be a line with the original y-intercept but with a slope that is smaller (more negative) by 9.92 × 10-5 V K-1. See blue line on the graph below.