[12%] A galvanic cell is constructed at 298 K with one half-cell consisting of a 10.00 g silver wire — Electrochemistry Chemistry Question
Problem Context
[12%] A galvanic cell is constructed at 298 K with one half-cell consisting of a 10.00 g silver wire immersed in 1.00 L of a 0.100 M solution of silver nitrate and the second half-cell consisting of a 20.00 g copper plate immersed in 1.00 L of a 0.200 M solution of copper(II) sulfate.
Half-reaction E°, V
Ag+(aq) + e– → Ag(s) 0.800
Cu2+(aq) + 2e– → Cu(s) 0.337
What voltage is measured for this galvanic cell?
Model Answer
The net cell reaction is
2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+(aq)
with E° = 0.800 V – 0.337 V = 0.463 V. The cell is not under standard conditions, so applying the Nernst equation gives:
E = E° – (RT/nF)ln([Cu2+]/[Ag+]2)
E = (0.463 V) – ((8.314 J mol-1 K-1)(298 K)/2(96500 J V-1 mol-1)ln([0.200]/[0.100]2)
E = 0.425 V
The cell is discharged at a constant current of 0.150 A until the mass of the silver electrode is equal to the mass of the copper electrode. How much time does this take?
Model Answer
Ag deposition will take place with (107.9 g mol-1)/(96500 C mol-1) = 1.118 × 10-3 g C-1.
Cu dissolution will require (63.55 g mol-1)/2(96500 C mol-1) = 3.293 × 10-4 g C-1.
So if x = the amount of charge passed in the discharge, then
10.00 g + (1.118 × 10-3 g C-1)x = 20.00 g – (3.293 × 10-4 g C-1)x
10.00 g = (1.447 × 10-3 g C-1)x
x = 6909 C
Since q = it,
6909 C = (0.150 A)t
t = 46100 s = 12.8 h
A chemist wishes to add sodium oxalate to one of the half-cells in the original cell to decrease the measured voltage. To which cell should the sodium oxalate be added, and why?
Model Answer
Decreasing the voltage means making the net reaction less favorable, which would involve either decreasing the concentration of the reactants or increasing the concentration of the products. Adding oxalate would cause precipitation of the insoluble oxalate salt, so one would need to do that to the silver-containing half-cell to decrease the voltage.
What mass of sodium oxalate would need to be added to the appropriate half-cell to cause the voltage in the original cell to become 0.200 V? You may assume that there is no change in the volume of the solution. The Ksp of Ag2C2O4 is 3.5 × 10-11 and the Ksp of CuC2O4 is 3.0 × 10-8.
Model Answer
Using the Nernst equation,
0.200 V = (0.463 V) – ((8.314 J mol-1 K-1)(298 K)/2(96500 J V-1 mol-1)ln([0.200]/[Ag+]2)
–0.263 V = 0.021 V + 0.02567ln[Ag+]
[Ag+] = 1.567 × 10-5 M
From the Ksp expression, [Ag+]2[C2O4 2-] = [1.567 × 10-5]2[C2O4 2-] = 3.5 × 10-11
[C2O4 2-] = 0.143 M
This requires the addition of 0.143 mol Na2C2O4 (to achieve this concentration in 1.00 L of solution) + 0.0500 mol Na2C2O4 (to precipitate 0.0500 mol Ag2C2O4) = 0.193 mol Na2C2O4. The molar mass of Na2C2O4 is 134.00 g mol-1, so this is 0.193 mol × 134.00 g mol-1 = 25.9 g Na2C2O4.