Unknown solutions A and B contain a measurable concentration of copper (II) ions along with other in — Electrochemistry Chemistry Question

Sample plan, up to 5 pts
1) Use provided cotton pad as an electrochemical cell
2) Polish copper strips with sandpaper.
3) Pull cotton pads apart on opposite ends and insert the copper strips on either end so that about half of the strip is between the cotton and the remaining half is sticking out; just like the cotton pad with copper strips that was provided.
4) Place pad with electrodes on watch glass.
5) Carefully saturate the cotton around the copper electrodes. One side with the 1.0M Cu²⁺ solution, the other side with the unknown Cu²⁺ solution.
6) Connect the two solutions using the 1.0 M KNO₃ to serve as the salt bridge.
7) Use the multimeter to record the cell potential. The red (+) lead should go to the cathode and the black (-) lead should go to the anode. (students may reverse this not knowing if the unknown solution is higher or lower than 1.0 M, which will lead to a negative Ecell)
8) Determine unknown concentration using the Nernst Equation.

Anode/Cathode may be flipped giving a negative voltage May show half reactions or cell notation
Anode/Oxidation Cu(s) → Cu²⁺ (aq, dilute) + 2 e^-
Cathode/Reduction. Cu²⁺ (aq, conc or 1.0M) + 2 e^- → Cu(s)
Cu(s) | Cu²⁺ (aq, dilute) || Cu²⁺ (aq, conc or 1.0M) | Cu(s)

Cell by be run in reverse/flipped, giving a negative voltage. Multiple trials in table should be expected. Student may take average for Ecell
Anode Cu²⁺ Unknown
Cathode Cu²⁺ Eo cell (May be measured non-zero value or assumed zero of a 1.0M/1.0M cell)
Ecell
A 1.0 M 0.000 V +0.030 V
A 1.0 M 0.000 V +0.039 V
A 1.0 M 0.000 V +0.029 V
B 1.0 M 0.000 V +0.044 V
B 1.0 M 0.000 V +0.043 V
B 1.0 M 0.000 V +0.059 V

a) Concentration of Cu²⁺ in unknown solution A: (0.100 M)
Q = [Cu²⁺]anode / [Cu²⁺]cathode
-RT / nF = -1 (8.314 J/mol K)(298 K) / (2 mol e^-)(96500 J/V mol) = -0.012837 V
For a Concentration Cell, Eo = 0.00 V. ln Q = (E - Eo) / -0.012837 V
(0.⁰³⁰V - 0.000 V) / -0.012837 V = -2.337
Q = e^-2.337 = 0.0966 Q = [Cu²⁺ anode] / [Cu²⁺ cathode] → 0.0966 = X / 1.0 M → X = 0.0966 M
Trial 2: X = 0.0480 M. Trial 3: X = 0.104 M. Unknown A avg = 0.0830 M students may have used avg Ecell

b) Concentration of Cu²⁺ in unknown solution B: (0.0100 M) shown as if +/- voltmeter leads were flipped.
-RT / nF = -0.012837 V
For a Concentration Cell, Eo = 0.00 V. ln Q = (E - Eo) / -0.012837 V
(-0.⁰⁴⁴V - 0.000 V) / -0.012837 V = 3.428
Q = e^3.428 = 30.801 Q = [Cu²⁺ anode] / [Cu²⁺ cathode] → 30.801 = 1.0 M / X → X = 0.0325 M
Trial 2: X = 0.0351 M. Trial 3: X = 0.0101 M Unknown B avg = 0.0259 M. Should show lower concentration than A