Copper sulfate has a solubility limit in water of 3.2 M. You make a concentratio
ID: 898356 • Letter: C
Question
Copper sulfate has a solubility limit in water of 3.2 M. You make a concentration cell of two half cells, one comprising 1L of 5M Na2SO4 and a copper (Cu, 63 g/mol) electrode weighing 250g, the other containing 1 L of saturated CuSO4 and a copper electrode weighing 100 g. They are connected by a salt bridge.
1. What is the miting reactant for this concentration cell, if you let it discharge completley?
2. What is the potential of this cell when it is 1% discharged?
3. Whar is the potential of this cell when it is 99 % discharged?
Explanation / Answer
1.The anode reaction:
Cu = Cu2+ + 2e n=2 electrons
The cathode reaction
Cu2+ + 2e = Cu
Cu(250 g)| Cu2+ (traces). Na2SO4|| Cu2+ 3.200 M |Cu(100g) initial state
Anode cathode
The copper electrode at anode will dissolve. The available quantity is
250 g/ 63.5 g/mol = 3.94 mol. Approximate it to 4.0 mol.
A complete discharge will be when the concentrations of Cu2+ will be the same in the two half cells. The two half cells have equal volumes (1L). Then the final/equilibrium concentration will be
3.2 M /2 = 1.6 mol/L
if Cu metal is not the limiting reactant.
1.6 mol Cu2+ will be reduced to Cu and 1.6 Cu will oxidized to Cu2+..
The limiting reactant is not Cu (4 moles available), but CuSO4(1.6 moles availables).
2.
1 % x 1.6 mol = 0.016 mol
3.2 – 0.016 = 3.184 mol
The cell is now
Cu| Cu2+0.016 M, Na2SO4|| Cu2+3.184 M |Cu
E = Eo – (0.0592/2) V log (0.016/3.184)
= 0 V - 0.0296 Vx (-2.30)=
= 0.068 V = 68 mV
Note. Eo for any concentration cell is 0 V.
3.
99 % x 1.6 mol =1.584 mol
3.200 – 1.584 = 1.616 mol
The cell is now
Cu| Cu2+1.584 M, Na2SO4|| Cu2+1.616M |Cu
E = Eo – (0.0592/2) V log (1.584 /1.616)
= 0 V - 0.0296 Vx (-0.0087)=
= 0.00026 V = 0.26 mV
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