Consider these reactions among copper and iodine compounds (all reactions at 298 K): CuI(s) Cu+ (aq) — Aqueous Equilibria and Electrochemistry Chemistry Question
Problem Context
Consider these reactions among copper and iodine compounds (all reactions at 298 K):
CuI(s) Cu+ (aq) + I– (aq) Ksp = 1.2 10–12
Cu+ (aq) + 2 I– (aq) CuI2– (aq) Kf = 7.1 108
Cu2+ (aq) + 2 e– Cu(s) Eº = +0.337 V
Cu2+ (aq) + e– Cu+ (aq) Eº = +0.159 V
I2(s) + 2 e– 2 I– (aq) Eº = +0.540 V
Calculate the number of moles of copper that dissolve if 1.00 10–3 mol CuI(s) is suspended in 1.00 L of solution.
Model Answer
[Cu+] = [I–] in this solution, and [Cu+][I–] = Ksp = 1.2 10-12
[Cu+] = 1.1 10-6 M, so 1.1 10-6 mol Cu dissolve in 1.00 L.
(Because the [I–] is so low, there is a negligible amount of CuI2– present:
[CuI2–]/([Cu+][I–]2) = 7.1 108, so [CuI2–]/[Cu+] = 8.6 10-4 if [I–] = 1.1 10-6 M)
Calculate the minimum number of moles of NaI that would need to be added to the mixture in (a) to fully dissolve the CuI. You may assume that the volume remains 1.00 L.
Model Answer
The major reaction that takes place is
CuI(s) + I– (aq) CuI2– (aq) Keq = Ksp•Kf = 8.5 10-4
Since almost all the Cu in solution is complexed, [CuI2–] = 1.00 10-3 M.
[CuI2–]/[I–] = (1.00 10-3)/[I–] = 8.5 10-4
[I–] = 1.18 M
Only 1.00 10-3 mol iodide has bonded to the copper, which is negligible compared to the amount found free in solution. So 1.18 mol NaI needs to be added to dissolve the CuI.
Calculate Keq for the (favorable) disproportionation of aqueous Cu+ ion in neutral solution:
2 Cu+ (aq) Cu2+ (aq) + Cu(s)
Model Answer
2 Cu+ (aq) 2 Cu2+ (aq) + 2 e– Eº = –0.159 V
Cu2+ (aq) + 2 e– Cu(s) Eº = +0.337 V
2 Cu+ (aq) Cu(s) + Cu2+ (aq) Eº = 0.178 V
∆Gº = –nFEº = –(2)(96.485 kJ V-1 mol-1)(0.178 V) = –34.3 kJ mol-1
Keq = e–∆Gº/RT = e(34300 J mol-1)/(8.314 J mol-1 K-1)(298 K) = 1.05 106
Copper(II) iodide, CuI2, is not stable. Write a reasonable chemical reaction that describes the decomposition of CuI2 in aqueous solution, and show that this is a spontaneous reaction under standard conditions.
Model Answer
Net reaction: Cu2+ (aq) + 2 I– (aq) CuI(s) + 0.5 I2(s)
This can be written as a sum of two reactions:
(I) A redox reaction:
Cu2+ (aq) + I– (aq) Cu+ (aq) + 0.5 I2
∆Gº = –(1)(96.485 kJ V-1 mol-1)(0.159 V – 0.540 V) = +36.8 kJ mol-1
(II) A precipitation reaction:
Cu+ (aq) + I– (aq) CuI(s)
∆Gº = –RTln(Keq) = –(0.008314 kJ mol-1 K-1)(298 K)ln(1/Ksp) = –68.0 kJ mol-1
Thus the net reaction has ∆Gº = +36.8 kJ mol-1 + (–68.0 kJ mol-1) = –31.2 kJ mol-1, and is spontaneous (under standard conditions).