A solution is formed by mixing 25 ml. of a solution containing 0.05M Ce(NO_3)4 a

Question

A solution is formed by mixing 25 ml. of a solution containing 0.05M Ce(NO_3)4 and 0.10M Ce(NO_3)_3 with 75ml. of 0.03333M SnCl_4 and 0.05M SnCl_2. A second solution is prepared by adding 15.961 milligrams of CuSO4 to 100 ml. of water. A platinum wire is inserted into the first solution and a copper wire is inserted into the second solution. A salt bridge is placed between the two solutions and the potential between the two wires is measured. Ce^4 +e Ce^3+ E degree =1.61V Cu^2 + 2e Cu degree E degree = 0.337 Sn^4+ 2e Sn^2+ E degree =0.153 Molecular weight of CuSO_4 = 159.61gms. What is the concentration of Cu^2+ in the copper solution? What are the equilibrium concentrations of all dominant species in the first solution? What is the potential of the first solution relative to a standard hydrogen electrode? What are the concentrations of all the minor species in the first solution? What is the potential that would be measured? Which of the solutions is the cathode

Explanation / Answer

a) moles of CuSO4= (15.961*10-3) / 159.61 = 1*10-4 moles
Molarity of CuSO4 in water = (1*10-4)moles / 0.1 lit = 10-3 M
Calculate Ecu+2/cu
Ecu+2/cu = E° - (RT / zF) * ln (1 / [Cu2])
Ecu+2/cu = E° =0.337 V
Ecu+2/cu = E° - (RT / zF) * ln [ox /red]
0.337 = 0.337 - (8.314 J K-1 (298.15K) / 2 *96485 JV-1) * ln (10-3 / [Cu2])
0.012845 *ln (10-3 / [Cu2+] ) =0
ln(10-3) - ln [Cu2+] = 0
2.303 log [Cu2+] = -6.907755
[Cu2+] = 1.0012*10-3 M

b)
First solution contains, two mixtures
calculate number of moles to determine dominant species,
For first mixture, 0.05 M of Ce(NO3)4 +0.10M of Ce(No3)3
0.05 mol/lit * (0.025 lit) + 0.10 mol/lit * (0.025 lit) = 3.75*10-3 moles
For second mixture 0.03333M SnCl4 + 0.05M SnCl2 ,
0.033 mol/lit * (0.075lit) + 0.05 mol/lit * (0.075lit) = 6.2475*10-3 moles
Since 3.75*10-3 moles < 6.2475*10-3 moles, Ce4+will be your minor
species and Sn4+ will be dominant species.

For dominant species equilibrium conc wil be
Esn4+/sn+2 =E0 = 0.153 V
0.153 =0.153 - (8.314 J K-1 (298.15K) / 2 *96485 JV-1) * ln (6.2475*10-3 / [Sn4])
2.303 log [Sn4] =-5.07557
[Sn4] =6.2532*10-3 moles(divide 0.075 to get M) =0.0835M

c)
Ce+4+e-    <------------> Ce+3 1.61 V
(+) Sn+2 <------------>Sn+4+2e- -0.153 V
------------------------------------------------------------------
Ce+4+ Sn+2<---------->Ce+3+Sn+4+e-1.457V (potential of the 1st solution)

d) Calculating equilibrium conc for minor species,
ECe4+/Ce+3 =E0 =1.61V
1.61 = 1.61 - (8.314 J K-1 (298.15K) / 1*96485 JV-1) * ln (3.75*10-3 / [Ce4]
0.025691 *ln (3.75*10-3 / [Ce4+] ) =0
2.303 log[Ce4+] = -5.58599
[Ce4+] =3.75377*10-3 moles(divide 0.025lit to get M) =0.1501 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.