Consider the titration of 100.0 mL of 0.010 0 M Ce in ! M HC10_4 by 0.040 0 M Cu
ID: 925240 • Letter: C
Question
Consider the titration of 100.0 mL of 0.010 0 M Ce in ! M HC10_4 by 0.040 0 M Cu^+ to give Ce^3+ and Cu^2+, using Pt and saturated Ag | AgCl electrodes to find the end point. Write a balanced titration reaction. Write two different half-reactions for the indicator electrode. Write two different Nernst equations for the cell voltage. Calculate E at the following volumes of Cu^+: 1.00, 12.5, 24.5, 25.0, 25.5, 30.0, and 50.0 mL. Sketch the titration curve. Consider the titration of 25.0 mL of 0.010 0 M Sn^2+ by 0.050 0 M Tl in I M HC1, using Pt and saturated calomel electrodes to find the end point. Write a balanced titration reaction. Write two different half-reactions for the indicator electrode. Write two different Nernst equations for the cell voltage. Calculate E at the following volumes of Tl^3+: 1.00, 2.50. 4.90, 5.00, 5.10, and 10.0 mL. Sketch the titration curve.Explanation / Answer
16-2 :
(a) Balanced equation : Ce4+ + Cu+ ---> Ce3+ + Cu2+
(b) Half reaction,
Reduction half cell : Ce4+ + 1e- ---> Ce3+
Oxidation half cell : Ce+ ----> Cu2+ + 1e-
(c) Nernst equation for the cell voltage
E = {1.70 - 0.0592 log([Ce3+]/[Ce4+])} - 0.197
E = {0.161 - 0.0592 log([Cu+]/[Cu2+])} - 0.197
(d) Calculation of E
(a) 1.0 ml of 0.040 M Cu+ added
moles of Ce4+ = 0.01 M x 0.1 L = 1 x 10^-3 mols
moles of Cu+ = 0.04 M x 0.001 L = 4 x 10^-5 mols
[Ce3+] = 4 x 10^-5/0.101 = 3.96 x 10^-4 M
[Ce4+] = 9.6 x 10^-4/0.101 = 9.505 x 10^-3 M
E = {1.70 - 0.0592 log([Ce3+]/[Ce4+])} - 0.197
= {1.70 - 0.0592 log(3.96 x 10^-4/9.505 x 10^-3} - 0.197
= 1.585 V
(b) 12.5 ml of 0.04 M Cu+ added
moles of Ce4+ = 0.01 M x 0.1 L = 1 x 10^-3 mols
moles of Cu+ = 0.04 M x 0.0125 L = 5 x 10^-4 mols
[Ce3+] = 5 x 10^-4/0.1125 = 4.44 x 10^-3 M
[Ce4+] = 5 x 10^-4/0.101 = 4.44 x 10^-3 M
[Ce4+] = [Ce3+]
E = 1.70 - 0.197 = 1.503 V
(c) 24.5 ml of 0.04 M Cu+ added
moles of Ce4+ = 0.01 M x 0.1 L = 1 x 10^-3 mols
moles of Cu+ = 0.04 M x 0.0245 L = 9.8 x 10^-4 mols
[Ce4+] = 2 x 10^-5/0.1245 = 1.61 x 10^-4 M
[Ce3+] = 9.8 x 10^-4/0.1245 = 7.87 x 10^-3 M
E = {1.70 - 0.0592 log([Ce3+]/[Ce4+])} - 0.197
= {1.70 - 0.0592 log(7.87 x 10^-3/1.61 x 10^-4} - 0.197
= 1.403 V
(d) 25 ml of 0.04 M Cu+ added
moles of Ce4+ = 0.01 M x 0.1 L = 1 x 10^-3 mols
moles of Cu+ = 0.04 M x 0.025 L = 1.0 x 10^-3 mols
this is the equivalence point
E = (1.70 + 0.161)/2 - 0.197 = 0.7335 V
(e) 25.5 ml of 0.04 M Cu+ added
E = {0.161 - 0.0592 log([Cu+]/[Cu2+])} - 0.197
= {0.161 - 0.0592 log(0.5/25} - 0.197
= 0.065 V
(f) 30 ml of 0.04 M Cu+ added
E = {0.161 - 0.0592 log([Cu+]/[Cu2+])} - 0.197
= {0.161 - 0.0592 log(5/25} - 0.197
= 0.005 V
(g) 50 ml of 0.04 M Cu+ added
E = {0.161 - 0.0592 log([Cu+]/[Cu2+])} - 0.197
= {0.161 - 0.0592 log(25/25} - 0.197
= -0.036 V
16-3.
(a) Sn2+ + Tl3+ ---> Sn4+ + Tl+
(b) Half cell reaction
Sn2+ ---> Sn4+ + 2e-
Tl3+ + 2e- ---> Tl+
(c) Nernst equation
E = {0.139 - 0.0592/2 log([Sn2+]/[Sn2+]} - 0.241
E = {0.77 - 0.0592/2 log([Tl+]/[Tl3+]} - 0.241
(d) Titration
(a) 1.0 ml of 0.05 M Tl3+ added
moles of Sn2+ = 0.01 M x 0.025 L = 2.5 x 10^-4 mols
moles of Tl3+ = 0.05 M x 0.001 L = 5 x 10^-5 mols
[Sn4+] = 5 x 10^-5/0.026 = 1.923 x 10^-3 M
[Sn2+] = 2 x 10^-4/0.026 = 7.692 x 10^-3 M
E = {0.139 - 0.0592/2 log([Sn2+]/[Sn2+]} - 0.241
= {0.139 - 0.0592/2 log(7.692 x 10^-3/1.923 x 10^-3) - 0241
= -0.120 V
(b) 2.50 ml of 0.05 M Tl3+ added
moles of Sn2+ = 0.01 M x 0.025 L = 2.5 x 10^-4 mols
moles of Tl3+ = 1.25 x 10^-4 mols
[Sn4+] = [Sn2+]
E = {0.139 - 0.241
= -0.102 V
(c) 4.90 ml of 0.05 M Tl3+ added
moles of Sn2+ = 0.01 M x 0.025 L = 2.5 x 10^-4 mols
moles of Tl3+ = 0.05 M x 0.0049 L = 2.45 x 10^-4 mols
[Sn4+] = 2.45 x 10^-4/0.0299 = 8.194 x 10^-3 M
[Sn2+] = 5 x 10^-6/0.0299 = 1.672 x 10^-4 M
E = {0.139 - 0.0592/2 log([Sn2+]/[Sn2+]} - 0.241
= {0.139 - 0.0592/2 log(1.672 x 10^-4/8.194 x 10^-3) - 0241
= -0.002 V
(d) 5 ml of 0.05 M Tl3+ added
moles of Tl3+ = 0.05 x 0.005 = 2.5 x 10^-4 mols
This is equivalence point
E = 0.45 - 0.241 = 0.21 V
(e) 10 ml of 0.05 M Tl3+ added
excess moles of Tl3+ = 0.05 x 0.005 = 2.5 x 10^-4 mols
[Tl3+] = [Tl+]
E = 0.77 - 0.241 = 0.529 V