saplingleaming-com 17-RECHSIG I Activities and Due Dates I Hws Redox Titrations
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saplingleaming-com 17-RECHSIG I Activities and Due Dates I Hws Redox Titrations Spring17 Gradebook 3/8/2017 11:55 PM S 10.8/100 3/4/2017 01:28 PM Print calaulator Periodic Table 3 of You are titrating 1000 mL of 00200 MFes in 1 M Hco. with 0.100 M cu to give Fe and cu using Mapit Pt and saturated AgIAgCl electrodes to find the endpoint. (a) Write the balanced titration reaction. (b) Complete the two half reactions for E 0.161 V cu e the Pt E -0.767 v Fe Fe (c) From the list in the column at the right, select the (A) E-0.767-0.059 16log 0.197 equations for the cell voltage Fe (Each applying at different points in the titration.) E of the Ag IAgCI electrode is 0.197 V. Fe (B) E-0.767-0.05916 log 0,197 O (A) Fe (C) Fe (C) E- 0.161 -0.059 16log 0.197 (D) Fe (E) Fe (D) E 0.161 -0.05916 0.197 Fe (H) Cu E) E-0.767-0.05916 log t 0.197 (d) Calculate the values of Elor the cell when Cu Previous Give Up & View Solution Check Answer 0Next Exit MacBook AirExplanation / Answer
For the given titration reaction,
(a) Balanced titration equation,
Fe3+ + Cu+ ---> Fe2+ + Cu2+
(b) Half-reactions,
Cu2+ + e- <==> Cu+
Fe3+ + e- <==> Fe2+
(c) The Nernst equations for the cell would be,
(A)
(G)
(d) E values when volume of Cu+ added
1.00 ml
[Fe2+] formed = 0.1 M x 1 ml/101 ml = 0.00099 M
[Fe3+] remained = (0.02 M x 100 ml - 0.1 M x 1 ml)/101 ml = 0.019 M
E = [0.767 - 0.05916 log(0.00099/0.02)] - 0.197 = 0.647 V
10.00 ml
[Fe2+] formed = 0.1 M x 10 ml/110 ml = 0.0091 M
[Fe3+] remained = (0.02 M x 100 ml - 0.1 M x 10 ml)/110 ml = 0.0091 M
E = 0.767 - 0.197 = 0.570 V
18.50 ml
[Fe2+] formed = 0.1 M x 18.5 ml/118.5 ml = 0.0156 M
[Fe3+] remained = (0.02 M x 100 ml - 0.1 M x 18.5 ml)/118.5 ml = 0.0013 M
E = [0.767 - 0.05916 log(0.0156/0.0013)] - 0.197 = 0.506 V
20.0 ml
[Fe2+] formed = 0.1 M x 20 ml/120 ml = 0.0167 M
This is equivalence point
E = [0.767 + 0.161]/2 - 0.197 = 0.267 V
21.0 ml
[Cu2+] formed = 0.02 M x 100 ml/121 ml = 0.0165 M
[Cu+] remained = (0.1 M x 21 ml - 0.02 M x 100 ml)/121 ml = 0.00083 M
E = [0.161 - 0.05916 log(0.00083/0.0165)] - 0.197 = 0.041 V
40.0 ml
[Cu2+] formed = 0.02 M x 100 ml/140 ml = 0.0143 M
[Cu+] remained = (0.1 M x 40 ml - 0.02 M x 100 ml)/140 ml = 0.0143 M
E = 0.161 - 0.197 = -0.036 V
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