Map E E-0.767- 0.05916log (d) Calculate the values of E for the cell when the following volumes of the Cu\' titrant have been added: (Activity coefficients may be ignored as they tend to cancel when calculating concentration ratios.) Cu2-0.197 F) E-0.767-0.05916log Cu0.197 Volume Number (G) E-0.161 -0.05916 log Cu ]/0.197 2.00 mL H) E-0.161 0.05916log Cu0.197 Number 10.0 mL Number 18.0 mL Number 20.0 mL Number 21.0 mL Solution Titration reaction (a) balanced equation, Fe3+ + Cu+ ---> Fe2+ + Cu2+ (b) half reactions Cu2+ + e- --> Cu+   Eo = 0.161 V Fe3+ + e- --> Fe2+ Eo = 0.767 V (c) Nernst equations to use, (A) and, (G) (d) calculating E value 2 ml Cu+ added formed Fe2+ = 0.1 M x 2 ml = 0.2 mmol remined Fe3+ = 0.02 M x 100 ml - 0.2 mmol = 1.8 mmol E = 0.767 - 0.05916 log([Fe2+]/[Fe3+]) - 0.197 = 0.767 - 0.05916 log(0.2/1.8) - 0.197 = 0.626 V 10 ml Cu+ added formed Fe2+ = 0.1 M x 10 ml = 1 mmol remined Fe3+ = 0.02 M x 100 ml - 1 mmol = 1 mmol E = 0.767 - 0.05916 log([Fe2+]/[Fe3+]) - 0.197 = 0.767 - 0.05916 log(1.0/1.0) - 0.197 = 0.570 V 18 ml Cu+ added formed Fe2+ = 0.1 M x 18 ml = 1.8 mmol remined Fe3+ = 0.02 M x 100 ml - 1.8 mmol = 0.2 mmol E = 0.767 - 0.05916 log([Fe2+]/[Fe3+]) - 0.197 = 0.767 - 0.05916 log(1.8/0.2) - 0.197 = 0.513 V 20 ml Cu+ added formed Fe2+ = 0.1 M x 20 ml = 2 mmol [Fe3+] = [Cu+] (equivalence point) E = (0.767 + 0.161)/2 - 0.197 = 0.267 V 21 ml Cu+ added Post equivalence point formed Cu2+ = 0.1 M x 20 ml = 2 mmol remaining Cu+ = 0.1 M x 1 ml = 0.1 mmol E = 0.161 - 0.05916 log([Cu+]/[Cu2+]) - 0.197 = 0.161 - 0.05916 log(0.1/2) - 0.197 = 0.041 V 40 ml Cu+ added Post equivalence point formed Cu2+ = 0.1 M x 20 ml = 2 mmol remaining Cu+ = 0.1 M x 20 ml = 2 mmol E = 0.161 - 0.05916 log([Cu+]/[Cu2+]) - 0.197 = 0.161 - 0.05916 log(2/2) - 0.197 = -0.036 V .

1. Map E E-0.767- 0.05916log (d) Calculate the values of E for the cell when the following
volumes of the Cu' titrant have been added: (Activity coefficients may be ignored as they tend to
cancel when calculating concentration ratios.) Cu2-0.197 F) E-0.767-0.05916log Cu0.197
Volume Number (G) E-0.161 -0.05916 log Cu ]/0.197 2.00 mL H) E-0.161 0.05916log Cu0.197
Number 10.0 mL Number 18.0 mL Number 20.0 mL Number 21.0 mL
Solution
Titration reaction
(a) balanced equation,
Fe3+ + Cu+ ---> Fe2+ + Cu2+
(b) half reactions
Cu2+ + e- --> Cu+Â Â Eo = 0.161 V
Fe3+ + e- --> Fe2+ Eo = 0.767 V
(c) Nernst equations to use,
(A)
and,
(G)
(d) calculating E value
2 ml Cu+ added
formed Fe2+ = 0.1 M x 2 ml = 0.2 mmol
remined Fe3+ = 0.02 M x 100 ml - 0.2 mmol = 1.8 mmol

2. E = 0.767 - 0.05916 log([Fe2+]/[Fe3+]) - 0.197
= 0.767 - 0.05916 log(0.2/1.8) - 0.197
= 0.626 V
10 ml Cu+ added
formed Fe2+ = 0.1 M x 10 ml = 1 mmol
remined Fe3+ = 0.02 M x 100 ml - 1 mmol = 1 mmol
E = 0.767 - 0.05916 log([Fe2+]/[Fe3+]) - 0.197
= 0.767 - 0.05916 log(1.0/1.0) - 0.197
= 0.570 V
18 ml Cu+ added
formed Fe2+ = 0.1 M x 18 ml = 1.8 mmol
remined Fe3+ = 0.02 M x 100 ml - 1.8 mmol = 0.2 mmol
E = 0.767 - 0.05916 log([Fe2+]/[Fe3+]) - 0.197
= 0.767 - 0.05916 log(1.8/0.2) - 0.197
= 0.513 V
20 ml Cu+ added
formed Fe2+ = 0.1 M x 20 ml = 2 mmol
[Fe3+] = [Cu+] (equivalence point)
E = (0.767 + 0.161)/2 - 0.197
= 0.267 V
21 ml Cu+ added
Post equivalence point
formed Cu2+ = 0.1 M x 20 ml = 2 mmol

3. remaining Cu+ = 0.1 M x 1 ml = 0.1 mmol
E = 0.161 - 0.05916 log([Cu+]/[Cu2+]) - 0.197
= 0.161 - 0.05916 log(0.1/2) - 0.197
= 0.041 V
40 ml Cu+ added
Post equivalence point
formed Cu2+ = 0.1 M x 20 ml = 2 mmol
remaining Cu+ = 0.1 M x 20 ml = 2 mmol
E = 0.161 - 0.05916 log([Cu+]/[Cu2+]) - 0.197
= 0.161 - 0.05916 log(2/2) - 0.197
= -0.036 V