sapling Learning You are titrating 100.0 mL of 0.0200 M Fear in 1 M Hcio. with o
ID: 530401 • Letter: S
Question
sapling Learning You are titrating 100.0 mL of 0.0200 M Fear in 1 M Hcio. with o 100 M cu' to give Fe? and cu using Pt and saturated Ag I AgCl electrodes to find the endpoint. (a) Write the balanced titration reaction. Fe''(aq) Cu (aq) Fe"(aq)+Ca (b) Complete the two half reactions for E -0.161 v the Pt electrode. E-0767 v Fe e- Fe (c) From the list in the column at the right, select the Fe 0.197 two correct Nernst equations for the cell A) E-0.767-0.05916 log Fe' (Each applying at different points in the titration.) EO of the Ag lAgCI e is 0.197 V. (B) E 0.767-0.05916log. 0.197 M (A) (B) Fe (C) E-0.161-0.059 16log Fe 0.197 (C) (D) (E) Fe (D) E-0.161-0.05916 log 0.197 M (F) (G) o Previous ® Give Up & View solution Try Again 0 Next JEit (H)Explanation / Answer
For the given cell,
(a) Balanced equation,
Fe3+(aq) + Cu+(aq) --> Fe2+(aq) + Cu2+(aq)
(b) half reactions,
Fe3+(aq) + e- <==> Fe2+(aq) Eo = 0.767 V
Cu2+(aq) + e- <==> Cu+(aq) Eo = 0.161 V
(c) Correct Nernst equations
A
G
(d) Ecell values,
(i) Volume Cu+ = 2.50 ml
moles Fe3+ = 0.02 M x 100 ml = 2 mmol
moles Cu+ = 0.1 M x 2.50 ml = 0.25 mmol
[Fe2+] formed = 0.25 mmol/102.5 ml = 0.00244 M
[Fe3+] remained = 1.75 mmol/102.5 ml = 0.0171 M
Ecell = 0.767 - 0.05916 log(0.00244/0.0171) - 0.197 = 0.62 V
(ii) Volume Cu+ = 10 ml
moles Fe3+ = 0.02 M x 100 ml = 2 mmol
moles Cu+ = 0.1 M x 10 ml = 1 mmol
[Fe2+] formed = [Fe3+] remained
Ecell = 0.767 - 0.197 = 0.57 V
(iii) Volume Cu+ = 18 ml
moles Fe3+ = 0.02 M x 100 ml = 2 mmol
moles Cu+ = 0.1 M x 18 ml = 1.8 mmol
[Fe2+] formed = 1.8 mmol/118 ml = 0.01525 M
[Fe3+] remained = 0.2 mmol/118 ml = 0.0017 M
Ecell = 0.767 - 0.05916 log(0.01525/0.0017) - 0.197 = 0.514 V
(iv) Volume Cu+ = 20 ml
moles Fe3+ = 0.02 M x 100 ml = 2 mmol
moles Cu+ = 0.1 M x 20 ml = 2 mmol
Equivalence point
Ecell = (0.767 + 0.161)/2 - 0.197 = 0.267 V
(v) Volume Cu+ = 21.5 ml
moles Fe3+ = 0.02 M x 100 ml = 2 mmol
moles Cu+ = 0.1 M x 21.5 ml = 2.15 mmol
[Cu2+] formed = 2 mmol/121.5 ml = 0.0165 M
[Cu+] remained = 0.15 mmol/121.5 ml = 0.00123 M
Ecell = 0.161 - 0.05916 log(0.00123/0.0165) - 0.197 = 0.031 V
(vi) Volume Cu+ = 40 ml
moles Fe3+ = 0.02 M x 100 ml = 2 mmol
moles Cu+ = 0.1 M x 4 ml = 4 mmol
[Cu2+] formed = [Cu+] remained
Ecell = 0.161 - 0.197 = -0.036 V
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