You are titrating 100.0 mL of 0.0200 M Fe3 in 1 M HClO4 with 0.100 M Cu to give

Question

You are titrating 100.0 mL of 0.0200 M Fe3 in 1 M HClO4 with 0.100 M Cu to give Fe2 and Cu2 using Pt and saturated Ag | AgCl electrodes to find the endpoint.

(a) Write the balanced titration reaction.

(b) Complete the two half reactions for the Pt electrode.

(c) From the list in the column at the right, select the two correct Nernst equations for the cell voltage. (Each applying at different points in the titration.)      Eo of the Ag |AgCl electrode is 0.197 V.

a. E = 0.767- 0.05916log([Fe^2+]/[Fe^3+]-0.197
b. E = 0.767- 0.05916log([Fe^3+]/[Fe^2+]-0.197
c. E = 0.161- 0.05916log([Fe^2+]/[Fe^3+]-0.197
d. E = 0.161- 0.05916log([Fe^3+]/[Fe^2+]-0.197
e. E = 0.767- 0.05916log([Cu^+]/[Cu^2+]-0.197
f. E = 0.767- 0.05916log([Cu^2+]/[Cu^+]-0.197
g. E = 0.161- 0.05916log([Cu^+]/[Cu^2+]-0.197
h. E = 0.161- 0.05916log([Cu^2+]/[Cu^+]-0.197

(d) Calculate the values of E for the cell when at the volume 21.0 mL of the Cu titrant has been added: (Activity coefficients may be ignored as they tend to cancel when calculating concentration ratios.)

Explanation / Answer

(a.) Balanced titration, reaction :

        Fe2+ + Cu2+ <====> Cu+ + Fe3+

(b). The two half reaction:

      Fe3+ + e-   <===> Fe2+     Eo = 0.767 V

       Cu2+ + e- <===> Cu+       Eo = 0.161 V

Reference half reaction:

   Ag (s) + Cl- <===> AgCl (s) + e-

Above two half reactions for Pt electrode are:

Fe3+ + Ag (s) + Cl- === Fe2+ + AgCl (s)

Cu2+ + Ag (s) + Cl- <==> Cu+ + AgCl (s)

(c). Two correct Nernst equations:

    a . E = 0.767- 0.05916log([Fe^2+]/[Fe^3+]-0.197

    g. E = 0.161- 0.05916log([Cu^+]/[Cu^2+]-0.197

d)

(d) Fe2+ + Cu2+ <====> Cu+ + Fe3+

At 21.0mL of Cu+ added, VFeMFe < VCuMCu

Therefore, after equivalence point

      [Cu+] = (100 mL)(0.02M)/(100+21)mL = 2/121 M

[Cu2+] = (21.0mL)(0.1M) – (100mL)(0.02M)/(100+21)mL = 0.1/121

Substituting the values in Nernst equation:

E = 0.161- 0.05916log([Cu+]/[Cu2+]-0.197

E = 0.161- 0.05916log([2/121]/[0.1/121]-0.197

E = 0.161- 0.05916log[20]-0.197

E = 0.161- 0.05916*1.30103-0.197

E = - 0.113 V

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