Question

A 100.0 mL solution of 0.0200 M Fe 3 + in 1 M HClO 4 is titrated with 0.100 M Cu + , resulting in the formation of Fe 2 + and Cu 2 + . A Pt indicator electrode and a saturated Ag ∣ ∣ AgCl electrode are used to monitor the titration. Write the balanced titration reaction. titration reaction: -> ⟶ Complete the two half‑reactions that occur at the Pt indicator electrode. Write the half‑reactions as reductions. half‑reaction: Cu +e^{-} <=> Cu + e − − ⇀ ↽ − E ∘ = 0.161 V half‑reaction: Fe +e^{-} <=> Fe + e − − ⇀ ↽ − E ∘ = 0.767 V Select the two equations that can be used to determine the cell voltage at different points in the titration. E of the Ag ∣ ∣ AgCl electrode is 0.197 V . E = 0.767 V − 0.05916 × log ( [ Fe 2 + ] [ Fe 3 + ] ) − 0.197 V E = 0.767 V − 0.05916 × log ( [ Cu + ] [ Cu 2 + ] ) − 0.197 V E = 0.161 V − 0.05916 × log ( [ Fe 2 + ] [ Fe 3 + ] ) − 0.197 V E = 0.161 V − 0.05916 × log ( [ Cu + ] [ Cu 2 + ] ) − 0.197 V E = 0.767 V − 0.05916 × log ( [ Cu 2 + ] [ Cu + ] ) − 0.197 V E = 0.161 V − 0.05916 × log ( [ Fe 3 + ] [ Fe 2 + ] ) − 0.197 V E = 0.767 V − 0.05916 × log ( [ Fe 3 + ] [ Fe 2 + ] ) − 0.197 V E = 0.161 V − 0.05916 × log ( [ Cu 2 + ] [ Cu + ] ) − 0.197 V Calculate the values of E for the cell after each of the given volumes of the Cu + titrant have been added. 1.50 mL E = V 10.0 mL E = V 18.5 mL E = V 20.0 mL E = V 21.0 mL E = V 40.0 mL E = V

Answer 1

(a) Balanced titration, reaction :

        Fe²⁺ + Cu²⁺ <====> Cu⁺ + Fe³⁺

(b) The two half reaction:

      Fe³⁺ + e-   <===> Fe²⁺     E^o = 0.767 V

       Cu²⁺ + e- <===> Cu+       E^o = 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:

E = 0.767- 0.05916 log [Fe²⁺]/[Fe³⁺] - 0.197

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

V_Fe3+ = 100 mL, M_Fe3+ = 0.02 M

V_Cu+ = 0.10 M

At equivalence point,

Moles of Fe³⁺ = Moles of Cu⁺

M_Fe3+ x V_Fe3+ = M_Cu+ x V_Cu+

V_Cu+ = 0.02 M x 100 mL/0.10 M

V_Cu+ = 20.0 mL

(d) After adding 1.5 mL of Cu⁺

[Fe³⁺] = (Initial moles of Fe³⁺ - moles of Cu⁺ added)/total volume

= (0.02 M x 100 mL - 0.10 M x 1.5 mL)/(100 + 1.5) mL

= 0.01823 M

[Fe²⁺] = Moles of Cu⁺ added/total volume

= 0.10 M x 1.5 mL / (100 + 1.5) mL

= 0.00148 M

E = 0.767- 0.05916 log [Fe²⁺]/[Fe³⁺] - 0.197

E = 0.767- 0.05916 log 0.00148/0.01823 - 0.197

E = 0.6345 V

(e) After adding 10.0 mL of Cu⁺

[Fe³⁺] = (Initial moles of Fe³⁺ - moles of Cu⁺ added)/total volume

= (0.02 M x 100 mL - 0.10 M x 10.0 mL)/(100 + 10.0) mL

= 0.00909 M

[Fe²⁺] = Moles of Cu⁺ added/total volume

= 0.10 M x 10.0 mL / (100 + 10.0) mL

= 0.00909 M

E = 0.767- 0.05916 log [Fe²⁺]/[Fe³⁺] - 0.197

E = 0.767- 0.05916 log 0.00909/0.00909 - 0.197

E = 0.570 V

(f) After adding 18.5 mL of Cu⁺

[Fe³⁺] = (Initial moles of Fe³⁺ - moles of Cu⁺ added)/total volume

= (0.02 M x 100 mL - 0.10 M x 18.5 mL)/(100 + 18.5) mL

= 0.001266 M

[Fe²⁺] = Moles of Cu⁺ added/total volume

= 0.10 M x 18.5 mL / (100 + 18.5) mL

= 0.015612 M

E = 0.767- 0.05916 log [Fe²⁺]/[Fe³⁺] - 0.197

E = 0.767- 0.05916 log 0.015612/0.001266 - 0.197

E = 0.505 V

(g) After adding 20.0 mL of Cu⁺ i.e. equivalence point

E_eq = E^o_Fe3+/Fe2+ + E^o_Cu2+/Cu+ / 2

E_eq = 0.767 + 0.161 / 2

E_eq = 0.464 V

(h) After adding 21.0 mL of Cu⁺

[Cu⁺] = (0.10 M x 21.0 mL - 0.02 M x 100.0 mL)/(100 + 21.0) mL

= 0.000826 M

[Cu²⁺] = 0.02 M x 100 mL / (100 + 21.0) mL

= 0.01653 M

E = 0.161- 0.05916 log 0.000826/0.01653 - 0.197

E = 0.041 V

(i) After adding 40.0 mL of Cu⁺

[Cu⁺] = (0.10 M x 40.0 mL - 0.02 M x 100.0 mL)/(100 + 40.0) mL

= 0.01429 M

[Cu²⁺] = 0.02 M x 100 mL / (100 + 40.0) mL

= 0.01429 M

E = 0.161- 0.05916 log 0.01429/0.01429 - 0.197

E = - 0.036 V