calculation of E

Question

A 100.0 mL solution of 0.0200 M Fe3+ in 1 M HClO4 is titrated with 0.100 M Cu+, resulting in the formation of Fe2+ and Cu2+. A Pt indicator electrode and a saturated Ag∣∣AgCl electrode are used to monitor the titration.

Calculate the values of 𝐸 for the cell after each of the given volumes of the Cu+Cu+ titrant have been added.

10.0 ml E = ? V

19.5 ml E = ? V

Answer 1

Fe³⁺ + Cu⁺ ==> Fe²⁺ + Cu²⁺

Reduction potentials as follows:

Fe³⁺ + e- ==> Fe²⁺ Eº = 0.77

Cu²⁺ ==> Cu⁺ Eº = 0.16 V

Eº_cell = 0.77 - 0.16 = 0.61 V

Using the Nernst equation:

E_cell = Eº_cell - 2.303RT/nF log Q and assuming 298K we have E_cell = Eº_cell - 0.0591/n log Q

Q = [Fe²⁺][Cu²⁺]/[Fe³⁺][Cu⁺]

Fe³⁺ + Cu⁺ ==> Fe²⁺ + Cu²⁺

0.002......0.001..........0..............0........Initial

-0.001....-0.,001.....+0.001......+0.001...Change

0.001......0.............0.001.........0.001....Final moles in volume of 0.110 L

0.0909....0...........0.0909........0.0909....Final M

Q = (0.0909)(0.0909)/(0.0909) = 0.0909

E_cell = 0.61 - 0.0591/1 log 0.0909

E_cell = 0.61 - 0.062

E_cell = 0.55 V

You can do the second part the same way using 19.5 ml (0.0195 L) of the 0.1 M Cu⁺