Question

Consider a cell made of one compartment with 0.2M Pd(NO3)2, and a palladium wire. The other compartment is filled with 0.15M Tl(NO3)3, and a thallium wire. The two compartments

are connected with a salt bridge. E° for Tl3+ + 3e- → Tl0 = 0.72V vs SHE, and E° for Pd2+ + 2e- → Pd0 = 0.915V vs SHE. What is the potential of this cell?

Answer 1

As from data construction of cell the criterion would be as:

Anode (because it should have less E^o value): Tl(NO₃)₃             E^o = 0.72 (given) left hand side ie. LHS

Cathode (because it should have more E^o value): Pd(NO₃)₂      E^o = 0.915 (given) right hand side i. RHS

Now we can calculate Standard cell potential as:

E^o_cell = [ E^o of RHS (cathode) - E^o of LHS (anode)]

         = [0.915 - 0.72]

         = 0.195 V

For calculation of potential of this cell (E_cell), lets see the balenced chemical change in terms of equation:

2Tl    + 3Pd²⁺ =   2Tl³⁺   + 3Pd

So, in this process total exchange of electrons are 6.

For a simle cell reaction like:     aA + bB    ==== xX + yY

Nernst eq can be given as:   E_cell = E^o_cell - 2.303RT / nF log[X]^x [Y]^y / [A]^a [B]^b

where, E_cell = EMF at given concentration,   E^o_cell = Standard EMF of cell, R = gas constant (8.314 J/Kmole), n = no. of electrons involved in cell reaction and F = 1 Faraday = 96500 coulombs (approx)

       After calculation of all constant values at STP condition net eq:     E_cell = E^o_cell - 0.0591/n log [Tl³⁺]² / [Pd²⁺]³

[Tl³⁺] = 0.15M and [Pd²⁺] = 0.2M                ( as given in problem)

Now, from eq. E_cell = 0.195 - 0.0591/6 log [0.15]² / [0.2]³              (here n = 6)

after final calculation:        E_cell = 0.191V