Question

Consider the cell described below at 261 K:

Sn | Sn²⁺ (0.777 M) || Pb²⁺ (0.957 M) | Pb

Given E^o_Pb²⁺→Pb = -0.131 V, E^o_Sn²⁺→Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn²⁺ to have changed by 0.379 mol/L.

I got an E standard of .17

I then used the nernst equation using concentrations that I got from setting up an ice table.

E=.17-((.059/2) ln(1.336/.398)) = .1343 V

Does anyone know where I went wrong?

Answer 1

we know that

Eo cell = Eo cathode - Eo anode

here

Sn is anode

and

Pb is cathode

so

Eo cell = -0.131 - ( -0.143)

Eo cell = 0.012

now

the cell reaction is

Sn + Pb+2 ---> Sn+2 + Pb

given

Sn+2 has been changed by 0.379 M

so

finally

[Sn+2] = 0.777 + 0.379 = 1.156

[Pb+2] = 0.957 - 0.379 = 0.578

now

E = Eo - (RT/zF) ln K

so

E = Eo - (8.314 x 261 / 2 x 96485) ln [Sn+2] / [Pb+2]

E = 0.012 - 0.011245 ln [ 1.156 / 0.578)

E = 0.0042

so

the cell potential is 0.0042 V