Question

Calculate the equilibrium constant K of the reaction Sn(s)|Sn²⁺(aq)||Ag⁺(aq)|Ag(s) at 25 °C.

Answer 1

The half cell reactions can be written as:

At cathode: Ag+(aq)|Ag(s)

2Ag+ (aq) + 2e- 2Ag (s)     , E0Red = 0.80 V ; E0 vale is unaffected if we multiply the reaction by 2

At Anode: Sn(s)|Sn2+(aq)

Sn (s) Sn2+ (aq) + 2e-   , E0Oxi = +0.14 V

Overall reaction is: (note that total 2 electrons are exchanged, n = 2)

2Ag+ (aq) + Sn (s) + 2e- 2Ag (s)+ Sn2+ (aq) + 2e-

E0 cell = E0Red + E0Oxi = 0.80 V + 0.14 V = 0.94 V

Note that electrode potential values (E0) are:

Ag+ (aq) + e- Ag (s)     , E0 = 0.80 V

Sn2+ (aq) + 2e- Sn (s) , E0 = -0.14 V

At equilibrium, Ecell is 0.

The standard free energy change () = -nFE0 cell = -RT ln Keq

So, RT ln Keq = nFE0 cell

Or, ln Keq = (1/ RT) x nFE0 cell

Putting values: T = 250 C = (25 + 273 ) K = 298 K

                         R = 8.314 J / mol. K

                        n = 2 (as 2 electrons are exchanged)

                       F = 96485 C mol-1

Note that C. V = J, so the expression on right hand side is dimensionless.

ln Keq = 73.21

Keq = e 73.21           ....... Answer