What is the value of the equilibrium constant at 25°C for the reaction between Al(s) and Cd^2+(aq)?

The half cell reactions and the standard electrode potentials at 25 degrees Celsius for the given species are:

`Al^(3+) + 3e^(-) -> Al(s)`

`Cd^(2+) + 2e^(-) -> Cd (s)`

Standard electrode potentials are -1.66 V and -0.4 V

The overall cell reaction between Cd2+ (aq) and Al (s) can be written as:

`Al(s) + Cd^(2+) -> Al^(3+) (aq) + Cd (s)`

and the balanced cell reaction can be written as:

`2Al(s) + 3Cd^(2+) (aq) -> 2Al^(3+) (aq) + 3Cd (s)`

At equilibrium, standard cell potential, E' = +1.66 + (-0.4) V = 1.26 V

Using Nernst equation: E' = (RT/nF) ln K

where, R = 8.314 J/k/mol, n = 6, F = 96500 C/mol, T = 298 K

we get: 1.26 = (8.314 x 298)/(6 x 96500)  ln K

solving the equation, we get, K = 7.6 x 10^127.

Thus, the equilibrium constant at 25 degrees C for the given cell is 7.3 x 10^127.

Hope this helps.