Calculate the e. m. f of a cell if 0.01M solution of each of Cd2 + and Ag + are used?

1.20 V

Explanation:

The concentration of both solutions as stated in the question is 0.01M

However, based of reduction potential; Cd^2 + = - 0.40V and Ag^ + = 0.80V. Silver will be the cathode while cadmium will be the anode.

Hence

E°cell = E°cathode - E°anode

E°cell = 0.80 - (-0.40) = 1.20 V

But n=2 moles of electrons

From Nernst Equation:

E = E°cell - 0.0592/n log [Red]/[Ox]

E = 1.20 - 0.0592/2 log (0.01/0.01)

E = 1.20 V