Answer:
1.20 V
Step-by-step 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