A cell is set up between copper and silver electrodes as follows:
`Cu(s)I Cu^(2+)(aq)II Ag^+(aq)Iag(S)`
If the two half cells work under standard conditions, calculate the EMF of the cell
`(Given E^(@)_(Cu^(2+)//Cu) =+0.34 V,E^(@)_(Ag^(+)//Ag)" " =+0.80 V)`

Explanation

From the `E^(@)` values it is clear that copper acts as the anode silver as the cathode.
`E^(@)_(cell)=E^(@)_(cathode-E^(@))_(Enode)=0.80-(0.34)=+0.46 "volt"`