A cell is set up between copper and silver as follows : `Cu(s)|Cu^(2+)(aq)||Ag^(+)(aq)|Ag(s)`
If two half cells work under standard conditions, calculate the emf of the cell.
Given `E_(Cu^(2+)|Cu)^(@)=0.34V,E_(Ag^(+)//Ag)^(@)=0.80V`

Explanation

The cell reaction is given as follows: `Cu(s)|Cu^(2+)(aq)||Ag^(+)(aq)|Ag(s)`
We are given that: `E_(Cu^(2+)|Cu)^(@)=0.34V,E_(Ag^(+)//Ag)^(@)=0.80V`
From the `E^(@)` values, it is clear that Cu acts as anode while Ag acts as cathode.
The emf of cell is given by `E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)`
Putting the values in above relation, we have `E_("cell")^(@)=0.80-0.34=+0.46V`