`E^(@)(Cu^(2+)//Cu) and E^(@)(Ag^(+)//AG)" is "0.337V and +0.799 V` respectively. Make a cell whose EMF is `+ve`. If the concentration of `Cu^(2+)" is 0.01M and "E_("cell")" at "25^(@)C` is zero, calculate the concentration of `Ag^(+)`.

Explanation

Cu is more reactive than silver, so that the cell is as `Cu//Cu_(2)^(+) (0.01M) || Ag^(+) (C )//Ag` or cell reaction
`Cu+2Ag^(+)rarrCu^(2+)+2Ag`
`E_("cell")=E_("cell")^(@)-(0.0591)/(n)log.([Cu^(2+)][Ag]^(2))/([Cu][Ag^(+)]^(2))`
`=E_("cell")^(@)-(0.0591)/(n)log.((0.01)xx1^(2))/(1xx[Ag^(+)]^(2))`
`"Or "[Ag^(+)]=1.47xx10^(-9)M`