Calculate the following cell reaction cell at 298 K.
`2Ag^(+)+Cd to 2Ag+Cd^(2+)`
`E^(@)`for `Ag^(+)//Ag` and `Cd^(2+)//Cd` are 0.80 V and -0.40 V respectively.
`E^(@)`for `Ag^(+)//Ag` and `Cd^(2+)//Cd` are 0.80 V and -0.40 V respectively.
(i) Write the cell representation.
(ii) What will be the emf of the cell if the concentration of `Cd^(2+)` is 0.1 M ?
(iii) Will the cell work spontaneously for the condition given above ?

Answer

1.18824V; Yes

Explanation

(i) The cell representation is :
`Cd(s)+2Ag^(+)(aq) to Cd^(2+)(aq)+2Ag`
(ii) `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Cd^(2+)])/([Ag^(+)]^(2))`
`E_(cell)=(1.2" V ")-((0.0591" V "))/(2)"log"(0.1)/(0.2xx0.2)`
`=1.2V-(0.02955" V")xx0.3979`
`=1.2-0.01176=1.18824" V "`
(iii) Yes, Since `E_(cell)` is positive.