Given that `E^@ (Zn^(2+) //Zn) =- 0.763 V` and `E^@ ( Cd^(2+) //Cd) = -0.403 V`, the emf of the following cell :
`Zn//Zn^(2+) (a = 0.04) || Cd^(2+) (a = 0.2)` ce is given by ,

Options

A. `E=- .36 + [0.059//2] [log (0.004 //0.2)]`
B. ` E= +36 + [0.059//2] [log (0.004 //0.2)]`
C. ` E=- 0.36 + [o.059 //2][log (0.2 //0.004)]`
D. ` E=- 0.36 + [o.059 //2][log (0.2 //0.004)]`

Answer

B

Explanation

Since ` E_(Cd^2 + //Cd)^(@)gt E_(Zn^(2+)//Nn)^(@)`, therefore Zn electrode acts at anode and Cd electrode as cathode .
Anode :` Zn (s)rarr Zn^(2+) (aq) + 2e^(-)`
`(Cathode :Ce^(2+) +2e^(-) rarr Cd (s))/(Zn(s) + Cd^(2+) (aq) rarr Zn^(2+) (aq) +Cd(s))`
` Q= [Zn^(2+) ]//[Cd^(2+) ] = 0.004 //02`
`E_("cell")^@ =E_("cathode")^@ E_(anode)^(@) - 0.403 V - (-0. 763 V)`
`=0.36 V`
`E_(cell) = 0.36 - (0.59)/2 log .(0.004)/(0.2)`.