In two vessels each containing `500mL` water, `0.05m` mol of aniline `(K_(b) = 10^(-9))` and `25m` mol of `HCI` are added separately .Two hydorgen electrodes are constructed using these solutions. Calculate the emf of cell made by connecting them appropriately.

Options

A. `0.395 V`
B. `0.792 V`
C. `0.452 V`
D. `0.312 V`

Answer

A

Explanation

`[OH^(-)]` in Aniline sol. `sqrt(k_(b)xxC)=sqrt(10^(-9)xx10^(-3))`
`=10^(6)M " " therefore [H^(+)]=10^(-8)M`
`[H^(+)]` in Hcl sol. `= (25xx10^(-3))/(0.5)=5xx10^(-2)M`
Cell: `H_((ap))^(+)(C_(1))|H_(2(g))(1.0 atm)|H_((ap))^(+)(c_(2))`
`E_("cell")=(0.059)/(1)"log"(c_(2))/(c_(1))=(0.059)/(1)"log"(5xx10^(-2))/(10^(-8))`
`=0.059(6+0.7)=0.395V`