Calculate the voltage `E` of the cell
`Ag(s)| AgIO_(3)(s)|Ag^(+) (x,M), HIO_(3) (0.300M)||Zn^(2+) (0.175M)|Zn(s)`
If `K_(SP) = 3.02 xx 10^(-8)` for `AgIO_(3)(s)` and `K_(a) = 0.162` for `HIO_(3), E^(@) (Zn^(+2)//Zn) =- 0.76 V, E^(@) (Ag//Ag^(+)) =- 0.8V`

Options

A. `-2.188V`
B. `-0.188V`
C. `-1.188V`
D. `-3.188V`

Answer

`-1.188V`

Explanation

`{:(HIO_(3),hArr,H^(+)+,IO_(3)^(-)),(C,,,),(C-X,,X,X):}`
`0.162 =(x^(2))/(0.3 -x)`
`x = 0.153`
`K_(sp) = [Ag^(+)] [IO_(3)^(-)]`
`[Ag^(+)] = 1.974 xx 10^(-7)`
`E_(cell) = 0.8 +(0.76) -(0.0591)/(2)log. (|Ag^(+)|^(2))/([Zn^(2+)])`
`E_(cell) - 1.188`