Question

Given the cell: `Cd(s)|Cd(OH)_2(s)|NaOH(aq,0.01M)|H_2(g,1"bar")|Pt(s)`
with `E_(cell)=0.0V."if"E_(Cd^(2+)|Cd)^(@)=-0.39V,"then"K_(sp)"of"Cd(OH_2)`is:
A. `0.1`
B. `10^(-13)`
C. `10^(-15)`
D. `10^(-18)`

Answer 1

Correct Answer - C
`E_(cell)=E_(H^(+)//H_(2))-E_(OH^(-)|Cd(OH)_(2)|Cd)`
`=E_(H^(+)|H_(2))-E_(OH^(-)|Cd(OH)_(2)|Cd)=E_(H^(+)|H_(2))-E_(OH_(Cd^(2+)|Cd))`
`E_(cell)=E^(@)-(0.06)/(2)log.([cd^(2+)])/([H^(+)]^(2))`
`E_(cell)="0 , E"_(cell)^(0)=0.31log.([cd^(2+)][OH^(-)]^(2))/(K_(w)^(2))`
`log.(K_(sp))/(K_(w)^(2))=(0.09)/(0.03)=13,K_(sp)=10^(13)xx(10^(-14))^(2)=10^(-15)`