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:

Calculate the e.m.f of the following cell: Cd|Cd^(2+)(0.01M)||H^(+)(0.02M)|Pt,H_2(0.8atm) Given: E^@(Cd^(2+)|Cd)=-0.40V

Calculate the standard Gibbs energy for the cell : Zn(s)|Zn^(2+)(aq)||Cd^(2+)(aq)|Cd(s) E_((Zn^(2+)//Zn))^(@)=-0.76V,E_((Cd^(2+)//Cd))^(@)=-0.403V,F=96500C .

For the galvanic cell Ag|Ag_((aq))^(+)(0.1M)"||"Cd_((aq))^(2+)(0.1M)|Cd (E_(Ag^(+)//Ag)^(0)=0.80V,E_(Cd^(2+)//Cd)^(0)=-0.40V) ,

Calculate the cell e.m.f. and Delta G for the cell reaction at 298 K for the cell. Zn(s)|Zn^(2+)(0.0004 M)|| Cd^(2+)(0.2 M)| Cd(s) Given E_(Zn^(2+)//Zn)^(@)=-0.763 V,E_(cd^(2+)//cd)^(@)=-0.403 V " at "298 K, F=96500 C " mol"^(-1) .

Calculate the cell e.m.f. and DeltaG for the cell reaction at 298K for the cell. Zn(s) | Zn^(2+) (0.0004M) ||Cd^(2+) (0.2M)|Cd(s) Given, E_(Zn^(2+)//Zn)^(@) =- 0.763 V, E_(Cd^(+2)//Cd)^(@) = - 0.403 V at 298K . F = 96500 C mol^(-1) .

Calculate the cell e.m.f. and DeltaG for the cell reaction at 298K for the cell. Zn(s) | Zn^(2+) (0.0004M) ||Cd^(2+) (0.2M)|Cd(s) Given, E_(Zn^(2+)//Zn)^(@) =- 0.763 V, E_(Cd^(+2)//Cd)^(@) = - 0.403 V at 298K . F = 96500 C mol^(-1) .

Calculate the cell e.m.f. and DeltaG for the cell reaction at 298K for the cell. Zn(s) | Zn^(2+) (0.0004M) ||Cd^(2+) (0.2M)|Cd(s) Given, E_(Zn^(2+)//Zn)^(@) =- 0.763 V, E_(Cd^(+2)//Cd)^(@) = - 0.403 V at 298K . F = 96500 C mol^(-1) .

Calculate the cell emf and triangle_(r)G^(@) for the cell reactin at 25^(@)C Zn(s)|Zn^(2+)(0.1M)||Cd^(2+)(0.01M)|Cd(s) (given E_(Zn^(2+)//Zn)^(@)=-0.763V,E_(Cd^(2+)//Cd)^(@)=-0.403V 1F=96500Cmol^(-1),R=8.314JK^(-1)mol^(-1)]

Calculate the standard cell potentials of the galvanic cells in which the following reactions take place. (a) " " 2Cr(s)+3Cd^(2+) to 2Cd^(3+)(aq)+3Cd(s) Given E_(Cr^(3+)//Cr)^(@)=-0.74" V" , E_(Cd^(2+)//Cd)^(@)=-0.40" V" (b) " " Fe^(2+)(aq)+Ag^(+)(aq) to Fe^(3+)(aq)+Ag(s) Gievn E_(Ag^(+)//Ag)^(@)=0.80" V" ,E_(Fe^(3+)//Fe^(2+))^(@)=0.77 " V" Also calculate DeltaG^(@) and equilibrium constant for the reaction.

If the equilibrium constant for the reaction Cd^(2+)(aq)+4NH_3(aq)iffCd(NH_3)_4^(2+)(aq) is 10^x then find the value of x. (Given: E_(Cd^(2+)"|"Cd)^(@)=-0.4V,E_(Cd(NH_3)_4^(2+)"|"Cd)^(@)=-0.61V )