For the cell, Pt|Cl_2(g,0.4"bar")|Cl^(-)(aq,0.1M)"||"Cl^(-)(aq),0.01M)|Cl_2(g,0.2"bar")|pt
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 .
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 )
Calculate the standard Gibbs energy for the cell : Zn(s)+Zn^(2+)(aq)||Cu^(2+)(aq)|Cu(s) E_((Zn^(2+)//Zn)^(@))=-0.76V,E_((Cu^(2+)//Cu))^(@)=0.34V , F=96500 C .
Give the Nernst equation of the cell: Ni(s)|NI^(2+)(aq)(0.1M)||Ag^(+)(aq)(0.1M)|Ag(s) and also find the cell potential. given that E^@(Ag^+|Ag)=0.80V, E^@(Ni^(2+)|Ni)=-0.25V
Write Nernst equation and calculate the e.m.f. of the following cell at 298 K: Cu(s)|Cu^(2+)(0.130M)||Ag^(+)(1.0xx10^(-4)M)|Ag(s) Given that: E_(Cu^(2+)//Cu)^(@)=+0.34V and E_(Ag^(+)//Ag)^(@)=+0.80V (log 0.130=-1.1139).
Calculate the standard Gibbs energy for the cell : Zn(s)|Zn^(2+)(aq)||Sn^(2+)(aq)|Sn(s) E_((Zn^(2+)//Zn))^(@)=-0.76V,E_((Sn^(2+)//Sn))^(@) = -0.16V , F=96500C .
Calculate the cell e.m.f for reaction. Zn_((s))//Zn_((0.0004M))^(2+)//Cd_((0.2M))^(2+)//Cd_((s)) . Given E_(Zn^(2+)//Zn)^(@)= -0.763V, E_(Cd^(2+)//Cd)^(@)= -0.403V