Given that `E^@ (Zn^(2+) //Zn) =- 0.763 V` and `E^@ ( Cd^(2+) //Cd) = -0.403 V`, the emf of the following cell :
`Zn//Zn^(2+) (a = 0.04) || Cd^(2+) (a = 0.2)` ce is given by ,

If E^@(Zn//Zn^(2+))=+0.76V , E^@ (Ag^+//Ag) =+0.80V , find the emf of the cell - Zn//Zn^(2+) (0.01M)abs"Ag+(0.1M)//Ag .

Cell reaction for the cell Zn|Zn^(2+)(1.0M)||Cd^(2+)(1.0M)|Cd is given by

E_(Cu^(2+)|Cu)^@ = +0.337 V, E_(Zn^(2+)| Zn)^@ = - 0.762 V. The EMF of the cell , Zn|Zn^(2+) (0.01M)||Cu^(2+) (0.01M)|Cu is

Calculate the cell emf and Delta_(r ) G^(@) for the cell reation at 25^(@)C . Zn(s) | Zn^(@+) ( 0.1 M )|| Cd^(2+) ( 0.01M) | Cd(s) Given, E_(Zn^(2+) //Zn)^(@) = - 0.763V , E_(Cd^(2+) //Cd) ^(@) = - 0.403 V 1F = 96500 C "mol"^(-1) R = 8.314 JK^(-1) "mol"^(-1) ], Find E_("cell")^(@) = E_("cathode")^(@) - E_("anode")^(@) then Delta _(r ) G^(@) by using formula, Delta _(r ) G^(@) = - n FE_("cell")^(@)

The standard reduction potential of Zn^(2+)|Zn and Cu^(2+)|Cu are -0.76V and +0.34 V respectively. What is the cell e.m.f (in V) of the following cell? ((RT)/(F)=0.059) Zn//Zn^(2+)" "(0.05M)//Cu^(2+)(0.005M)//Cu

The standard potential (E^(0)) for the half reaction are as Zn to Zn^(2+) + 2e^(-) , E^(0) = + 0.76 V Fe to Fe^(2+) + 2e^(-) , E^(0) = + 0.41 V . The emf for the cell reaction Fe^(2+) + Zn to Zn^(2+) + Fe is