Find `E_(cell)^(@)` for the cell :
`Zn | Zn^(2+)(1M) || Ag^(+)(1M) | Ag`
[Given that : `E_(Zn//Zn^(2+))^(@)=0.76" V" , E_(Ag^(+)//Ag)^(@)=0.80" V"`.

Calculate the emf of the following cell. Zn(s)|Zn^(2+) (0.01M)|| Ag^+ (0.001M) | Ag(s) Given E_(Zn|Zn^(2+))^@ = 0.76V and E_(Ag|Ag^+)^@ = -0.80V

The cell potential of the galvanic cell, Zn | Zn^(2+) || Ag^(+) | Ag, where E_(Zn^(2-)pin)^(@)= -0.76V,E_(Ag'//Ag)^(@) =+8.0V is

Find the E_(cell)^(@) for the following cell reaction Fe^(+2)+Zn rarr Zn^(+2)+Fe Given E_(Zn//Zn^(+2))^(@)=0.76V,E_(Fe//Fe^(+2))^(@)=+0.41 V

Write the cell reaction and cell potential of the cell Zn//Zn^(2+)(0.01 M)abs"Ag^+(0.1M)//Ag. Given E^0(Zn^(2+)//Zn)=-0.76V E^0(Ag^+//Ag)=+0.80V .

For the galvanic cell Zn(s)[Zn^2+ (aq)||Ag^+(aq) Ag(s)] , G° of the reaction in joule will be E°_ (Zn^(2+) / Zn )= - 0.76V , E°_( Ag_+/Ag )= 0.80 V

For the galvanic cell Zn(s)[Zn^2+ (aq)||Ag^+(aq) Ag(s)] , G° of the reaction in joule will be E°_ (Zn^(2+)) /( Zn )= - 0.76V , E°_( Ag_+)/(Ag )= 0.80 V

For the cell : Zn (s) |Zn^(2+)(a M)||Ag^(o+)(bM)|Ag(s) . a. Write Nernst equation to show how E_(cell) vary with concentration of Zn^(2+) and Ag^(o+) ions. Given E^(c-)._((Zn^(2+)|Zn))=0.76V,E^(-)._((Ag^(o+)|Ag))=0.80V. b. Find E_(cell) for [Zn^(2+)]=0.01M and [Ag^(c-)]=0.02M . c. For what values of Q will the cell EMF be i. 0.0V" "ii. 0.97 V