`E_(Zn^(2+)|Zn)^(@)=-0.76V and E_(Ag^(+)|Ag)^(@)=+0.799V` and a galvanic cell has `Zn^(2+)|Zn and Ag^(+)|Ag` electrodes. The Zn - electrode will be ___________ and Ag - electrode will be _____________ .

For Zn^(2+) //Zn, E^(@) =- 0.76 , for Ag^(+)//Ag, E^(@) = -0.799V . The correct statement is

For Zn^(2+) //Zn, E^(@) =- 0.76 , for Ag^(+)//Ag, E^(@) = -0.799V . The correct statement is

For Zn^(2+) //Zn, E^(@) =- 0.76 , for Ag^(+)//Ag, E^(@) = -0.799V . The correct statement is

At 25^(@)C temperature, E_(Zn^(2+)|Zn)^(@)=-0.76V and E_(Ag^(2+)|Ag)^(@)=+0.80V. If the E_("cell")^(@) for a galvanic cell formed by combining the two half - cells, Zn^(2+)(1M)|Zn and Cu^(2+)(1M)|Cu is 1.1V , then E_("cell")^(@) for a galvanic cell formed by combining the two half - cells, Cu^(2+)(1M)|Cu and Ag^(+)(1M)|Ag, will be -

Given that, E_(Zn^(2+)|Zn)^(@)=-0.76V, E_(Cu^(2+)|Cu)^(@)=+0.34V , E_(Ni^(2+)|Ni)^(@)=-0.25V and E_(Ag^(+)|Ag)^(@)=+0.80V Which metal ions will be oxidised by Ag^(+) ion?

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 .

Given that, E_(Zn^(2+)|Zn)^(@)=-0.76V, E_(Cu^(2+)|Cu)^(@)=+0.34V , E_(Ni^(2+)|Ni)^(@)=-0.25V and E_(Ag^(+)|Ag)^(@)=+0.80V Which metal ions will be oxidised by Zn^(2+) ion?

When E_(Ag^(+) | Ag)^@ = 0.80 volt and E_(Zn^(2+)|Zn)^@ = 0.76 volt, which of the following is correct?

Given that, E_(Zn^(2+)|Zn)^(@)=-0.76V, E_(Cu^(2+)|Cu)^(@)=+0.34V , E_(Ni^(2+)|Ni)^(@)=-0.25V and E_(Ag^(+)|Ag)^(@)=+0.80V Which metal ions are reduced by Ag^(+) ?

For Zn^(2+) //Zn, E^@ =- 0. 76 V , for Ag^+ //Af E^@ =0.799 V . The correct statement is .