`E^@` value of `Ni^(2+)//Ni` is -0.25 V and `Ag^+ //Ag` is +0.80 V . If a cell is made by taking the two electrodes what is the feasibility of the reaction?

Standard electrode potential of (Ni^(2+))/(Ni) couple is - 0.25 V and that of (Ag^+)/(Ag) is 0.80 V. These couples in their standard state are connected to make a cell. The cell potential will be

Standard electrode potential of (Ni^(2+))/(Ni) couple is - 0.25 V and that of (Ag^+)/(Ag) is 0.80 V. These couples in their standard state are connected to make a cell. The cell potential will be

The SRP values of Ag^(+)//Ag and Zn^(2+)//Zn electrodes are 0.80 V and -0.7 v. In the cell built with these two electrodes

Calculate the emf of the cell at 298K E_(NI^(2+)//NI)^0 is -0.25V and E_(Ag^+//Ag)^0 is +0.80V .

E^(@) values of Ni^(2+)//Ni and Cu^(2+)//Cu are -0.25 V and +0.34 V respectively. Is the reaction: Ni+Cu^(2+)toNi^(2+)+Cu spontaneous or not ? Give reason also.

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 metals are displaced by Ni from the solutions of their salts ?

The standard electrode potential of the htwo half cells are given below Ni^(+2) + 2e^(-) to Ni , E^(@) = -0.25 V , Zn^(+2) + 2e^(-) to Zn , E^(0) = -0.77 V The voltage of cell formed by combining the two half cells would be

A cell is constructed Pb^(2+) Pb and Ni^(2+) , Ni electrodes. If E^(@) values of Pb and Ni electrodes are respectively -0.13 and -0.25V, write (a) the cell reaction and (b) cell notation.

Given that, Ni^(2+)//Ni=0.25 V, Cu^(2+)//Cu=0.34 V , Ag^(+)//Ag=0.80 V and Zn^(2+)//Zn=-0.76 V Which of the following reaction under standard condition will not take place in the specified direction ?