If the concentration of `Zn^(2+)` ions is halved, then calculate change in potential of cell at `25^(o)`C zn` ` `Zn(+2)`

What is the effect of increase in concentration of zinc ions on the electrode potential of zinc electrode for which E_(Zn^(2+)//Zn) equals - 0.76 V

A change of Zn to Zn^(2+) is a accompanied by a decrease in:

A change of Zn to Zn^(2+) is a accompanied by a decrease in:

A change of Zn to Zn^(2+) is a accompanied by a decrease in:

To increase significantly the concentration of free Zn^(2+) ion is a solution of the complex ion [Zn(NH_(3))_(4)]^(2+) Zn^(2+) (aq)+4NH_(3) (aq) hArr [Zn(NH_(3))_(4)]^(2+) (aq) Add to the solution some:

Using the date in the preceding problem, calculate the equilibrium constant of the reaction at 25^(@)C Zn +Cu^(++) hArr Zn^(++) +Cu, K ([Zn^(2+)])/([Cu^(2+)])

Statement-I: In the Daniel cell, if concentration of Cu^(2+) and Zn^(2+) ions are doubled the emf of the cell will not change. Because Statement-II: If the concentration of ions in contact with the metals is doubled, the electrode potential is doubled.

What are the coordination number of Zn^(2+) and S^(2-) ions in ZnS?

E_(Cu^(+2)//Cu)^(@)=0.34V, E_(Zn//Zn^(+2))^(@)=0.76V A cell formed by the conbination of Cu and Zn (a) when CuSO_(4) is added to Cu^(+2) compartment what is the effect on emf of cell (b) when NH_(3) is added to Cu^(+2) compartment what is the effect on emf of cell (c ) When ZnSO_(4) is added to Zn^(+2) compartment is the effect on emf of cell (d) When Zn^(+2) is diluted what is the effect on emf of cell ?

Zn^(2+) ions are....... and.......