For a reaction,$A + B^{2+} \to B + A^{2+}; E^{\circ} = 0.2955 \ V$. Hence,the equilibrium constant of the reaction at $25 \ ^oC$ is:

The cell potential for $Zn | Zn^{2+}_{(aq)} || Sn^{x+}| Sn$ is $0.801 \ V$ at $298 \ K$. The reaction quotient for the above reaction is $10^{-2}$. The number of electrons involved in the given electrochemical cell reaction is .... (Given $E^{0}_{Zn^{2+}|Zn} = -0.763 \ V, E^{0}_{Sn^{x+}|Sn} = +0.008 \ V$ and $\frac{2.303 \ RT}{F} = 0.06 \ V$)

On diluting the concentration of $Cl^{-}$ ions by ten times in a calomel electrode,its reduction potential at $298 \ K$ is -

Derive the Nernst equation for calculating $E_{cell}$ of a Daniell cell and explain the effect on $E_{cell}$ when there is a change in the concentration of $Zn^{2+}$ and $Cu^{2+}$ ions.

If the $E^{\circ}_{cell}$ of an equilibrium reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ at $298 \ K$ is $0.59 \ V$,the equilibrium constant $K_c$ is

Calculate the $EMF$ of the cell: $Cr | Cr^{+3}(0.1 \, M) || Fe^{+2}(0.01 \, M) | Fe$
(Given: $E^o_{Cr^{+3}|Cr} = -0.75 \, V$,$E^o_{Fe^{+2}|Fe} = -0.45 \, V$) (in $, V$)