For a cell reaction involving a two-electron change,the standard $EMF$ of the cell is $0.295 \ V$ at $25 \ ^\circ C$. The equilibrium constant of the reaction at $25 \ ^\circ C$ will be:

For a $Daniel$ cell $Zn | ZnSO_{4(0.01 \ M)} || CuSO_{4(1 \ M)} | Cu$ at $298 \ K$,the cell potential is $E_1$. When the concentrations of $ZnSO_4$ and $CuSO_4$ are changed to $1 \ M$ and $0.01 \ M$ respectively,the cell potential becomes $E_2$. Determine the relationship between $E_1$ and $E_2$.

For a cell,$Cu_{(s)} \mid Cu^{2+}(0.001\,M) \mid\mid Ag^{+}(0.01\,M) \mid Ag_{(s)}$,the cell potential is found to be $0.43\,V$ at $298\,K$. The magnitude of standard electrode potential for $Cu^{2+}/Cu$ is $......... \times 10^{-2}\,V$. $[\text{Given}: E^{\Theta}_{Ag^{+}/Ag} = 0.80\,V \text{ and } \frac{2.303RT}{F} = 0.06\,V]$

Consider the following half-cell reaction: $Cr_2O_7^{2-}{_{\text{(aq)}}} + 14H^{+}{_{\text{(aq)}}} + 6e^{-} \rightarrow 2Cr^{3+}{_{\text{(aq)}}} + 7H_2O_{\text{(l)}}$. The reaction was conducted with the ratio $\frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}]} = 10^{-6}$. The $pH$ value at which the $EMF$ of the half-cell will become zero is $............$ (nearest integer value). [Given: standard half-cell reduction potential $E^{o}_{Cr_2O_7^{2-}, H^{+} / Cr^{3+}} = 1.33 \ V$,$\frac{2.303 RT}{F} = 0.059 \ V$]

For the electrochemical cell,$Mg_{(s)} \mid Mg^{2+}(aq, 1 \ M) \parallel Cu^{2+}(aq, 1 \ M) \mid Cu_{(s)}$,the standard emf of the cell is $2.70 \ V$ at $300 \ K$. When the concentration of $Mg^{2+}$ is changed to $x$,the cell potential changes to $2.67 \ V$ at $300 \ K$. The value of $x$ is.
(Given: $\frac{F}{R} = 11500 \ K \ V^{-1}$,where $F$ is the Faraday constant and $R$ is the gas constant; $\ln(10) = 2.30$)

For a galvanic cell reaction at $25^{\circ}C$ with $n = 4$,the standard $emf$ is $0.295 \ V$. What is the equilibrium constant for the reaction? $(F = 96500 \ C \ mol^{-1}; R = 8.314 \ J \ K^{-1} \ mol^{-1})$