For a cell involving a one-electron change at $25^o C$,$E^{o}_{cell} = 0.591 \ V$. The equilibrium constant for the reaction is .....

If $E^{\circ}(Cu^{2+}_{(aq)} \mid Cu_{(s)}) = +0.34 \ V$. What is the potential for $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} (0.1 \ M) + 2e^-$ at $298 \ K$?

If the standard electrode potential for a cell is $2 \ V$ at $300 \ K,$ the equilibrium constant $(K)$ for the reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}$ at $300 \ K$ is approximately $(R = 8 \ J \ K^{-1} \ mol^{-1}, F = 96000 \ C \ mol^{-1})$

The equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is:

What must be the concentration of $Ag^{+}$ in an aqueous solution containing $Cu^{2+} = 1.0 \ M$ so that both the metals can be deposited on the cathode simultaneously? Given that $E^0_{Cu^{2+}/Cu} = 0.34 \ V$ and $E^0_{Ag^{+}/Ag} = 0.812 \ V$ at $T = 298 \ K$.

Consider the following electrochemical cell,$Zn_{(s)} + 2Ag^{+}(0.04\, M) \longrightarrow Zn^{2+}(0.28\, M) + 2Ag_{(s)}$. If $E_{\text{cell}}^{\circ} = 2.57\, V$,then the emf of the cell at $298\, K$ is $......\, V$. (in $.5$)