Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)

For a cell reaction involving a two-electron change,the standard emf of the cell is found to be $0.295 \ V$ at $25 \ ^oC$. The equilibrium constant of the reaction at $25 \ ^oC$ will be

Calculate cell potential at $298 \ K$ for the following cell.
$Ag_{(s)} | Ag^{+}(0.25 \ M) || Ag^{+}(0.75 \ M) | Ag_{(s)}$ $\left[ E_{Ag^{+} \mid Ag}^{o} = 0.80 \ V \right]$

The equilibrium constant for the following general reaction is $10^{30}$. Calculate $E^{o}$ for the cell at $298 \ K$ ............ $V$
$2X_{2(s)} + 3Y^{2+}_{(aq)} \to 2{X_{2}}^{3+}_{(aq)} + 3Y_{(s)}$

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 $298 \ K$?

At $298 \ K$,the standard reduction potentials are $1.51 \ V$ for $MnO_4^- \ | \ Mn^{2+}$,$1.36 \ V$ for $Cl_2 \ | \ Cl^{-}$,$1.07 \ V$ for $Br_2 \ | \ Br^{-}$ and $0.54 \ V$ for $I_2 \ | \ I^{-}$. At $pH = 3$,permanganate is expected to oxidize $\left( \frac{RT}{F} = 0.059 \ V \right)$