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 $V$)

The cell potential for the following cell notation is approximately
$M_{(s)} | M^{3+}(aq, 0.01 \ M) || N^{2+}(aq, 0.1 \ M) | N_{(s)}$
$E_{M^{3+} / M}^0 = 0.6 \ V$ and $E_{N^{2+} / N}^0 = 0.1 \ V$ (in $V$)

If $E^{\circ}(Mg^{+2}_{(aq)} \mid Mg_{(s)}) = -2.37 \ V$. What is the potential for $Mg_{(s)} \rightarrow Mg^{+2}_{(0.01 \ M)} + 2 \overline{e}$ at $298 \ K$?

For the given cell $Cu(s) | Cu^{2+}(C_1 \ M) || Cu^{2+}(C_2 \ M) | Cu(s)$,the change in Gibbs energy $(\Delta G)$ is negative,if:

Calculate the $E.M.F.$ of the following cell at $298 \ K$: $Zn_{(s)} | ZnSO_4(0.01 \ M) || CuSO_4(1.0 \ M) | Cu_{(s)}$ if $E^o_{cell} = 2.0 \ V$. (in $V$)

The cell,$Zn\ |\ Zn^{2+} \,(1\ M)\ ||\ Cu^{2+}\ (1\ M)\ |\ Cu$ $(E^o_{cell} = 1.10\ V)$ was allowed to be completely discharged at $298\ K.$ The relative concentration of $Zn^{2+}$ to $Cu^{2+}$ $\left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$ is