Determine the electrode potential of $Sn^{2+}(0.01 \text{ M}) | Sn(s)$ at $25^{\circ} \text{C}$ if $E^0_{Sn^{2+}/Sn}$ is $-0.136 \text{ V}$. (in $\text{ V}$)

For the cell reaction $Sn_{(s)} + 2Ag^+_{(aq)} \rightarrow Sn^{2+}_{(aq)} + 2Ag_{(s)}$,what happens when the cell voltage increases?

For a cell having a standard $emf$ of $0.295 \ V$ at $25^o \ C$ involving a two-electron change,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 $298 \ K$?

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 the disproportionation reaction $2 Cu ^{+}( aq ) \rightleftharpoons Cu ( s ) + Cu ^{2+}( aq )$ at $298 \ K$,$\ln K$ (where $K$ is the equilibrium constant) is....... $\times 10^{-1}$.
Given: $(E _{ Cu ^{2+} / Cu ^{+}}^{0} = 0.16 \ V, E _{ Cu ^{+} / Cu }^{0} = 0.52 \ V, \frac{ RT }{ F } = 0.025 \ V)$