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 the equilibrium constant of the reaction,$Cu_{(s)} + 2 Ag^{+}_{(aq)} \longrightarrow Cu^{2+}_{(aq)} + 2 Ag_{(s)}$,given that for the reaction $E^{\circ}_{cell} = 0.46 \ V$.

For an electrochemical cell
$Sn_{(s)} | Sn^{2+}(aq, 1 \ M) || Pb^{2+}(aq, 1 \ M) | Pb_{(s)}$
the ratio $\frac{[Sn^{2+}]}{[Pb^{2+}]}$ when this cell attains equilibrium is
(Given $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

For the cell,$Mn_{(s)} | Mn^{2+}_{(aq)} (0.4 \, M) || Sn^{2+}_{(aq)} (0.04 \, M) | Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298 \, K$ in $kJ$. Given: $E^o_{Mn^{2+}/Mn} = -1.18 \, V$; $E^o_{Sn^{2+}/Sn} = -0.14 \, V$; $\frac{2.303RT}{F} = 0.06$.

The potential of a hydrogen electrode at $pH = 10$ is (in $V$)

$Cu_{(s)} | Cu^{+2}(aq, 10^{-3} M) || Ag^{+}(aq, 10^{-5} M) | Ag_{(s)}$
If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$
$E_{cell}$ will be