Calculate $E_{cell}$ of the reaction in $V$:
$Mg_{(s)} + 2Ag^{+}(0.0001 \ M) \to Mg^{+2}(0.100 \ M) + 2Ag_{(s)}$
Given that $E_{cell}^o = 3.17 \ V$.

If a cell has a standard electrode potential of $0.295 \ V$ and $n = 2$,calculate its equilibrium constant at $298 \ K$.

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:

The $E^o$ for the cell $Zn|Zn^{2+}_{(aq)}||Cu^{2+}_{(aq)}|Cu$ is $1.10 \ V$ at $25 \ ^oC$. The equilibrium constant for the reaction $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$ is of the order of:

At $298 \ K$ the equilibrium constant for the reaction $M_{(s)} + 2 Ag^{+}_{(aq)} \rightarrow M^{2+}_{(aq)} + 2 Ag_{(s)}$ is $10^{15}$. What is the $E_{cell}^{\ominus}$ (in $V$) for this reaction? $\left(\frac{2.303 RT}{F}\right) = 0.06 \ V$

$A$ copper electrode is dipped in a $0.1 \, M$ copper sulfate solution at $25 \, ^\circ C$. Calculate the reduction potential of the copper electrode. $(E^o_{Cu^{2+}/Cu} = 0.34 \, V)$ (in $, V$)