Calculate $E_{cell}$ of the reaction $Mg_{(s)} + 2Ag^{+}_{(aq)} (0.0001 \ M) \to Mg^{2+}_{(aq)} (0.100 \ M) + 2Ag_{(s)}$ in $V$. If $E^o_{cell} = 3.17 \ V$.

If $E^{\circ}(Ag^{+}_{(aq)} \mid Ag_{(s)}) = +0.80 \ V$,what is the potential developed for $Ag_{(s)} \rightarrow Ag^{+}_{(aq)} (0.01 \ M) + e^{-}$ at $298 \ K$?

The cell reaction involving the quinhydrone electrode is given by the following equation:
$C_6H_4(OH)_2 \rightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$E^{\circ} = 1.30 \ V$
What will be the electrode potential at $pH = 3$ (in $V$)?

Consider the cell:
$Pt_{(s)} | H_{2(g)}(1 \ atm) | H^{+}_{(aq)}, [H^{+}]=1 \ M || Fe^{3+}_{(aq)}, Fe^{2+}_{(aq)} | Pt_{(s)}$
Given: $E^0_{Fe^{3+}/Fe^{2+}} = 0.771 \ V$ and $E^0_{H^{+}/\frac{1}{2}H_2} = 0 \ V$ at $T = 298 \ K$.
If the potential of the cell is $0.712 \ V$,the ratio of concentration of $Fe^{2+}$ to $Fe^{3+}$ is $........$. (Nearest integer)

$Pt_{(s)} | H_{2(g)}(1 \ bar) | H^{+}_{(aq)}(1 \ M) || M^{3+}_{(aq)}, M^{+}_{(aq)} | Pt_{(s)}$
The $E_{cell}$ for the given cell is $0.1115 \ V$ at $298 \ K$ when $\frac{[M^{+}_{(aq)}]}{[M^{3+}_{(aq)}]} = 10^{a}$.
The value of $a$ is.
Given : $E^{\circ}_{M^{3+}/M^{+}} = 0.2 \ V$
$\frac{2.303 \ RT}{F} = 0.059 \ V$

The standard electrode potential of $Cu^{2+}/Cu$ is $0.34 \text{ V}$ at $298 \text{ K}$. Calculate its electrode potential at the same temperature when the $Cu^{2+}$ ion concentration is $0.1 \text{ M}$. (in $\text{ V}$)