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 $\text{ V}$)

$Pt_{(s)} | Fe^{+2}(10^{-2} \ M, aq), Fe^{+3}(10^{-3} \ M, aq) || MnO_4^-(10^{-3} \ M, aq), Mn^{+2}(10^{-2} \ M, aq) | Pt_{(s)}$
At $298 \ K$,$E^o_{cell}$ for the cell is $-2.31 \ V$. What will be the $E_{cell}$ at $pH = 1$? .......... $Volt$

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 $E^{0}_{Sn^{2+}/Sn} = -0.14 \ V$,$E^{0}_{Pb^{2+}/Pb} = -0.13 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

Calculate the cell potential for $Ag_{(s)}|Ag^{+} \, (0.01 \ M)||Ag^{+} \, (0.1 \ M)|Ag_{(s)}$ at $298 \ K$. (in $V$)

The standard Gibbs energy change in $kJ \ mol^{-1}$ for a galvanic cell $A_{(s)} + B_{(aq)}^{3+} \longrightarrow A_{(aq)}^{3+} + B_{(s)}$ that has a standard emf of $0.5 \ V$ is: $\left(F = 96500 \ C \ mol^{-1}\right)$

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)