Calculate the value of $\mathrm{\Delta G}$ and ${\mathrm{E}}_{\mathrm{cell}}$ at $298 \mathrm{K}$ for the following cell.

$\mathrm{Al}\left|{\mathrm{AI}}_{(0.01\mathrm{M})}^{3+}\right|\left|{\mathrm{Fe}}_{(0.02\mathrm{M})}^{2+}\right|\mathrm{Fe}$

Given that ${\mathrm{E}}_{{\mathrm{Al}}^{3+}/\mathrm{Al}}^{\mathrm{\Theta }}=-1.66 \mathrm{V}$ and ${\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\mathrm{\Theta }}=0.80 \mathrm{V}$

Calculate the maximum work that can be obtained from the following cell

$\mathrm{Zn}\left|{\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)\right|\left|{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\right|\mathrm{Ag}$

Given that ${\mathrm{E}}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}^{°}=-0.76 \mathrm{V}$ and ${\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}^{°}=0.80 \mathrm{V}$

The standard electrode potential for Daniel cell is $1.1 \mathrm{V}.$ Calculate the standard Gibbs energy for the reaction:

$\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)⟶{\mathrm{Zn}}^{2+}+\mathrm{Cu}\left(\mathrm{s}\right)$

Calculate ${\mathrm{\Delta G}}^{\mathrm{\Theta }}$ and the equilibrium constant for the cell reaction:

${\mathrm{Cl}}_2+2{\mathrm{I}}^{-}⟶2{\mathrm{Cl}}^{-}+{\mathrm{I}}_2$

Given that ${\mathrm{E}}^{\mathrm{\Theta }}\left({\mathrm{Cl}}_2, {\mathrm{Cl}}^{-}\right)=1.36 \mathrm{V}; {\mathrm{E}}^{\mathrm{\Theta }}\left({\mathrm{I}}_2, {\mathrm{I}}^{-}\right)=0.536 \mathrm{V}$

$$Determine the equilibrium constant of the reaction at $298 \mathrm{K},$ 

given $$\begin{gathered} {{\mathrm{E}}^{°}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}=0.771 \mathrm{V} \\ {{\mathrm{E}}^{°}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn} =}+0.150 \mathrm{V} \end{gathered}$$

for chemical reaction $2{\mathrm{Fe}}^{3+}+{\mathrm{Sn}}^{2+}\rightleftharpoons 2{\mathrm{Fe}}^{2+}+{\mathrm{Sn}}^{4+}$.

From the obtained value of the equilibrium constant, predict whether ${\mathrm{Sn}}^{2+}$ ions can reduce ${\mathrm{Fe}}^{3+}$ to ${\mathrm{Fe}}^{2+}$ quantitatively or not     

Calculate the resistivity, conductivity and molar conductivity of a $0.05 \mathrm{M} \mathrm{NaOH}$ solution if the resistance of its column having diameter $1 \mathrm{cm}$ and length $50 \mathrm{cm}$ is $5.55\times {10}^3 \mathrm{ohm}.$

Hint. $\mathrm{A}={\mathrm{\pi r}}^2=3.14\times {\left(0.5\right)}^2{ \mathrm{cm}}^2=0.785{ \mathrm{cm}}^2$ or $0.785\times {10}^{-4}{\mathrm{m}}^2; \mathrm{l}=50 \mathrm{cm}$ or $0.5\mathrm{m}, \mathrm{\rho }=\frac{\mathrm{R}·\mathrm{A}}{\mathrm{l}}$