Calculate the potential of a cell in which hydrogen electrode is immersed, in a solution with a $\mathrm{pH}$of $3.5$ and in a solution with a $\mathrm{pH}$ of $10.7.$ Round off your answer to three decimal places.

For the cell, $\mathrm{Pt}| {\mathrm{H}}_2 \left(1 \mathrm{atm}\right) | {\mathrm{H}}^{+}\left(\mathrm{pH} \mathrm{unknown}\right)\left|\right|{\mathrm{H}}^{+} \left(\mathrm{pH}=1\right)\left|{\mathrm{H}}_2\left(1 \mathrm{atm}\right)\right|\mathrm{Pt}$. The measured cell potential at $25°\mathrm{C}$ is $0.16 \mathrm{volt}$  Calculate the unknown $\mathrm{pH}$.

Calculate the potential of the cell,

${\mathrm{Cl}}_2 \left(\mathrm{p}=0.9 \mathrm{atm}\right)\left|\mathrm{NaCl}\text{ solution }\right|{\mathrm{Cl}}_2\left(\mathrm{p}=0.1 \mathrm{atm}\right)$

Answer correct up to three decimal places with proper sign on the potential.

The $\mathrm{emf}$ of a galvanic cell composed of two hydrogen electrodes is $272 \mathrm{mV}$. What is the $\mathrm{pH}$ of the solution in which the anode is immersed if, the cathode is in a solution with a $\mathrm{pH}$ of $3$?

Round off your answer to the first place of decimal.

Calculate the standard emf, standard free energy change and equilibrium constant of a cell in which the following reaction takes place at $25°\mathrm{C}$. $\frac{1}{2}\mathrm{Cu} \left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{Cl}}_2 \left(\mathrm{g}\right)\rightleftharpoons \frac{1}{2}{\mathrm{Cu}}^{2+}+{\mathrm{Cl}}^{-}$.

${\mathrm{E}}_{\mathrm{Cl}2/{\mathrm{Cl}}^{-}}^{°}=+1.36 \mathrm{V}$ and  ${\mathrm{E}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}^{°}=+0.34 \mathrm{V}$.

Show by calculation that, ${\mathrm{E}}^0=-1·662 \mathrm{V}$ for the reduction of ${\mathrm{Al}}^{3+}$ to $\mathrm{Al} \left(\mathrm{s}\right),$ regardless of whether the equation for the reaction is written as

$\frac{1}{3}{{\mathrm{Al}}^{3+}}_{\left(\mathrm{aq}\right)}+{\mathrm{e}}^{-}\to \frac{1}{3}\mathrm{Al} \left(\mathrm{s}\right); {\mathrm{\Delta G}}^{°}=160·4 \mathrm{kJ}/\mathrm{mole}$

${{\mathrm{Al}}^{3+}}_{\left(\mathrm{aq}\right)}+3{\mathrm{e}}^{-}\to \mathrm{Al} \left(\mathrm{s}\right); \mathrm{\Delta G}°=481·2 \mathrm{kJ}/\mathrm{mole}$.