Consider the following equation for an electrochemical cell reaction. Which of the following changes in condition will increase the cell voltage? ${\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{PbCl}}_2\left(\mathrm{s}\right)\to \mathrm{Pb}\left(\mathrm{s}\right)+2\mathrm{HCl}\left(\mathrm{aq}\right)$
(I) Dissolve concentrated ${\mathrm{HClO}}_4$ in the cell solution
(II) Increase the pressure of ${\mathrm{H}}_2\left(\mathrm{g}\right)$
(III) Increase the amount of $\mathrm{Pb}\left(\mathrm{s}\right)$

For the cell $\mathrm{Zn} \left(\mathrm{s}\right)\left|{\mathrm{ZnSO}}_4 \left(\mathrm{aq}\right)\right|\left|{\mathrm{CuSO}}_4 \left(\mathrm{aq}\right)\right|\mathrm{Cu} \left(\mathrm{s}\right)$, when the concentration of ${\mathrm{Zn}}^{2+}$ is $10$ times the concentration of ${\mathrm{Cu}}^{2+}$, the expression for $\mathrm{\Delta G} \left(\mathrm{in}\mathrm{ }\mathrm{J}\mathrm{ }{\mathrm{mol}}^{-1}\right)$ is

$[\mathrm{F}=$Faraday's constant, $\mathrm{R}=$universal gas constant, $\mathrm{T}=$temperature, ${\mathrm{E}}^{\mathrm{o}}\mathrm{cell}=1.1\mathrm{V}]$

Depict the galvanic cell in which the reaction takes place. Further show, which of the electrode is negatively charged.

$\mathrm{Zn}\left(\mathrm{s}\right)+2{\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)+2\mathrm{Ag}\left(\mathrm{s}\right)$.

At $298 \mathrm{K}$, the $\mathrm{e}.\mathrm{m}.\mathrm{f}.$ of the galvanic cell mentioned below is $\left({\mathrm{E}}^0=1.1 \mathrm{V}\right)$

$\mathrm{Zn}\left(\mathrm{s}\right)\left|{\mathrm{Zn}}^{2+}\left({10}^{-3}\mathrm{M}\right)\Vert {\mathrm{Cu}}^{2+}\left(0.1\mathrm{M}\right)\right|\mathrm{Cu}\left(\mathrm{s}\right)$

Represent a cell consisting of ${\mathrm{Mg}}^{2+}$ | $\mathrm{Mg}$ half cell and ${\mathrm{Ag}}^{+}$ | $\mathrm{Ag}$ half cell and write the cell reaction.

$(\mathrm{E}{°}_{\mathrm{Ag}}=0.799 \mathrm{V}, \mathrm{E}{°}_{\mathrm{Mg}}=-2.37 \mathrm{V})$

The following reaction takes place at $298 \mathrm{K}$ in an electrochemical cell involving two metals $\mathrm{A}$ and $\mathrm{B}$,

${\mathrm{A}}^{2+}\left(\mathrm{aq}\right)+\mathrm{B}\left(\mathrm{s}\right)\to {\mathrm{B}}^{2+}\left(\mathrm{aq}\right)+\mathrm{A}\left(\mathrm{s}\right)$

with $\left[{\mathrm{A}}^{2+}\right]=4\times {10}^{-3} \mathrm{M}$ and $\left[{\mathrm{B}}^{2+}\right]=2\times {10}^{-3} \mathrm{M}$ in the respective half-cells, the cell EMF is $1.091 \mathrm{V}$. The equilibrium constant of the reaction is closest to;