Explain the working of Daniel cell with a neat diagram.

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

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

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;

Daniell cell converts the chemical energy liberated during the redox reaction to electrical energy.

$\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)\stackrel{ }{\to }{\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Cu}\left(\mathrm{s}\right); {{\mathrm{E}}^{\circ }}_{\mathrm{cell}}=1.1 \mathrm{V}$

Identify the anode and cathode in Daniell cell.

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, ${\mathrm{E}}^{\mathrm{o}}\mathrm{cell}=1.1\mathrm{V}]$

In the given figure the experiment of conductivity solution is shown. Label against $\left[1\right],\left[2\right],\left[3\right],\left[4\right]$ and $\left[5\right]$.

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

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