Explain the working of the Daniell cell when an external voltage is applied.

 

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 $\left({\mathrm{E}}^0=1.1 \mathrm{V}\right)$

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