The ${{\mathrm{E}}^{º}}_{\mathrm{cell}}$for the $\mathrm{Pt},{\mathrm{H}}_{\mathrm{2}}\left|{\mathrm{H}}^{\mathrm{+}}\right|\left|{\mathrm{Ag}}^{\mathrm{+}}\right|\mathrm{Ag}$ is ______. $[{\mathrm{E}}_{\mathrm{Ag}/{\mathrm{Ag}}^{\mathrm{+}}}^{\mathrm{o}} = -0.799 \mathrm{V}]$:

In the cell $\mathrm{Zn}| {\mathrm{Zn}}^{2+} (1\mathrm{M}) || {\mathrm{H}}^{\mathrm{+}} (1 \mathrm{M}) {\mathrm{H}}_{2\left(\mathrm{g}\right)}| \mathrm{Pt}, {\mathrm{H}}_2$ electrode is replaced by $\text{Cu}$ electrode. The cell potential will:

In the electrochemical cell:

$\mathrm{Zn} | {\mathrm{SnSO}}_{\mathrm{4(0.01}\mathrm{M})}|| {\mathrm{CuSO}}_{\mathrm{4(1.0} \mathrm{M})} | \mathrm{Cu}$, the $\mathrm{e}.\mathrm{m}.\mathrm{f}.$ of this Daniel cell is ${\mathrm{E}}_1$. When the concentration of ${\mathrm{ZnSO}}_{\mathrm{4}}$ is changed to $1.0 \mathrm{M}$ and that of ${\mathrm{CuSO}}_{\mathrm{4}}$ changed to $0.01 \mathrm{M}$, the $\mathrm{e}.\mathrm{m}.\mathrm{f}.$changes to ${\mathrm{E}}_{\mathrm{2}}$. From the following, which one is the relationship between ${\mathrm{E}}_{\mathrm{1}} \text{and} {\mathrm{E}}_{\mathrm{2}}$?
(Given, $\frac{2.303 \mathrm{RT}}{\mathrm{F}} = 0.059$)

Standard electrode potentials are: 
$$\begin{gathered} {\mathrm{Fe}}^{\mathrm{2+}}|\mathrm{Fe}, \mathrm{E}º=-0.44 \mathrm{V}; \\ {\mathrm{Fe}}^{\mathrm{3+}}|{\mathrm{Fe}}^{\mathrm{2+}}, \mathrm{E}º=0.77 \mathrm{V}; \end{gathered}$$

If ${\mathrm{Fe}}^{\mathrm{2+}}, {\mathrm{Fe}}^{\mathrm{3+}} \text{and }\mathrm{Fe}$ blocks are kept together, it will lead to _______.

Nernst equation of ${\mathrm{Zn}}_{\left(\mathrm{aq}\right)}^{2+}+ 2{\mathrm{e}}^{\mathrm{-}}\mathrm{ }\stackrel{}{\to }{\mathrm{Zn}}_{\left(\mathrm{s}\right)} \mathrm{is} {\mathrm{E}}_{({\mathrm{Zn}}^{2+}/\mathrm{Zn})} = \mathrm{\_\_\_\_\_\_\_}.$

Consider the following well reaction:
$2\mathrm{Fe}{ }_{\left(\mathrm{s}\right)}+{\mathrm{O}}_{2 \left(\mathrm{g}\right)}+4{\mathrm{H}}_{ \left(\mathrm{aq}\right)}^{+}\to 2{\mathrm{Fe}}_{ \left(\mathrm{aq}\right)}^{2+} + 2{\mathrm{H}}_{\mathrm{2}}{\mathrm{O}}_{ \left(\mathrm{l}\right)} \mathrm{E}º = 1.67 \mathrm{V}$

At ${\mathrm{Fe}}^{2+}{ = 10}^{-3} \mathrm{M}, {\mathrm{P}}_{({\mathrm{O}}_{\mathrm{2}}\mathrm{)}} = 0.1$ $\mathrm{atm}$ and $\mathrm{pH} = 3$. The cell potential at $25 º\mathrm{C}$ is:

Calomel is _______.