Two identical beakers contain concentrated $\mathrm{HCl}$ solution at $298 \mathrm{K}$. In the first beaker solid $\mathrm{Cu}$ is added, while in the second beaker solid $\mathrm{Zn}$ is added. Which of the following will be observed?

$\left[{\mathrm{E}}^0\left({\mathrm{Zn}}^{2+}/\mathrm{Zn}\right)=-0.76\mathrm{V};{\mathrm{E}}^0\left({\mathrm{Cu}}^{2+}/\mathrm{Cu}\right)=+0.34\mathrm{V}\right]$

For the cell $\mathrm{Cu}\left(\mathrm{s}\right)\left|{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)(0.1\mathrm{M})\Vert {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)(0.01\mathrm{M})\right|\mathrm{Ag}\left(\mathrm{s}\right)$ the cell potential ${\mathrm{E}}_1=0.3095 \mathrm{V}$. For the cell $\mathrm{Cu}\left(\mathrm{s}\right)\left|{\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right)(0.01\mathrm{M})\Vert {\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)(0.001\mathrm{M})\right|\mathrm{Ag}\left(\mathrm{s}\right)$ the cell potential $=\mathrm{x}\times {10}^{-2} \mathrm{V}.$ Find value of x (Round off the Nearest Integer).

[ Use $:\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{J}$ ]

Upon dipping a copper rod, the aqueous solution of the salt that can turn blue is:

At $298 \mathrm{K}$, the standard electrode potentials of ${\mathrm{Cu}}^{2+}/\mathrm{Cu}, {\mathrm{Zn}}^{2+}/\mathrm{Zn}, {\mathrm{Fe}}^{2+}/\mathrm{Fe}$ and ${\mathrm{Ag}}^{+}/\mathrm{Ag}$ are $0.34 \mathrm{V}$, $-0.76 \mathrm{V}$, $-0.44 \mathrm{V}$ and $0.80 \mathrm{V}$, respectively.

On the basis of standard electrode potential, predict which of the following reaction cannot occur?

Given that,

${\mathrm{E}}_{{\mathrm{O}}_2/{\mathrm{H}}_2\mathrm{O}}^{\mathrm{o}}=+1.23 \mathrm{V};$
${\mathrm{E}}_{{\mathrm{S}}_2{\mathrm{O}}_8^{2-}/{\mathrm{SO}}_4^{2-}}^{\mathrm{o}}=2.05 \mathrm{V}$
${\mathrm{E}}_{{\mathrm{Br}}_2/{\mathrm{Br}}^{-}}^{\mathrm{o}}=+1.09 \mathrm{V};$
${\mathrm{E}}_{{\mathrm{Au}}^{3+}/\mathrm{Au}}^{\mathrm{o}}=1.4 \mathrm{V}$

The strongest oxidizing agent is