$\mathrm{EMF}$ of cell depends on :-

What will be the oxidation potential for the following hydrogen half cell at $1$bar pressure and $25º\mathrm{C}$ temperature?

$\mathrm{Pt}\left|\underset{ 1\mathrm{bar} }{{\mathrm{H}}_{2\left(g\right)}}\right|{\mathrm{HCl}}_{\left(\mathrm{aq}\right)}\mathrm{pH}=3$

Given the standard half-cell potentials $\left(\mathrm{E}°\right)$ of the following as

$\mathrm{Zn}⟶{\mathrm{Zn}}^{2+}+2{\mathrm{e}}^{-}; \mathrm{E}°=+0.76 \mathrm{V}$

$\mathrm{Fe}⟶{\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-}; \mathrm{E}°=+0.41 \mathrm{V}$

Then the standard e.m.f. of the cell with the reaction ${\mathrm{Fe}}^{2+}+\mathrm{Zn}⟶{\mathrm{Zn}}^{2+}+\mathrm{Fe}$ is

Calculate the standard cell potential (in V) of the cell in which the following reaction takes place:

${\mathrm{F}\mathrm{e}}^{2+}\mathrm{ }\left(\mathrm{a}\mathrm{q}\right)+\mathrm{A}{\mathrm{g}}^{+}\left(\mathrm{a}\mathrm{q}\right)\to \mathrm{F}{\mathrm{e}}^{3+}\left(\mathrm{a}\mathrm{q}\right)+\mathrm{A}\mathrm{g}\left(\mathrm{s}\right)$

Given that
${\mathrm{E}}_{{\mathrm{Ag}}^{+}/\mathrm{A}\mathrm{g}}^{\mathrm{o}}=\mathrm{x}\mathrm{ }\mathrm{V}$
${\mathrm{E}}_{{\mathrm{F}\mathrm{e}}^{2+}/\mathrm{F}\mathrm{e}}^{\mathrm{o}}=\mathrm{y}\mathrm{ }\mathrm{V}$
${\mathrm{E}}_{{\mathrm{F}\mathrm{e}}^{3+}/\mathrm{F}\mathrm{e}}^{\mathrm{o}}=\mathrm{z}\mathrm{ }\mathrm{V}$

In which metal container, the aqueous solution of ${\mathrm{CuSO}}_4$ can be stored?

${\mathrm{E}}_{{\mathrm{Cu}}^{3+}/\mathrm{Cu}}^0=0.34 \mathrm{V}$

$E_{Fe/F{e}^{2+}}^0=0.44V,E_{Al/A^{3+}}^0=1.66V$

$E_{Ni/N^{2+}}^0=0.25V,E_{A{g}^{+}/Ag}^0=0.80V$

The correct order of reduction potentials of the following pairs is

A. ${\mathrm{Cl}}_2/{\mathrm{Cl}}^{-}$

B. ${\mathrm{I}}_2/{\mathrm{I}}^{-}$

C. ${\mathrm{Ag}}^{+}/\mathrm{Ag}$

D. ${\mathrm{Na}}^{+}/\mathrm{Na}$

E. ${\mathrm{Li}}^{+}/\mathrm{Li}$

Choose the correct answer from the options given below.

Consider the following electrodes

$\mathrm{P}={\mathrm{Zn}}^{2+}(0.0001\mathrm{M})/\mathrm{Zn} \mathrm{Q}={\mathrm{Zn}}^{2+}(0.1\mathrm{M})/\mathrm{Zn}$

$\mathrm{R}={\mathrm{Zn}}^{2+}(0.01\mathrm{M})/\mathrm{Zn} \mathrm{S}={\mathrm{Zn}}^{2+}(0.001\mathrm{M})/\mathrm{Zn}$

${\mathrm{E}}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}^{\mathrm{o}}=-0.76 \mathrm{V}$ electrode potentials of the above electrodes in volts are in the order