Nernst Equation and Its Application

$\left(\mathrm{i}\right)$ The  ${\mathrm{E}}_{\mathrm{cell}}^{°}$ of the electrochemical cell representing the reaction is

$2\mathrm{Cr}\left(\mathrm{s}\right)+3{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)\to 2{\mathrm{Cr}}^{3+}\left(\mathrm{aq}\right)+3\mathrm{Fe}\left(\mathrm{s}\right)$

$\left(\mathrm{ii}\right)$ The ${\mathrm{E}}_{\mathrm{cell}}$ at  ${25}^{°}\mathrm{C}$ when $\left[{\mathrm{Cr}}^{+3}\right]=0.1 \mathrm{M}$ and $\left[{\mathrm{Fe}}^{2+}\right]=0.01 \mathrm{M}$ will be?

$$\begin{gathered} {\mathrm{E}}_{\raisebox{1ex}{${\mathrm{Cr}}^{+3}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Cr}$}\right.}^{°}=-0.74\mathrm{\hspace{0.17em}}\mathrm{V} \\ {\mathrm{E}}_{\raisebox{1ex}{${\mathrm{Fe}}^{+2}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Fe}$}\right.}^{°}=-0.44\mathrm{\hspace{0.17em}}\mathrm{V} \end{gathered}$$

The EMF of a cell corresponding to the reaction:

$\mathrm{Zn}\left(\mathrm{s}\right)+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)(0.1\hspace{0.17em}\hspace{0.17em}\mathrm{M})+{\mathrm{H}}_2\left(\mathrm{g}\right)\hspace{0.17em}\hspace{0.17em}(1\hspace{0.17em}\hspace{0.17em}\mathrm{atm}.)\hspace{0.17em}\hspace{0.17em}\mathrm{is}\hspace{0.17em}\hspace{0.17em}0.28\hspace{0.17em}\hspace{0.17em}\mathrm{volt}\hspace{0.17em}\hspace{0.17em}\mathrm{at}\hspace{0.17em}\hspace{0.17em}25°\mathrm{C}.$

The pH of the solution at the hydrogen electrode?
${\text{E}}_{\text{Z}{\text{n}}^{2+}/\text{Z}\text{n}}^{\text{o}}=-0.76\text{v}\text{o}\text{l}\text{t};{\text{E}}_{{\text{H}}^{+}/{\text{H}}_2}^{\text{o}}=0$

If $6.539\times {10}^{-2} \mathrm{g}$ of metallic zinc is added to $100 \mathrm{mL}$ saturated solution of $\mathrm{AgCl}$, then $\mathrm{Ag} \left(\mathrm{s}\right)$ precipitates in the above reaction.

$\left[{\mathrm{K}}_{\mathrm{sp}} \left(\mathrm{AgCl}\right)={10}^{-10}\right]$

(Atomic mass of $\mathrm{Zn}=65.39$)
What is the value of $\log \frac{\left[{\mathrm{Zn}}^{2+}\right]}{{\left[{\mathrm{Ag}}^{+}\right]}^2}$?

$$\begin{gathered} {{\mathrm{E}}^{°}}_{\mathrm{Ag}}=0\mathrm{.}80 \mathrm{V} \\ {{\mathrm{E}}^{°}}_{\mathrm{Zn}}=-0\mathrm{.}76 \mathrm{V} \end{gathered}$$

Find the number of moles of $\mathrm{Ag}$ formed.

What will be the emf of the given cell  ${\text{Pt | H}}_{\text{2}}{\text{(P}}_{\text{1}}{\text{) | H}}^{\text{+}}{\text{(aq) | H}}_{\text{2}}{\text{(P}}_{\text{2}}\text{) | Pt ?}$

The reduction potential of hydrogen electrode at $25°\mathrm{C}$ in a neutral solution is $\left({\mathrm{P}}_{{\mathrm{H}}_2}=1\text{ atm}\right)$

Using the Nernst equation, calculate emf of the following cell at $298 \mathrm{K}$:

$\mathrm{Cr}\left(\mathrm{s}\right) \left| {\mathrm{Cr}}^{3+}\left(0.1 \mathrm{M}\right) \right|\left| {\mathrm{Fe}}^{2+}\left(0.01\mathrm{M}\right) \right|\mathrm{Fe}\left(\mathrm{s}\right)$

Given that,

$$\begin{gathered} {\mathrm{E}}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\circ }=-0.75 \mathrm{V} \\ {\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\circ }=-0.45 \mathrm{V} \end{gathered}$$

$\frac{2.303\mathrm{RT}}{\mathrm{F}}=0.0591\mathrm{V}$

A hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $\mathrm{pH}=9$ and passing hydrogen gas around the platinum wire at $1.2$ atm pressure. The oxidation potential of such an electrode equals_______ $\mathrm{V}.$

At $25°\mathrm{C}$, the standard reduction potential for the cell reaction

$\mathrm{Zn}\left(\mathrm{s}\right)+2{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{H}}_2( \mathrm{g})$

is $0.28 \mathrm{V}$. What is the reduction potential of the cell reaction in $10 \mathrm{M} \left[{\mathrm{H}}^{+}\right]$ concentration, assuming all other species to be at unit concentration? [Universal gas constant, $\mathrm{R}=8.314 \mathrm{J}{ \mathrm{K}}^{-1}{ \mathrm{mol}}^{-1}$; Faraday constant, $\mathrm{F}=96500{\mathrm{Cmol}}^{-1}]$

In a cell that utilizes the reaction
$\text{Zn(s)}\text{+}{\text{2H}}^{\text{+}}\text{(aq)}\to {\text{Zn}}^{\text{2+}}\text{(aq)}\text{+}{\text{H}}_{\text{2}}\text{(g)}$
addition of H₂SO₄ to cathode compartment,

At $298 \mathrm{K},$ the $\mathrm{EMF}$ of the given cell is

${\mathrm{Zn}}_{\left(\mathrm{s}\right)}\left|{\mathrm{ZnSO}}_4(0.1\mathrm{M})\Vert {\mathrm{ZnSO}}_4\left(1\mathrm{M}\right)\right|{\mathrm{Zn}}_{\left(\mathrm{s}\right)}$

Calculate the $\mathrm{pH}$ of the following half cell:
$\mathrm{Pt},{\mathrm{H}}_2\mid {\mathrm{H}}_2{\mathrm{SO}}_4$
(Given the oxidation electrode potential of the half cell is $+0.3 \mathrm{V}$ )

Consider the cell $\mathrm{Pt}\left|{\mathrm{Cl}}_2\right(\mathrm{g}\left)\right|\mathrm{HCl}\left(\mathrm{aq}\right)\Vert \mathrm{HBr}\left(\mathrm{aq}\right)\left|{\mathrm{Br}}_2\left(\mathrm{l}\right)\right|\mathrm{Pt}$ If concentration of $\mathrm{HCl}$ is increased, the cell potential will

The value of constant in Nernst equation $\mathrm{E}={\mathrm{E}}^0-\frac{\text{ constant }}{\mathrm{n}}\mathrm{lnQ}$ at ${25}^0\mathrm{C}$ is

A cell consists of two hydrogen electrodes. The negative electrode is in contact with a solution having $\mathrm{pH}=6$. The positive electrode is in contact with a solution of $\mathrm{pH}=\mathrm{x}$. Calculate the value of $\mathrm{x}$ if the emf of the cell is found to be $0.118 \mathrm{V}$ at $298 \mathrm{K}$.