Thermodynamics of Cell

An excess of liquid mercury is added to an acidified solution of  It is found that 5% of  $F{e}^{3+}$remains at equilibrium at $\mathrm{}\mathrm{}\mathrm{}25°C.$ Calculate  $E_{\left(H{g}_2^{2+}/Hg\right)}^0,$  assuming that the only reaction that occurs is  $\mathrm{}\mathrm{}\mathrm{}2Hg+\mathrm{}\mathrm{}\mathrm{}2F{e}^{3+}\stackrel{}{\to }H{g}_2^{2+}+\mathrm{}\mathrm{}\mathrm{}2F{e}^{2+}.$ (Given  $E_{\left(F{e}^{3+}/F{e}^{2+}\right)}^0=0.77V)$:

What is the equilibrium constant for the reaction,  $2{\mathrm{Fe}}^{3+}+3{\mathrm{I}}^{-}\rightleftharpoons 2{\mathrm{Fe}}^{2+}+{{\mathrm{I}}_3}^{-}$, if the standard reduction potentials in acidic conditions are $0.77 \mathrm{V}$ and $0.54 \mathrm{V}$ for  $\raisebox{1ex}{${\mathrm{Fe}}^{3+}\mathrm{\hspace{0.17em}}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{\hspace{0.17em}}{\mathrm{Fe}}^{2+}$}\right.$ and $\raisebox{1ex}{$\hspace{0.17em}{{\mathrm{I}}_3}^{-}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{I}}^{-}$}\right.$ couples, respectively?

For the reaction

Given:

Calculate  $at298K.$

$\Delta {\mathrm{G}}_{\mathrm{reac}}^{\mathrm{o}}$  represent the reaction as cell calculate ${\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}$ & find ${\log }_{10} \mathrm{Ksp}$ for $\mathrm{AgCl}$.

The rusting of iron takes place as follows
$\begin{array}{l}2{\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}+2{\mathrm{e}}^{-}+\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\stackrel{}{\to }{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}\end{array}$; $\mathrm{E}°=+1.23\hspace{0.17em}\hspace{0.17em}\mathrm{V}$
${\mathrm{Fe}}_{\left(\mathrm{aq}\right)}^{2+}+2{\mathrm{e}}^{ -}\to {\mathrm{Fe}}_{\left(\mathrm{s}\right)} ;\hspace{0.17em}\hspace{0.17em}\mathrm{E}°=-0.44 \mathrm{V}$

Calculate  $\mathrm{\Delta G}°$for the net process:

The emf of the cell

$Zn/Z{n}^{2+}\left(0.01\right)\left|\right|F{e}^{2+}\left(0.001M\right)/Fe$ at 298 K is 0.2905 then the value of equilibrium  constant for the cell reaction is

The emf of the cell:

$\mathrm{Zn}\left|{\mathrm{Zn}}^{2+}(0.01\mathrm{M})\right|\left|{\mathrm{Fe}}^{2+}(0.001\mathrm{M})\right|\mathrm{Fe}$ at $298 \mathrm{K}$ is $0.2905 \mathrm{V}.$ Then the value of equilibrium constant for the cell reaction is:

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potentials are given below:

${\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+8{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+5{\mathrm{e}}^{-}\to {\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\mathrm{ };\mathrm{ }{\mathrm{E}}^{\mathrm{o}}=1.51\mathrm{V}$

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\left(\mathrm{aq}\right)+14{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+6{\mathrm{e}}^{-}\to 2{\mathrm{Cr}}^{3+}\mathrm{ }\left(\mathrm{aq}\right)+7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\mathrm{ };\mathrm{ }{\mathrm{E}}^{\mathrm{o}}=1.38\mathrm{V}$

${\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{e}}^{-}\to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)\mathrm{ };\mathrm{ }{\mathrm{E}}^{\mathrm{o}}=0.77\mathrm{V}$

${\mathrm{Cl}}_2\left(\mathrm{g}\right)+2{\mathrm{e}}^{-}\to 2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)\mathrm{ };\mathrm{ }{\mathrm{E}}^{\mathrm{o}}=1.41\mathrm{V}$

Identify the only incorrect statement regarding the quantitative estimation of aqueous $\mathrm{Fe}{\left({\mathrm{NO}}_3\right)}_2$.

The voltage of the cell: $\mathrm{Pb}\left|{\mathrm{PbSO}}_4\right|{\mathrm{Na}}_2{\mathrm{SO}}_4·10{\mathrm{H}}_2\mathrm{O} \left(\mathrm{salt}\right) \left|{\mathrm{Hg}}_2{\mathrm{SO}}_4\right|\mathrm{Hg}$ is $+0.9647$ at $25°\mathrm{C}$. The temperature coefficient is $1.74\times {10}^{-4} \mathrm{V} {\mathrm{K}}^{-1}$. Calculate the value of $\mathrm{\Delta S}$ $\left(\mathrm{cal} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}\right)$.

For a cell $\mathrm{Ag}\left|{\mathrm{Ag}}^{+},{\mathrm{Cl}}^{-}\right|\mathrm{AgCl}\left(\mathrm{s}\right)\mid \mathrm{Ag}, {\mathrm{E}}_{\mathrm{Red}}°$ for the half cells are given as: ${\mathrm{E}}_{{\mathrm{Ag}}^{+}\mid \mathrm{Ag}}°=0.79 \mathrm{V}$, ${\mathrm{E}}_{{\mathrm{Cl}}^{-}\left|\mathrm{AgCl}\right(\mathrm{s}\left)\right|\mathrm{Ag}}°=0.22 \mathrm{V}$ at $25{}^{\circ }\mathrm{C}.$ What is the value of $\ln \mathrm{K}$, where $\mathrm{K}$ is the equilibrium constant for the reaction $\mathrm{AgCl}\left(s\right)\rightleftharpoons {\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-}?$ Given, $1 \mathrm{F}=96485\mathrm{C}$.

The standard e.m.f. of a cell, involving one electron change is found to be $0.591\mathrm{V}$ at $25°\mathrm{C}$. The equilibrium constant of the reaction is
$\left(\mathrm{F}=96500\mathrm{C} {\mathrm{mol}}^{-1},R=8.314{\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1}\right)$

For a cell involving one electron ${\mathrm{E}}_{\mathrm{c}\mathrm{e}\mathrm{l}\mathrm{l}}^{⊝}=0.59\mathrm{ }\mathrm{V}$ at $298\mathrm{ }\mathrm{K}$, the equilibrium constant for the cell reaction is:
$\left[\mathrm{G}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{ }\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\mathrm{ }\frac{2.303\mathrm{ }\mathrm{R}\mathrm{T}}{\mathrm{F}}=0.059\mathrm{ }\mathrm{V}\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{ }\mathrm{T}=298\mathrm{ }\mathrm{K}\right]\mathrm{ }$

An oxidation-reduction reaction in which $3$ electrons are transferred has a $\mathrm{\Delta G}°$ of $-17.37 \mathrm{kJ} {\mathrm{mol}}^{-1}$ at $25°\mathrm{C}$. The value of $\mathrm{E}{°}_{\mathrm{cell} }$ (in $\mathrm{V}$) is $\mathrm{y}\times {10}^{-2}$. The value of $\mathrm{y}$ is 

Give your answer as the nearest integer.

$\left(1 \mathrm{F}=96,500 \mathrm{C} {\mathrm{mol}}^{-1}\right)$ 

For the cell reaction $3{\mathrm{Sn}}^{4+}+2\mathrm{Cr}\to 3{\mathrm{Sn}}^{2+}+2{\mathrm{Cr}}^{3+},\mathrm{E}{°}_{\mathrm{cell}}$ is $0.89 \mathrm{V}$. Then $\mathrm{\Delta G}°$ for the reaction is

$\mathrm{\Delta G}°$ for the reaction ${\mathrm{Ag}}^{+}\left(\mathrm{aq}\right)+\frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)⟶{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+\mathrm{Ag}\left(\mathrm{s}\right),$ where standard potential for silver half cell reaction is $0.8 \mathrm{V}$, will be