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  $E_{\left(F{e}^{3+}/F{e}^{2+}\right)}^0=0.77V)$:

Copper sulphate solution (250 mL) was electrolysed using a platinum anode and a copper cathode. A constant current of 2 mA was passed for 16 minutes. It was found that after electrolysis the absorbance of the solution was reduced to 50% of its original value. The concentration of copper sulphate in the solution at the beginning would be:

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

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 standard reduction potential of ${\mathrm{TiO}}^{2+}$ and ${\mathrm{Ti}}^{3+}$ are given by

${\mathrm{TiO}}^{2+}+2{\mathrm{H}}^{+}+{\mathrm{e}}^{-}\stackrel{}{\to }{\mathrm{Ti}}^{3+} {\mathrm{E}}^{\mathrm{o}}=0.10\mathrm{V}$

${\mathrm{Ti}}^{3+}+3{\mathrm{e}}^{-}\stackrel{}{\to }\mathrm{Ti} {\mathrm{E}}^{\mathrm{o}}=-1.21 \mathrm{V}$

Find the magnitude of standard reduction potential of ${\mathrm{TiO}}^{2+}$ to $\mathrm{Ti}$(Round off to the nearest integer)

The correct option(s) about entropy (S) is(are)

[$\mathrm{R}=$ gas constant, $\mathrm{F}=$ Faraday constant, $\mathrm{T}=$ Temperature]

The reduction potential $\left({\mathrm{E}}^0\right.$, in $\left.\mathrm{V}\right)$ of ${\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)/\mathrm{Mn}\left(\mathrm{s}\right)$ is

[Given: ${\mathrm{E}}_{\left({\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)/{\mathrm{MnO}}_2\left(\mathrm{s}\right)\right)}^{\mathrm{o}}=1.68 \mathrm{V}; {\mathrm{E}}_{\left({\mathrm{MnO}}_2\left(\mathrm{s}\right)/{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)\right)}^{\mathrm{o}}=1.21 \mathrm{V}; {\mathrm{E}}_{\left({\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)/\mathrm{Mn}\left(\mathrm{s}\right)\right)}^{\mathrm{o}}=-1.03 \mathrm{V}$]

Truncate/round-off the value to TWO decimal places.

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 $\mathrm{\Delta S}$ $\left(\mathrm{cal} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}\right)$.

The reaction is spontaneous if the cell potential is _____.(Positive/Negative)