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The cell reaction in which the following reactions occur:
$F{e^{3 + }}_{\left( {{\text{aq}}} \right)} + 2{I^{ - 1}}_{\left( {{\text{aq}}} \right)} \to F{e^{2 + }}_{\left( {{\text{aq}}} \right)} + {I_2}_{\left( s \right)}$ has ${\text{E}}_{{\text{cell}}}^o = 0 \cdot 236{\text{ V}}$ at ${\text{298 K}}$.
The equilibrium constant of the cell reaction is:
A. $6 \cdot 69 \times {10^{ - 7}}$
B. $7 \cdot 69 \times {10^{ - 7}}$
C. $9 \cdot 69 \times {10^7}$
D. $6 \cdot 69 \times {10^7}$

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Last updated date: 13th Jun 2024
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Answer
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Hint: Cell reaction is the overall cell reaction of an electrochemical cell. Equilibrium constant is the value of the reaction quotient at a state where no further change occurs in the chemical reaction (equilibrium state).

Complete step by step answer:
-Electric potential in volt of a cell is related by coulomb, which is a unit of measurement of current in Joule. In an electrochemical cell, we require a relation of coulomb with the charge of one mole of electrons. We get this relation by multiplying the charge on the electron by the Avogadro’s Number. This result gives a defined quantity called Faraday $\left( {\text{F}} \right)$, and it is considered as the charge transferred in an electrochemical cell.
-$\Delta {\text{G}}$ is called free energy change, and it is the maximum amount of which can be performed in an electrochemical reaction.
-There is a relationship between the free energy change, and the cell potential of the electrochemical cell in standard state. It is given by $\Delta G^\circ = - {\text{nFE}}_{{\text{cell}}}^o{\text{ }}..........{\text{(1)}}$,
where, $\Delta G^\circ $ is the change in free energy in standard state,
${\text{n}}$ is the number of moles transferred,
${\text{F}}$ is the charge in one Faraday, which has a constant value of ${\text{96500 C}}$,
${\text{E}}_{{\text{cell}}}^o$ is the standard cell potential of the electrochemical cell.
And, the negative sign indicates the spontaneous reaction.
-Given equation is: $F{e^{3 + }}_{\left( {{\text{aq}}} \right)} + 2{I^{ - 1}}_{\left( {{\text{aq}}} \right)} \to F{e^{2 + }}_{\left( {{\text{aq}}} \right)} + {I_2}_{\left( s \right)}$
${\text{E}}_{{\text{cell}}}^o$ is given $0 \cdot 236{\text{ V}}$. Then using equation ${\text{(1)}}$:
$\Delta G^\circ = - {\text{nFE}}_{{\text{cell}}}^o$
$ \Rightarrow \Delta G^\circ = - 2 \times 0 \cdot 236 \times 96500$
$ \Rightarrow \Delta G^\circ = - 45548{\text{ J }}.........{\text{(2)}}$
-Now, the equilibrium constant ${K_{\text{C}}}$ is related with $\Delta G^\circ = - \ln {K_{\text{C}}}{\text{RT}}$
where, ${K_{\text{C}}}$ is the equilibrium constant,
${\text{R}}$ is the Universal Gas Constant having constant value $8 \cdot 314{\text{ J }}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$
And, ${\text{T}}$ is the temperature which is given as ${\text{298 K}}$.
Or, ${\text{ln}}{{\text{K}}_{\text{C}}} = \dfrac{{ - \Delta G^\circ }}{{{\text{RT}}}}{\text{ }}..........{\text{(3)}}$
Putting the value of ${K_{\text{C}}}{\text{, R, }}\Delta {\text{G}}^\circ {\text{ and T}}$ in equation (3), we get,
$ \Rightarrow {\text{ln}}{{\text{K}}_{\text{C}}} = \dfrac{{45548}}{{8 \cdot 314 \times 298}}$
Converting the ${\text{ln to log}}$ , we get,
$ \Rightarrow {\text{log }}{{\text{K}}_{\text{C}}} = \dfrac{{45548}}{{8 \cdot 314 \times 298 \times 2 \cdot 303}}$
$ \Rightarrow {\text{log }}{{\text{K}}_{\text{C}}} = {\text{7}} \cdot {\text{9854}}$
$ \Rightarrow {{\text{K}}_{\text{C}}} = {\text{antilog 7}} \cdot {\text{9854}}$
$ \Rightarrow {{\text{K}}_{\text{C}}} = 9 \cdot 68 \times {10^7}$
Hence, option (d) is the correct answer of equilibrium constant.

Note: ${\text{E}}_{{\text{cell}}}^o$, and ${{\text{E}}_{{\text{cell}}}}$ are not same. ${\text{E}}_{{\text{cell}}}^o$ is the cell potential at standard conditions, where the concentration of the reactants is ${\text{1 M}}$, where ${\text{M}}$ is the Molarity. ${{\text{E}}_{{\text{cell}}}}$ is in non-standard conditions. ${{\text{E}}_{{\text{cell}}}}$ can be calculated using ${\text{E}}_{{\text{cell}}}^o$ in the Nernst equation.