
The emf of the cell, $Zn|Z{n^{2 + }}$(0.01 M) || $F{e^{2 + }}$(0.001 M)|Fe at 298 K is 0.2905 V then the value of equilibrium constant for the cell reaction is :
(A) ${e^{\dfrac{{0.32}}{{0.0295}}}}$
(B) ${10^{\dfrac{{0.32}}{{0.0295}}}}$
(C) ${10^{\dfrac{{0.26}}{{0.0295}}}}$
(D) ${10^{\dfrac{{0.32}}{{0.0591}}}}$
Answer
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Hint: The equilibrium constant is defined as the value of reaction quotient at the chemical equilibrium. It can be calculated as-
${E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{2}\log {K_{eq}}$
Where ${E_{cell}}$ is the potential difference between two half cells
${K_{eq}}$ is the equilibrium constant
Complete step by step solution:
In the question, we have to find the value of the equilibrium constant for the cell reaction.
First, let us write the cell in the form of a cell reaction. It can be written as-
$Zn + F{e^{2 + }} \to Fe + Z{n^{2 + }}$
We have been given that cell potential (${E_{cell}}$) = 0.2905
Temperature = 298 K
Concentration of Fe(+2) = 0.001 M
Concentration of Zn(+2) = 0.01 M
For finding the ${E_{cell}}$, we have formula as-
${E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{2}\log \dfrac{{product}}{{reac\tan t}}$
Thus, ${E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{2}\log \dfrac{{Z{n^{2 + }}}}{{F{e^{2 + }}}}$
By putting the values of all these, we can have value of $E_{cell}^0$ as-
$0.2905 = E_{cell}^0 - \dfrac{{0.059}}{2}\log \dfrac{{(0.01)}}{{(0.001)}}$
Solving the equation, we get
$E_{cell}^0$ = 0.32
Now, the question is to find equilibrium constant. We know that at equilibrium point value of ${E_{cell}}$ = 0.
Thus, again writing the Nernst equation as-
${E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{2}\log {K_{eq}}$
Putting the values, we get
$0 = 0.32 - \dfrac{{0.059}}{2}\log {K_{eq}}$
Now, solving it for $\log {K_{eq}}$; we have
$\log {\text{ }}{{\text{K}}_{eq}} = \dfrac{{0.32 \times 2}}{{0.059}}$
Thus, ${K_{eq}} = {10^{\dfrac{{0.32}}{{0.0295}}}}$
Thus, the option (B) is the correct answer.
Note: The equilibrium is the point when the rate of forwarding reaction is equal to the rate of backward reaction. At the point of equilibrium, the concentration of reactant and products is equal. The equilibrium constant tells about the relationship between products and reactants of a reaction at equilibrium.
${E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{2}\log {K_{eq}}$
Where ${E_{cell}}$ is the potential difference between two half cells
${K_{eq}}$ is the equilibrium constant
Complete step by step solution:
In the question, we have to find the value of the equilibrium constant for the cell reaction.
First, let us write the cell in the form of a cell reaction. It can be written as-
$Zn + F{e^{2 + }} \to Fe + Z{n^{2 + }}$
We have been given that cell potential (${E_{cell}}$) = 0.2905
Temperature = 298 K
Concentration of Fe(+2) = 0.001 M
Concentration of Zn(+2) = 0.01 M
For finding the ${E_{cell}}$, we have formula as-
${E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{2}\log \dfrac{{product}}{{reac\tan t}}$
Thus, ${E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{2}\log \dfrac{{Z{n^{2 + }}}}{{F{e^{2 + }}}}$
By putting the values of all these, we can have value of $E_{cell}^0$ as-
$0.2905 = E_{cell}^0 - \dfrac{{0.059}}{2}\log \dfrac{{(0.01)}}{{(0.001)}}$
Solving the equation, we get
$E_{cell}^0$ = 0.32
Now, the question is to find equilibrium constant. We know that at equilibrium point value of ${E_{cell}}$ = 0.
Thus, again writing the Nernst equation as-
${E_{cell}} = E_{cell}^0 - \dfrac{{0.059}}{2}\log {K_{eq}}$
Putting the values, we get
$0 = 0.32 - \dfrac{{0.059}}{2}\log {K_{eq}}$
Now, solving it for $\log {K_{eq}}$; we have
$\log {\text{ }}{{\text{K}}_{eq}} = \dfrac{{0.32 \times 2}}{{0.059}}$
Thus, ${K_{eq}} = {10^{\dfrac{{0.32}}{{0.0295}}}}$
Thus, the option (B) is the correct answer.
Note: The equilibrium is the point when the rate of forwarding reaction is equal to the rate of backward reaction. At the point of equilibrium, the concentration of reactant and products is equal. The equilibrium constant tells about the relationship between products and reactants of a reaction at equilibrium.
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