(D) For the reaction,$A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$Using the relation $\Delta G^{\circ} = -RT \ln K_c = -nFE^{

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(D) For the reaction,$A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$Using the relation $\Delta G^{\circ} = -RT \ln K_c = -nFE^{\circ}_{cell}$Here,$n = 2$ (number of electrons transferred).At $298 \ K$,the relation simplifies to $\log K_c = \frac{n E^{\circ}_{cell}}{0.0591}$.Substituting the values: $\log K_c = \frac{2 \times 0.59}{0.059} = 20$.Therefore,$K_c = 10^{20} = 1.0 \times 10^{20}$.Hence,option $(D)$ is the correct answer.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

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If the $E^{\circ}_{cell}$ of an equilibrium reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ at $298 \ K$ is $0.59 \ V$,the equilibrium constant $K_c$ is

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A
$1.0 \times 10^{10}$
B
$1.0 \times 10^2$
C
$1.0 \times 10^{-20}$
D
$1.0 \times 10^{20}$

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For the cell $Zn_{(s)} | Zn^{2+}_{(1\,M)} || Cu^{2+}_{(1\,M)} | Cu_{(s)}$,the $E^o_{cell} = 1.10 \, V$. When the cell is fully discharged at $298 \, K$,the ratio of the concentrations $[Zn^{2+}] / [Cu^{2+}]$ is:

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Which of the following will increase the voltage of the cell represented by the equation
$Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$

Medium

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Given the equilibrium constant $K_c$ of the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$ is $10 \times 10^{15}$,calculate the $E_{cell}^o$ of the reaction at $298 \ K$. [Given: $2.303 \ \frac{RT}{F} \text{ at } 298 \ K = 0.059 \ V$]

DifficultJEE MAIN 2019

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Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)

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For the cell $Fe | Fe^{+2} (x \, M) || Cu^{+2} (0.01 \, M) | Cu$,if $E_{cell} = 0.78 \, V$,$E^{\circ}_{Fe^{+2} | Fe} = -0.44 \, V$,and $E^{\circ}_{Cu^{+2} | Cu} = +0.34 \, V$,determine the value of $x$.

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If the $E^{\circ}_{cell}$ of an equilibrium reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ at $298 \ K$ is $0.59 \ V$,the equilibrium constant $K_c$ is

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For the cell $Zn_{(s)} | Zn^{2+}_{(1\,M)} || Cu^{2+}_{(1\,M)} | Cu_{(s)}$,the $E^o_{cell} = 1.10 \, V$. When the cell is fully discharged at $298 \, K$,the ratio of the concentrations $[Zn^{2+}] / [Cu^{2+}]$ is:

Which of the following will increase the voltage of the cell represented by the equation$Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$

Given the equilibrium constant $K_c$ of the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$ is $10 \times 10^{15}$,calculate the $E_{cell}^o$ of the reaction at $298 \ K$. [Given: $2.303 \ \frac{RT}{F} \text{ at } 298 \ K = 0.059 \ V$]

Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)

For the cell $Fe | Fe^{+2} (x \, M) || Cu^{+2} (0.01 \, M) | Cu$,if $E_{cell} = 0.78 \, V$,$E^{\circ}_{Fe^{+2} | Fe} = -0.44 \, V$,and $E^{\circ}_{Cu^{+2} | Cu} = +0.34 \, V$,determine the value of $x$.

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Which of the following will increase the voltage of the cell represented by the equation
$Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$

Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)