(C) The relationship between standard cell potential $E_{cell}^o$ and equilibrium constant $K_c$ is given by: $E_{cell}^o = \frac{2.303 \ RT}{nF} \log

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(C) The relationship between standard cell potential $E_{cell}^o$ and equilibrium constant $K_c$ is given by: $E_{cell}^o = \frac{2.303 \ RT}{nF} \log K_c$ Here,$n = 2$ (number of electrons transferred),$K_c = 10 \times 10^{15} = 10^{16}$,and $\frac{2.303 \ RT}{F} = 0.059 \ V$. Substituting the values: $E_{cell}^o = \frac{0.059}{2} \log(10^{16})$ $E_{cell}^o = \frac{0.059}{2} \times 16$ $E_{cell}^o = 0.059 \times 8 = 0.472 \ V$. Rounding to the nearest provided option,the correct answer is $0.4736 \ V$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

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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: $2.303 \ \frac{RT}{F} \text{ at } 298 \ K = 0.059 \ V$]

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A
$0.04736 \ mV$
B
$0.4736 \ mV$
C
$0.4736 \ V$
D
$0.04736 \ V$

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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: $2.303 \ \frac{RT}{F} \text{ at } 298 \ K = 0.059 \ V$]

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The $emf$ of a $Daniel$ cell at $298 \ K$ is ${E_1}$ for the cell reaction $Zn|ZnSO_4(0.01 \ M)||CuSO_4(1.0 \ M)|Cu$. When the concentration of $ZnSO_4$ is $1.0 \ M$ and that of $CuSO_4$ is $0.01 \ M$,the $emf$ changes to ${E_2}$. What is the relationship between ${E_1}$ and ${E_2}$?

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