Calculate the equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
Given $E^{\Theta}_{cell} = 0.46 \ V$
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
Given $E^{\Theta}_{cell} = 0.46 \ V$
(N/A) The relationship between the standard cell potential and the equilibrium constant is given by the Nernst equation at $298 \ K$:
$E^{\Theta}_{cell} = \frac{0.059 \ V}{n} \log K_C$
Here,$n = 2$ (number of electrons transferred in the reaction).
Substituting the values:
$0.46 \ V = \frac{0.059 \ V}{2} \log K_C$
$\log K_C = \frac{0.46 \times 2}{0.059} = \frac{0.92}{0.059} \approx 15.593$
$K_C = \text{antilog}(15.593) = 3.92 \times 10^{15}$
$E^{\Theta}_{cell} = \frac{0.059 \ V}{n} \log K_C$
Here,$n = 2$ (number of electrons transferred in the reaction).
Substituting the values:
$0.46 \ V = \frac{0.059 \ V}{2} \log K_C$
$\log K_C = \frac{0.46 \times 2}{0.059} = \frac{0.92}{0.059} \approx 15.593$
$K_C = \text{antilog}(15.593) = 3.92 \times 10^{15}$
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Similar Questions
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
View SolutionFor the cell,$Zn_{(s)} | Zn^{2+} (1 \ M) || Ag^{+} (1 \ M) | Ag_{(s)}$. If the concentration of $Zn^{2+}$ decreases to $0.1 \ M$ at $298 \ K$,then the $EMF$ of the cell:
$1 \ F$ electricity was passed through $Cu^{2+} (1.5 \ M, 1 \ L) / Cu$ and $0.1 \ F$ was passed through $Ag^{+} (0.2 \ M, 1 \ L) / Ag$ electrolytic cells. After this,the two cells were connected to make an electrochemical cell. The $emf$ of the cell thus formed at $298 \ K$ is:
Given: $E^0_{Cu^{2+} / Cu} = 0.34 \ V$,$E^0_{Ag^{+} / Ag} = 0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$ (in $V$)
Given: $E^0_{Cu^{2+} / Cu} = 0.34 \ V$,$E^0_{Ag^{+} / Ag} = 0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$ (in $V$)
Represent the cell in which the following reaction takes place:
$Mg_{(s)} + 2Ag^{+}(0.0001 \, M) \rightarrow Mg^{2+}(0.130 \, M) + 2Ag_{(s)}$
Calculate its $E_{cell}$ if $E^{\Theta}_{cell} = 3.17 \, V$.
$Mg_{(s)} + 2Ag^{+}(0.0001 \, M) \rightarrow Mg^{2+}(0.130 \, M) + 2Ag_{(s)}$
Calculate its $E_{cell}$ if $E^{\Theta}_{cell} = 3.17 \, V$.
Medium
View SolutionThe logarithm of the equilibrium constant for the reaction $Pd^{2+}{(aq)} + 4Cl^{-}{(aq)} \rightleftharpoons PdCl_4^{2-}{(aq)}$ is (Nearest integer).
Given: $\frac{2.303 RT}{F} = 0.06 \ V$
$Pd^{2+}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} \quad E^{\circ} = 0.83 \ V$
$PdCl_4^{2-}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} + 4Cl^{-}{(aq)} \quad E^{\circ} = 0.65 \ V$
Given: $\frac{2.303 RT}{F} = 0.06 \ V$
$Pd^{2+}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} \quad E^{\circ} = 0.83 \ V$
$PdCl_4^{2-}{(aq)} + 2e^{-} \rightleftharpoons Pd_{(s)} + 4Cl^{-}{(aq)} \quad E^{\circ} = 0.65 \ V$
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