(A) The relationship between standard $emf$ $(E^{\circ}_{cell})$ and equilibrium constant $(K_{eq})$ is given by the Nernst equation at $298 \ K$:$E^{

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(A) The relationship between standard $emf$ $(E^{\circ}_{cell})$ and equilibrium constant $(K_{eq})$ is given by the Nernst equation at $298 \ K$:$E^{\circ}_{cell} = \frac{0.0591}{n} \log K_{eq}$Given:$E^{\circ}_{cell} = 0.295 \ V$$n = 4$Substituting the values:$0.295 = \frac{0.0591}{4} \log K_{eq}$$\log K_{eq} = \frac{0.295 \times 4}{0.0591}$$\log K_{eq} \approx 20$$K_{eq} = 10^{20}$

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

For a galvanic cell reaction at $25^{\circ}C$ with $n = 4$,the standard $emf$ is $0.295 \ V$. What is the equilibrium constant for the reaction? $(F = 96500 \ C \ mol^{-1}; R = 8.314 \ J \ K^{-1} \ mol^{-1})$

A
$1.0 \times 10^{20}$
B
$2.0 \times 10^{11}$
C
$4.0 \times 10^{12}$
D
$1.0 \times 10^{2}$

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Electrochemistry

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Nernst equation and ECS

Calculate the cell potential at $298 \ K$ for the following cell:
$Cu_{(s)} | Cu^{2+}(0.1 \ M) || Cu^{2+}(1 \ M) | Cu_{(s)}$
Given: $E_{Cu^{2+}|Cu}^o = 0.34 \ V$ (in $V$)

Medium

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$Pt_{(s)} | H_{2(g)} (1 \ atm) | H^{+} (pH = 2) || H^{+} (pH = 3) | H_{2(g)} (1 \ atm) | Pt_{(s)}$ cell reaction will be

Easy

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The $emf$ of the cell $Tl_{(s)} | Tl^{+}_{(aq)} (0.0001 \ M) || Cu^{2+}_{(aq)} (0.01 \ M) | Cu_{(s)}$ is $0.83 \ V$. How can the $emf$ of this cell be increased?

Medium

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The change in electrode potential of $Cr^{3+}/Cr_2O_7^{2-}$ electrode at $25 \ ^oC$ due to change in $pH$ of its electrolytic solution from $1$ to $3$ is ........... $V$ (Assume: $[Cr_2O_7^{2-}]$ and $[Cr^{3+}]$ remain constant). Use: $\frac{2.303RT}{F} = 0.06$.

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What will be the reduction potential of $Cu$ in an aqueous solution with $pH = 12$? Given that the $K_{sp}$ of $Cu(OH)_2$ is $1 \times 10^{-19}$ and $E^o_{Cu^{2+}/Cu} = 0.34 \ V$.

Medium

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For a galvanic cell reaction at $25^{\circ}C$ with $n = 4$,the standard $emf$ is $0.295 \ V$. What is the equilibrium constant for the reaction? $(F = 96500 \ C \ mol^{-1}; R = 8.314 \ J \ K^{-1} \ mol^{-1})$

Similar Questions

Calculate the cell potential at $298 \ K$ for the following cell:$Cu_{(s)} | Cu^{2+}(0.1 \ M) || Cu^{2+}(1 \ M) | Cu_{(s)}$Given: $E_{Cu^{2+}|Cu}^o = 0.34 \ V$ (in $V$)

$Pt_{(s)} | H_{2(g)} (1 \ atm) | H^{+} (pH = 2) || H^{+} (pH = 3) | H_{2(g)} (1 \ atm) | Pt_{(s)}$ cell reaction will be

The $emf$ of the cell $Tl_{(s)} | Tl^{+}_{(aq)} (0.0001 \ M) || Cu^{2+}_{(aq)} (0.01 \ M) | Cu_{(s)}$ is $0.83 \ V$. How can the $emf$ of this cell be increased?

The change in electrode potential of $Cr^{3+}/Cr_2O_7^{2-}$ electrode at $25 \ ^oC$ due to change in $pH$ of its electrolytic solution from $1$ to $3$ is ........... $V$ (Assume: $[Cr_2O_7^{2-}]$ and $[Cr^{3+}]$ remain constant). Use: $\frac{2.303RT}{F} = 0.06$.

What will be the reduction potential of $Cu$ in an aqueous solution with $pH = 12$? Given that the $K_{sp}$ of $Cu(OH)_2$ is $1 \times 10^{-19}$ and $E^o_{Cu^{2+}/Cu} = 0.34 \ V$.

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Calculate the cell potential at $298 \ K$ for the following cell:
$Cu_{(s)} | Cu^{2+}(0.1 \ M) || Cu^{2+}(1 \ M) | Cu_{(s)}$
Given: $E_{Cu^{2+}|Cu}^o = 0.34 \ V$ (in $V$)