(B) The relationship between standard cell potential $E^o_{cell}$ and equilibrium constant $K$ is given by the Nernst equation at $25^o \ C$ (or $298

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(B) The relationship between standard cell potential $E^o_{cell}$ and equilibrium constant $K$ is given by the Nernst equation at $25^o \ C$ (or $298 \ K$): $E^o_{cell} = \frac{0.059}{n} \log K$ Here,$n = 2$ (number of electrons transferred in the redox reaction). Given $E^o_{cell} = 1.10 \ V$. Substituting the values: $1.10 = \frac{0.059}{2} \log K$ $\log K = \frac{1.10 \times 2}{0.059} = \frac{2.20}{0.059} \approx 37.288$ $K = 10^{37.288} \approx 10^{+37}$

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

For the cell $Zn | Zn^{2+}_{(aq)} || Cu^{2+}_{(aq)} | Cu$,the standard cell potential $E^o$ is $1.10 \ V$ at $25^o \ C$. What is the order of magnitude of the equilibrium constant $K$ for the reaction $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$?

A
$10^{+28}$
B
$10^{+37}$
C
$10^{+18}$
D
$10^{+17}$

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Class 12

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Electrochemistry

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Nernst equation and ECS

Calculate the $emf$ of the cell: $Cr | Cr^{3+}(0.1 \ M) || Fe^{2+}(0.01 \ M) | Fe$. Given: $E^0_{Cr^{3+}/Cr} = -0.75 \ V$; $E^0_{Fe^{2+}/Fe} = -0.45 \ V$. Cell reaction: $2 \ Cr_{(s)} + 3 \ Fe^{2+}_{(aq)} \rightarrow 2 \ Cr^{3+}_{(aq)} + 3 \ Fe_{(s)}$.

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For the galvanic cell,
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$

EasyJEE MAIN 2021

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$EMF$ of the following concentration cell will be ............. $V$
$Cl_2 \ (1 \ atm) | Cl^{-} \ (1 \ M) || Cl^{-} \ (0.1 \ M) | Cl_2 \ (2 \ atm)$

Medium

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The reduction potential of a hydrogen electrode at $25^{\circ} C$ in a neutral solution is $(P_{H_2} = 1 \ atm)$. (in $V$)

MediumAP EAMCET 2022

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If the standard electrode potential for a cell is $2 \ V$ at $300 \ K,$ the equilibrium constant $(K)$ for the reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}$ at $300 \ K$ is approximately $(R = 8 \ J \ K^{-1} \ mol^{-1}, F = 96000 \ C \ mol^{-1})$

DifficultJEE MAIN 2019

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For the cell $Zn | Zn^{2+}_{(aq)} || Cu^{2+}_{(aq)} | Cu$,the standard cell potential $E^o$ is $1.10 \ V$ at $25^o \ C$. What is the order of magnitude of the equilibrium constant $K$ for the reaction $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$?

Similar Questions

Calculate the $emf$ of the cell: $Cr | Cr^{3+}(0.1 \ M) || Fe^{2+}(0.01 \ M) | Fe$. Given: $E^0_{Cr^{3+}/Cr} = -0.75 \ V$; $E^0_{Fe^{2+}/Fe} = -0.45 \ V$. Cell reaction: $2 \ Cr_{(s)} + 3 \ Fe^{2+}_{(aq)} \rightarrow 2 \ Cr^{3+}_{(aq)} + 3 \ Fe_{(s)}$.

For the galvanic cell,$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$$[\frac{2.303 \ RT}{F} = 0.059 \ V]$

$EMF$ of the following concentration cell will be ............. $V$$Cl_2 \ (1 \ atm) | Cl^{-} \ (1 \ M) || Cl^{-} \ (0.1 \ M) | Cl_2 \ (2 \ atm)$

The reduction potential of a hydrogen electrode at $25^{\circ} C$ in a neutral solution is $(P_{H_2} = 1 \ atm)$. (in $V$)

If the standard electrode potential for a cell is $2 \ V$ at $300 \ K,$ the equilibrium constant $(K)$ for the reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}$ at $300 \ K$ is approximately $(R = 8 \ J \ K^{-1} \ mol^{-1}, F = 96000 \ C \ mol^{-1})$

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For the galvanic cell,
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$

$EMF$ of the following concentration cell will be ............. $V$
$Cl_2 \ (1 \ atm) | Cl^{-} \ (1 \ M) || Cl^{-} \ (0.1 \ M) | Cl_2 \ (2 \ atm)$