(B) The equation that relates the electrode potential $(E)$,the standard electrode potential $(E^o)$,and the concentration of ions in the solution is

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(B) The equation that relates the electrode potential $(E)$,the standard electrode potential $(E^o)$,and the concentration of ions in the solution is known as the Nernst equation.For a general reaction $M^{n+} + ne^- \rightarrow M(s)$,the Nernst equation is given by $E = E^o - (RT/nF) \ln([M]/[M^{n+}])$.Thus,the correct option is $(B)$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

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The name of the equation showing the relation between electrode potential $(E)$,standard electrode potential $(E^o)$,and concentration of ions in solution is

A
Kohlrausch's equation
B
Nernst's equation
C
Ohm's equation
D
Faraday's equation

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Nernst equation and ECS

For the cell reaction $Zn_{(s)} + Ni^{2+}_{(aq)} (a = 0.1) \rightarrow Zn^{2+}_{(aq)} (a = 1.0) + Ni_{(s)}$ at $298 \ K$,the $emf$ is observed to be $0.5105 \ V$. What is the standard $emf$ $(E^{\circ}_{cell})$ of the cell in $V$?

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The potential of the cell for the reaction $M_{(s)} + 2H^{+}(1 \ M) \to H_2(1 \ atm) + M^{2+}(0.1 \ M)$ is $1.5 \ V$. The standard reduction potential for $M^{2+}/M_{(s)}$ couple is ............. $V$.

Medium

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Calculate the $e.m.f.$ of the half-cell given below:
$Fe | FeSO_4$ $(a = 0.1 \ M)$
Given: $E^o_{OP} = 0.44 \ V$ (in $V$)

Medium

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For the given cell $Cu(s) | Cu^{2+}(C_1 \ M) || Cu^{2+}(C_2 \ M) | Cu(s)$,the change in Gibbs energy $(\Delta G)$ is negative,if:

MediumJEE MAIN 2020

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The $EMF$ of the cell $Cr|Cr^{3+}(0.1 \ M) \| Fe^{2+}(0.01 \ M)|Fe$ is: $(E^{\circ}_{Cr^{3+}|Cr} = -0.75 \ V, E^{\circ}_{Fe^{2+}|Fe} = -0.45 \ V)$ (in $V$)

Medium

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The name of the equation showing the relation between electrode potential $(E)$,standard electrode potential $(E^o)$,and concentration of ions in solution is

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For the cell reaction $Zn_{(s)} + Ni^{2+}_{(aq)} (a = 0.1) \rightarrow Zn^{2+}_{(aq)} (a = 1.0) + Ni_{(s)}$ at $298 \ K$,the $emf$ is observed to be $0.5105 \ V$. What is the standard $emf$ $(E^{\circ}_{cell})$ of the cell in $V$?

The potential of the cell for the reaction $M_{(s)} + 2H^{+}(1 \ M) \to H_2(1 \ atm) + M^{2+}(0.1 \ M)$ is $1.5 \ V$. The standard reduction potential for $M^{2+}/M_{(s)}$ couple is ............. $V$.

Calculate the $e.m.f.$ of the half-cell given below: $Fe | FeSO_4$ $(a = 0.1 \ M)$ Given: $E^o_{OP} = 0.44 \ V$ (in $V$)

For the given cell $Cu(s) | Cu^{2+}(C_1 \ M) || Cu^{2+}(C_2 \ M) | Cu(s)$,the change in Gibbs energy $(\Delta G)$ is negative,if:

The $EMF$ of the cell $Cr|Cr^{3+}(0.1 \ M) \| Fe^{2+}(0.01 \ M)|Fe$ is: $(E^{\circ}_{Cr^{3+}|Cr} = -0.75 \ V, E^{\circ}_{Fe^{2+}|Fe} = -0.45 \ V)$ (in $V$)

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Calculate the $e.m.f.$ of the half-cell given below:
$Fe | FeSO_4$ $(a = 0.1 \ M)$
Given: $E^o_{OP} = 0.44 \ V$ (in $V$)