(C) The relationship between standard Gibbs free energy change and equilibrium constant is $\Delta G^o = -RT \ln K = -2.303 RT \log K$.Also,$\Delta G^

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(C) The relationship between standard Gibbs free energy change and equilibrium constant is $\Delta G^o = -RT \ln K = -2.303 RT \log K$.Also,$\Delta G^o = -nF E^o$.Equating the two: $-nF E^o = -2.303 RT \log K$.Rearranging gives $\log K = \frac{nF E^o}{2.303 RT}$,which is statement $I$.Since $\frac{1}{2.303} \approx 0.4342$,we can write $\log K = 0.4342 \frac{nF E^o}{RT}$,which is statement $IV$.From $\ln K = \frac{nF E^o}{RT}$,we get $K = e^{\frac{nF E^o}{RT}}$,which is statement $II$.Therefore,statements $I$,$II$,and $IV$ are correct.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

Which of the following relations represents the correct relation between standard electrode potential and equilibrium constant?
$I$. $\log K = \frac{nF E^o}{2.303 RT}$
$II$. $K = e^{\frac{nF E^o}{RT}}$
$III$. $\log K = -\frac{nF E^o}{2.303 RT}$
$IV$. $\log K = 0.4342 \frac{nF E^o}{RT}$
Choose the correct statement$(s)$.

AIIMS 2017 All AIIMS

A
$I$,$II$ and $III$ are correct
B
$II$ and $III$ are correct
C
$I$,$II$ and $IV$ are correct
D
$I$ and $IV$ are correct

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Nernst equation and ECS

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$E^0 = \frac{RT}{nF} \ln K_{eq}$. This is called

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The cell potential for the following cell
$Pt \mid H_{2(g)} \mid H^{+}_{(aq)} \parallel Cu^{2+}(0.01 \, M) \mid Cu_{(s)}$
is $0.576 \, V$ at $298 \, K$. The $pH$ of the solution is $......$ (Nearest integer)

MediumJEE MAIN 2022

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If the $E^{\circ}_{cell}$ of an equilibrium reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ at $298 \ K$ is $0.59 \ V$,the equilibrium constant $K_c$ is

EasyAP EAMCET 2019

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Consider the following $4$ electrodes:
$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$ $B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$ $D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

$E^{\circ}_{Ag^{+} / Ag} = +0.80 \ V$
Arrange the reduction potential of these electrodes in decreasing order.

EasyKCET 2023

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For the cell $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$ $(E^{\circ}_{cell} = 1.10 \ V)$ at $298 \ K$,when the cell is completely discharged,what is the ratio of concentrations $\frac{[Zn^{2+}]}{[Cu^{2+}]}$?

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$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$	$B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$	$D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

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$E^0 = \frac{RT}{nF} \ln K_{eq}$. This is called

The cell potential for the following cell$Pt \mid H_{2(g)} \mid H^{+}_{(aq)} \parallel Cu^{2+}(0.01 \, M) \mid Cu_{(s)}$is $0.576 \, V$ at $298 \, K$. The $pH$ of the solution is $......$ (Nearest integer)

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For the cell $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$ $(E^{\circ}_{cell} = 1.10 \ V)$ at $298 \ K$,when the cell is completely discharged,what is the ratio of concentrations $\frac{[Zn^{2+}]}{[Cu^{2+}]}$?

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The cell potential for the following cell
$Pt \mid H_{2(g)} \mid H^{+}_{(aq)} \parallel Cu^{2+}(0.01 \, M) \mid Cu_{(s)}$
is $0.576 \, V$ at $298 \, K$. The $pH$ of the solution is $......$ (Nearest integer)

Consider the following $4$ electrodes:
$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$ $B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$ $D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

$E^{\circ}_{Ag^{+} / Ag} = +0.80 \ V$
Arrange the reduction potential of these electrodes in decreasing order.