(A) The Nernst equation for the cell reaction is given by: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log Q$,where $n = 2$ and $Q = \frac{[Zn^{2

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$[Zn^{2+}] = 1 \ M$ and $[Ag^{+}] = 0.1 \ M$

(A) The Nernst equation for the cell reaction is given by: $E_{cell} = E^{\circ}_{cell} - \frac{0.0592}{n} \log Q$,where $n = 2$ and $Q = \frac{[Zn^{2+}]}{[Ag^{+}]^2}$.Given that $E_{cell} = E^{\circ}_{cell} - 0.0592 \ V$,we have:$E^{\circ}_{cell} - \frac{0.0592}{2} \log Q = E^{\circ}_{cell} - 0.0592$$\frac{1}{2} \log Q = 1 \implies \log Q = 2 \implies Q = 10^2 = 100$.Checking the options:For option $A$: $Q = \frac{1}{(0.1)^2} = \frac{1}{0.01} = 100$.Thus,option $A$ is correct.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

For the cell reaction,$Zn_{(s)} + 2 Ag_{(aq)}^{+} \longrightarrow Zn_{(aq)}^{2+} + 2 Ag_{(s)}$,the cell potential is less than $E^{\circ}_{cell}$ by $0.0592 \ V$ at $298 \ K$ when:

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A
$[Zn^{2+}] = 1 \ M$ and $[Ag^{+}] = 0.1 \ M$
B
$[Zn^{2+}] = 1 \ M$ and $[Ag^{+}] = 0.01 \ M$
C
$[Zn^{2+}] = 0.1 \ M$ and $[Ag^{+}] = 1 \ M$
D
$[Zn^{2+}] = 0.01 \ M$ and $[Ag^{+}] = 1 \ M$

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Find the $emf$ of the cell in which the following reaction takes place at $298 \ K$ (in $V$):
$Ni_{(s)} + 2Ag^{+}(0.001 \ M) \rightarrow Ni^{2+}(0.001 \ M) + 2Ag_{(s)}$
(Given that $E_{cell}^{\circ} = 10.5 \ V$,$\frac{2.303 RT}{F} = 0.059$ at $298 \ K$)

MediumNEET 2022

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The standard electrode potential of $Cu^{2+}/Cu$ is $0.34 \text{ V}$ at $298 \text{ K}$. Calculate its electrode potential at the same temperature when the $Cu^{2+}$ ion concentration is $0.1 \text{ M}$. (in $\text{ V}$)

DifficultMHT CET 2026

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Fill in the blanks :
$1.$ The ratio of concentration of products to concentration of reactants is ........
$2.$ $\ln(\log(x)) =$ ............
$3.$ At equilibrium,between $E_{cell}$ and $E_{cell}^{o}$,......... will be zero.

Difficult

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Write the Nernst equation for the $E_{cell}$ reaction in the Daniell cell. How will the $E_{cell}$ be affected when the concentration of $Zn^{2+}$ ions is increased?

Medium

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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 $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$?

Medium

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For the cell reaction,$Zn_{(s)} + 2 Ag_{(aq)}^{+} \longrightarrow Zn_{(aq)}^{2+} + 2 Ag_{(s)}$,the cell potential is less than $E^{\circ}_{cell}$ by $0.0592 \ V$ at $298 \ K$ when:

Similar Questions

Find the $emf$ of the cell in which the following reaction takes place at $298 \ K$ (in $V$):$Ni_{(s)} + 2Ag^{+}(0.001 \ M) \rightarrow Ni^{2+}(0.001 \ M) + 2Ag_{(s)}$(Given that $E_{cell}^{\circ} = 10.5 \ V$,$\frac{2.303 RT}{F} = 0.059$ at $298 \ K$)

The standard electrode potential of $Cu^{2+}/Cu$ is $0.34 \text{ V}$ at $298 \text{ K}$. Calculate its electrode potential at the same temperature when the $Cu^{2+}$ ion concentration is $0.1 \text{ M}$. (in $\text{ V}$)

Fill in the blanks :$1.$ The ratio of concentration of products to concentration of reactants is ........ $2.$ $\ln(\log(x)) =$ ............$3.$ At equilibrium,between $E_{cell}$ and $E_{cell}^{o}$,......... will be zero.

Write the Nernst equation for the $E_{cell}$ reaction in the Daniell cell. How will the $E_{cell}$ be affected when the concentration of $Zn^{2+}$ ions is increased?

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 $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$?

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Find the $emf$ of the cell in which the following reaction takes place at $298 \ K$ (in $V$):
$Ni_{(s)} + 2Ag^{+}(0.001 \ M) \rightarrow Ni^{2+}(0.001 \ M) + 2Ag_{(s)}$
(Given that $E_{cell}^{\circ} = 10.5 \ V$,$\frac{2.303 RT}{F} = 0.059$ at $298 \ K$)

Fill in the blanks :
$1.$ The ratio of concentration of products to concentration of reactants is ........
$2.$ $\ln(\log(x)) =$ ............
$3.$ At equilibrium,between $E_{cell}$ and $E_{cell}^{o}$,......... will be zero.