(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

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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:

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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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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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The standard cell potential of a cell reaction involving the transfer of $2$ electrons at $25^{\circ}C$ is $0.295 \ V$. What is its equilibrium constant at $25^{\circ}C$?

Derive the Nernst equation for calculating $E_{cell}$ of a Daniell cell and explain the effect on $E_{cell}$ when there is a change in the concentration of $Zn^{2+}$ and $Cu^{2+}$ ions.

$2Ag^{+}_{(aq)} + Cu_{(s)} \longrightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$The standard potential for this reaction is $0.46 \ V$. Which change will increase the potential the most?

At $298 \ K$ the equilibrium constant for the reaction $M_{(s)} + 2 Ag^{+}_{(aq)} \rightarrow M^{2+}_{(aq)} + 2 Ag_{(s)}$ is $10^{15}$. What is the $E_{cell}^{\ominus}$ (in $V$) for this reaction? $\left(\frac{2.303 RT}{F}\right) = 0.06 \ V$

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$2Ag^{+}_{(aq)} + Cu_{(s)} \longrightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
The standard potential for this reaction is $0.46 \ V$. Which change will increase the potential the most?