2Ag^{+}_{(aq)} + Cu_{(s)} longrightarrow Cu^{ | vedclass

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(A) The Nernst equation for the reaction is $E = E^{\circ} - \frac{0.059}{2} \log \frac{[Cu^{2+}]}{[Ag^{+}]^2}$.$1$. If $[Ag^{+}]$ is doubled,the term

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Doubling the $[Ag^{+}]$

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(A) The Nernst equation for the reaction is $E = E^{\circ} - \frac{0.059}{2} \log \frac{[Cu^{2+}]}{[Ag^{+}]^2}$.$1$. If $[Ag^{+}]$ is doubled,the term $\frac{[Cu^{2+}]}{[Ag^{+}]^2}$ becomes $\frac{1}{4}$ of its initial value,which significantly increases $E$.$2$. If $[Cu^{2+}]$ is halved,the term $\frac{[Cu^{2+}]}{[Ag^{+}]^2}$ becomes $\frac{1}{2}$ of its initial value,which increases $E$ but by a smaller magnitude than doubling $[Ag^{+}]$.$3$. Changing the size of solid electrodes ($Cu$ or $Ag$) does not affect the reaction quotient $Q$ or the cell potential $E$.Therefore,doubling the $[Ag^{+}]$ causes the largest increase in potential.

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

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

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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?