Consider the following 4 electrodes:A. Ag^{+} | vedclass

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(A) The Nernst equation for the reduction of $Ag^{+}$ to $Ag_{(s)}$ is given by:$E_{Ag^{+} / Ag} = E^{\circ}_{Ag^{+} / Ag} + 0.059 \log [Ag^{+}]$Since

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$B > C > D > A$

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(A) The Nernst equation for the reduction of $Ag^{+}$ to $Ag_{(s)}$ is given by:$E_{Ag^{+} / Ag} = E^{\circ}_{Ag^{+} / Ag} + 0.059 \log [Ag^{+}]$Since $E^{\circ}_{Ag^{+} / Ag}$ is constant $(+0.80 \ V)$,the reduction potential $E_{Ag^{+} / Ag}$ depends directly on the concentration of $Ag^{+}$ ions.As the concentration of $Ag^{+}$ increases,the value of $\log [Ag^{+}]$ increases,and thus the reduction potential $E_{Ag^{+} / Ag}$ increases.Comparing the concentrations:$B: 0.1 \ M$$C: 0.01 \ M$$D: 0.001 \ M$$A: 0.0001 \ M$Since $0.1 > 0.01 > 0.001 > 0.0001$,the order of reduction potential is $B > C > D > A$.

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

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.

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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.