Consider the following four electrodes: $\mathrm{P}=\mathrm{Cu}^{2+}(0.0001 \mathrm{M}) / \mathrm{Cu}(s)$…

Consider the following four electrodes:
$\mathrm{P}=\mathrm{Cu}^{2+}(0.0001 \mathrm{M}) / \mathrm{Cu}(s)$
$\mathrm{Q}=\mathrm{Cu}^{2+}(0.1 \mathrm{M}) / \mathrm{Cu}(s)$
$\mathrm{R}=\mathrm{Cu}^{2+}(0.01 \mathrm{M}) / \mathrm{Cu}(s)$
$\mathrm{S}=\mathrm{Cu}^{2+}(0.001 \mathrm{M}) / \mathrm{Cu}(s)$
If the standard reduction potential of $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is $+0.34 \mathrm{~V}$, the reduction potentials in volts of the above electrodes follow the order.
  1. $\mathrm{P}>\mathrm{S}>\mathrm{R}>\mathrm{Q}$
  2. $\mathrm{S}>\mathrm{R}>\mathrm{Q}>\mathrm{P}$
  3. $\mathrm{R}>\mathrm{S}>\mathrm{Q}>\mathrm{P}$
  4. $\mathrm{Q}>\mathrm{R}>\mathrm{S}>\mathrm{P}$

Solution

$\mathrm{E}_{\mathrm{red}}=\mathrm{E}_{\mathrm{red}}^{\mathrm{o}}+\frac{0.591}{\mathrm{n}} \log \left[\mathrm{M}^{\mathrm{n}+}ight]$
Lower the concentration of $\mathrm{M}^{\mathrm{n}+}$, lower is the reduction potential. Hence order of reduction potential is :
$\mathrm{Q}>\mathrm{R}>\mathrm{S}>\mathrm{P}$ ^

Asked in: JEE-TOPICTESTS-CHEMISTRY

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