Electrode potentials $\left(\mathrm{E}^{\circ}ight)$ are given below : \(\begin{aligned} & \mathrm{Cu}^{+} /…

Electrode potentials $\left(\mathrm{E}^{\circ}ight)$ are given below :
\(\begin{aligned} & \mathrm{Cu}^{+} / \mathrm{Cu}=+0.52 \mathrm{V} \\ & \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=+0.77 \mathrm{V} \\ & \frac{1}{2} \mathrm{I}_2(\mathrm{s}) / \mathrm{I}^{-}=+0.54 \mathrm{V} \\ & \mathrm{Ag}^{+} / \mathrm{Ag}=+0.88 \mathrm{V}\end{aligned}\)
Based on the above potentials, strongest oxidizing agent will be:
  1. $\mathrm{Cu}^{+}$
  2. $\mathrm{Fe}^{3+}$
  3. $\mathrm{Ag}^{+}$
  4. $I_{2}$

Solution

Higher the value of reduction potential stronger will be the oxidising hence based on the given values $\mathrm{Ag}^{+}$ will be strongest oxidizing agent. /

Asked in: JEE-TOPICTESTS-CHEMISTRY

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