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

Electrode potentials $\left(\mathrm{E}^{\circ}\right)$ 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. $\mathrm{I}_2$

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 Main 2013 (09 Apr Online)

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