1. Tardigrade
  2. Question
  3. Chemistry
  4. Electrode potentials (E°) are given below: Cu+/Cu=+0.52 V, Fe3+/Fe2+ = +0.77 V, (1/2)I2(s)/I- = 40.54 V, Ag+/Ag =+ 0.88 V. Based on the above potentials, strongest oxidizing agent will be :

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Q. Electrode potentials (E$^{\circ}$) are given below :
$Cu^{+}/Cu=+0.52\, V,$
$Fe^{3+}/Fe^{2+} = +0.77 \,V,$
$\frac{1}{2}I_{2}\left(s\right)/I^{-} = 40.54 \,V,$
$Ag^{+}/Ag =+ 0.88\,V.$
Based on the above potentials, strongest oxidizing agent will be :

JEE MainJEE Main 2013Electrochemistry

Solution:

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