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  • I have a doubt, kindly clarify. - Redox Reaction and Electrochemistry - JEE Main-2

Given that E_{O_{2}/H_{2}O}^{\circleddash}=+1.23V;

E_{S_{2}O_{8}^{2-}/SO_{4}^{2-}}^{\circleddash}=2.05V;

E_{Br_{2}/Br^{-}}^{\circleddash}=+1.09 \; V;

E_{Au^{3+}/Au}^{\circleddash}=+1.4\; V;

The strongest oxidizing agent is :

  • Option 1)

    O_{2}

  • Option 2)

    Br_{2}

  • Option 3)

    S_{2}O_{8}^{2-}

     

  • Option 4)

    Au^{3+}

Answers (1)

best_answer

For strongest oxidising agent, standard reduction potential should be highest and so, it is highest for S_{2}O_{8}^{2-}.

E_{S_{2}O_{8}^{2-}SO_{4}^{2-}/}^{\circleddash}=2.05V;

 


Option 1)

O_{2}

Option 2)

Br_{2}

Option 3)

S_{2}O_{8}^{2-}

 

Option 4)

Au^{3+}

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