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Given

E_{Cr^{3+}/Cr}^{0}= -0.74 V;E_{MnO_{4}^{-}/Mn^{2+}}^{0}= 1.51 V

E_{Cr_{2}O_{7}^{2-}/Cr^{3+}}^{0}= 1.33 V;E_{Cl/Cl^{-}}^{0}= 1.36 V

Based on the data given above ,strongest oxidising agent will be:

Option: 1

MnO_{4}^{-}


Option: 2

Cl^{-}


Option: 3

Cr^{3+}


Option: 4

Mn^{2+}


Answers (1)

best_answer

As learned in

Standard reduction potential -

It measures the tendency of a chemical species to acquire electrons and thereby be reduced. It is measured in volts or mv.

-

 

   Given that               E_{Cr^{3+}/Cr}^{0}= -0.74 V;                  ...........1

                                E_{MnO_{4}^{-}/Mn^{2+}}^{0}= 1.51 V                ...........2

                                E_{Cr_{2}O_{7}^{2-}/Cr^{3+}}^{0}= 1.33 V;               ..........3

                                E_{Cl/Cl^{-}}^{0}= 1.36 V                          ...........4

All the above are oxidizing agents undergoing reduction.

Higher the value of E° (ERP or Standard Reduction Potential) of the reaction, greater will be the oxidizing power of the constituents.

Clearly, 1.51V>1.36V>1.33V>-0.74V

Therefore, the order of oxidizing power is :

{MnO_{4}}^{-}> Cl^{-}> {Cr_{2}O_{7}}^{2-}> Cr^{3+}

Posted by

Kuldeep Maurya

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