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  • Help me understand! - Redox Reaction and Electrochemistry - JEE Main

Given

E_{Cl_{2}/Cl^{-1}}^{0}=1.36V,E_{Cr^{3 +}/Cr}^{0}=-0.74V

E_{Cr_{2}O_{7}^{2-}/Cr^{3+}}^{0}=1.33V,E_{MnO_{4}^{-}/Mn^{2+}}^{0}=1.1V

Among the following, the strongest reducing agent is :

  • Option 1)

    Cr3+

  • Option 2)

    Cl−

  • Option 3)

    Cr

  • Option 4)

     Mn2+

     

 

Answers (1)

As we learnt in 

Reducing Agent -

A substance that act as a donor of electron is called reducing agent. Reducing agent itself gets oxidised.

- wherein

e.g.2Na(s)+ O_{2}(g)\rightarrow (Na_{2}^{+}) O^{2-}(s)
Na is reducing agent

 

 E^{\circ}cl_{2}/cl^{-}=1.36V

E^{\circ}cl_{2}/cl{_{2}}=1.36V

Similarly, E^{\circ}G/G^{3+}=0.74V

E^{\circ}Cr^{3+}/Cr_{2}O_{7}^{2}=-1.33V and E^{\circ}m_{n}^{2+}/M_{n}O_{4}^{-}=-1.1V

Clearly E^{\circ}G/G^{3+}>E^{\circ}Cr^{3+}/Cr_{2}O_{7}^{-2}

                                >E_{Cl^{-}/Cl_{2}}^{o}                                  

                            >E_{Mn^{2+}/MnO_{4}^{-}}^{o}

So G is the strongest reducing agent


Option 1)

Cr3+

This option is incorrect

Option 2)

Cl−

This option is incorrect

Option 3)

Cr

This is option is correct

Option 4)

 Mn2+

 

This is option is incorrect

Posted by

Vakul

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