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  • Can someone help me with this, - Electrochemistry - NEET-3

Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :

BrO^{-}_{4} \overset{1.82 V}{\rightarrow} BrO^{-}_{3} \overset{1.5 V}{\rightarrow} HBrO \overset{1.595 V}{\rightarrow} Br_{2} \overset{1.0652 V}{\rightarrow} Br^{-}

Then the species undergoing disproportionation is

  • Option 1)

    Br_{2}

  • Option 2)

    BrO^{-}_{4}

  • Option 3)

    BrO^{-}_{3}

  • Option 4)

    HBrO

 

Answers (1)

best_answer

As we learnt that

Gibbs energy of the reaction -

\Delta_{r}G=-nFE_{cell}

- wherein

\Delta _{r}G = gibbs energy of the reaction

E_{cell}= emf of the cell

nF = amount of charge passed

 

 Calculate E^{o}_{cell} corresponding to each compound undergoing disproportionation reaction. The reaction for which E^{o}_{cell} comes out +ve is spontaneous.

HBrO\rightarrow Br_{2} \; \; E^{o} = 1.595 V \; (Cathode)

HBrO\rightarrow BrO_{3}^{-} \; \; E^{o} = -1.5 V \; (Anode)

2HBrO\rightarrow Br_{2} = BrO_{3}^{-}

E^{o}_{cell} = E^{o}_{Cathode} - E^{o}_{Anode} = 1.595 -1.5 = 0.095 V

E^{o}_{cell} > 0 , \Delta G^{o} < 0 , spontaneous \; process

 


Option 1)

Br_{2}

This is incorrect.

Option 2)

BrO^{-}_{4}

This is incorrect.

Option 3)

BrO^{-}_{3}

This is incorrect.

Option 4)

HBrO

This is correct.

Posted by

prateek

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