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← JEE Main 2014 · 11 Apr · Shift 1
ChemistrySingle correct+4 / -1Question 37 of 90

JEE Main 2014 · 11 Apr · Shift 1, Chemistry question 37

Given

FeX3+(aq)+eX−→FeX2+(aq)\ce{Fe^3+(aq) + e- -> Fe^2+(aq)}; E∘=+0.77E^\circ=+0.77 V

AlX3+(aq)+3 eX−→Al(s)\ce{Al^3+(aq) + 3e- -> Al(s)}; E∘=−1.66E^\circ=-1.66 V

BrX2(aq)+2 eX−→2 BrX−\ce{Br2(aq) + 2e- -> 2Br-}; E∘=+1.09E^\circ=+1.09 V

Considering the electrode potentials, which of the following represents the correct order of reducing power ?

Choose one option
Answer and worked solution

Official answer: B

Reducing power rises as the oxidation potential rises (lower reduction potential of the couple):

  • BrX−\ce{Br-}: −1.09-1.09 V;
  • FeX2+\ce{Fe^2+}: −0.77-0.77 V;
  • Al: +1.66+1.66 V.

So BrX−<FeX2+<Al\ce{Br- < Fe^2+ < Al}.

Topics:Galvanic cells and EMFOxidation number
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