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JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

दिया गया है - \(Fe ^{3+}\)(जलीय) \(+ e ^{-} \rightarrow Fe ^{2+}\)(जलीय);\(E ^{\circ}=+0.77 V\) \(Al ^{3+}\)(जलीय) \(+3 e ^{-} \rightarrow Al ( s ) ; E ^{\circ}=-1.66 V\) \(Br _{2}\)((जलीय) \(+2 e ^{-} \rightarrow 2 Br ^{-} ; E ^{\circ}=+1.09 V\) इलैक्ट्रोड विभवों के आधार पर निम्नों में से कौन क्रम अपचयन शक्तियों को सही प्रस्तुत करता है ?

  1. A \(Fe^{2+} < Al < Br^-\)­
  2. B \(Br^- < Fe^{2+} < Al\)
  3. C \(Al < Br^- < Fe^{2+}\)
  4. D \(Al < Fe^{2+} < Br^-\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(Al < Fe^{2+} < Br^-\)

Step-by-step Solution

Detailed explanation

A negative \(E^o\) means that the redox couple is a stronger reducing agent. Hence, we can say that Aluminium is strongest reducing agent. Reducing character decreases down the series. Hence the correct order is \(Al < Fe^{2+} < Br^-\)
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