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JEE Mains · Chemistry · STD 12 - 2. Electrochemistry

નીચે વિધુતધ્રુવ પોટેન્શિયલ આપેલા છે. \(Cu^+ /Cu = + 0.52\, V\), \(Fe^{3+} /Fe^{2+} = +0.7 7\, V\), \(\frac{1}{2}{I_2}\left( s \right)/{I^ - }\, =  + 0.54\,V,\) \(Ag^+ /Ag = + 0.88\,V\). ઉપરના પોટેન્શિયલને આધારે, સૌથી પ્રબળ ઓક્સિડેશનકર્તા જણાવો.

  1. A \(Cu^+\)
  2. B \(Fe^{3+}\)
  3. C \(Ag^+\)
  4. D \(I_2\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(Ag^+\)

Step-by-step Solution

Detailed explanation

Higher the value of reduction potential stronger will be the oxidising hence based on the given values \(Ag^+\) will be strongest oxidizing agent
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