JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Given \(F{e^{ + 3}}\left( {aq} \right) + {e^ - } \to F{e^{2 + }}\left( {aq} \right);{E^o} = + 0.77\,V\) \(A{l^{ + 3}}\left( {aq} \right) + 3{e^ - } \to Al\left( s \right);{E^o} = - 1.66\,V\) \(B{r_2}\left( {aq} \right) + 2{e^ - } \to 2B{r^ - };{E^o} = + 1.09\,V\) Considering the electrode potentials, which of the following represents the correct order of reducing power?
- A \(Fe^{2+} < Al < Br^-\)
- B \(Br^- < Fe^{2+} < Al\)
- C \(Al < Br^- < Fe^{2+}\)
- D \(Al < Fe^{2+} < Br^-\)
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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