JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
The correct order of reduction potentials of the following pairs is \(A.\) \(Cl _{2} / Cl^{-}\) \(B.\) \(I _{2} / I^{-}\) \(C.\) \(Ag ^{+} / Ag\) \(D.\) \(Na ^{+} / Na\) \(E.\) \(Li ^{+} / Li\) Choose the correct answer from the options given below.
- A \(A > C > B > D > E\)
- B \(A > B > C > D > E\)
- C \(A > C > B > E > D\)
- D \(A > B > C > E > D\)
Answer & Solution
Correct Answer
(A) \(A > C > B > D > E\)
Step-by-step Solution
Detailed explanation
\(E _{ Cl _{2} / Cl ^{-}}^{\circ}=+1.36\, V\) \(E _{ I _{2} / I ^{-}}^{\circ}=+0.54\, V\) \(E _{ Ag ^{+} / Ag }^{\circ}=+0.80\, V\) \(E _{ Na ^{+} / Na }^{\circ}=-2.71 \,V\) \(E _{ Li ^{+} / Li }^{\circ}=-3.05\, V\)
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