ExamBro
ExamBro
CUET · CHEMISTRY · PYQ PAPER 2023

Given :
\(\begin{array}{l}
E_{K^{+} / K}^{\circ}=-2.93 V \\
E_{Ag^{+} / Ag}^{\circ}=0.80 V \\
E_{Hg^{2+} / Hg}^{\circ}=0.79 V \\
E_{Mg^{2+} / Mg^{-}}^{\circ}=-2.37 V \\
E_{Cr^{3+} / Gr}^{\circ}=-0.74 V
\end{array}\)
Choose the correct order :
(A) \(Ag < Hg < Cr < Mg < K\) (increasing reducing power)
(B) \(Ag < Hg < Cr < Mg < K\) (decreasing oxidising power)
(C) \(Ag > Hg > Cr > Mg > K\) (decreasing oxidising power)
(D) \(Hg > Cr > Ag > K > Mg\) (increasing reactivity )
(E) \(K > Mg > Cr > Hg > Ag\) (decreasing reactivity order)
Choose the correct answer from the options given below :

  1. A (A), (C) and (E) only
  2. B (B), (C) and (E) only
  3. C (C), (D) and (E) only
  4. D (A) and (B) only
Verified Solution

Answer & Solution

Correct Answer

(A) (A), (C) and (E) only

Step-by-step Solution

Detailed explanation

Order of standard reduction potentials (\(E^{\circ}\)): \(E_{K^{+} / K}^{\circ} = -2.93 V\) \(E_{Mg^{2+} / Mg}^{\circ} = -2.37 V\) \(E_{Cr^{3+} / Cr}^{\circ} = -0.74 V\) \(E_{Hg^{2+} / Hg}^{\circ} = 0.79 V\) \(E_{Ag^{+} / Ag}^{\circ} = 0.80 V\) Increasing reducing power…
Same subject
Explore more questions on app