Question

Match the items given in List-I and List-II.

 

• A. A-IV, B-I, C-III, D-II
• B. A-III, B-I, C-IV, D-II
• C. A-I, B-II, C-III, D-IV
• D. A-II, B-IV, C-I, D-III

Hint:

Standard electrode potentials are crucial for understanding the reactivity of metals and the feasibility of redox reactions. More negative \(E^\ominus\) values indicate a greater tendency for the metal to be oxidized (i.e., it is a stronger reducing agent) and a lower tendency for its ion to be reduced. Transition metals generally have negative \(E^\ominus\) values for \(M}^{2+}/M}\) couples, indicating they are stronger reducing agents than hydrogen.

Answer 1

The Correct Option is B

Step 1: Recall the standard electrode potentials for the given transition metals.
The standard electrode potential (\(E^\ominus\)) for the reduction of the divalent metal ion to its metallic form (\(M^{2+}/M\)) are experimentally determined values.
Let's list the approximate standard reduction potentials for each metal:
\(Ni^{2+}/Ni\): -0.25 V
\(Mn^{2+}/Mn\): -1.18 V
\(Fe^{2+}/Fe\): -0.44 V
\(Cr^{2+}/Cr\): -0.91 V
Step 2: Match the metals from List-I with their corresponding electrode potentials from List-II.
A) Ni: The standard electrode potential for \(Ni^{2+}/Ni\) is -0.25 V.
This matches III) -0.25 V in List-II.
\(\rightarrow\) A - III
B) Mn: The standard electrode potential for \(Mn^{2+}/Mn\) is -1.18 V.
This matches I) -1.18 V in List-II.
\(\rightarrow\) B - I
C) Fe: The standard electrode potential for \(Fe^{2+}/Fe\) is -0.44 V.
This matches IV) -0.44 V in List-II.
\(\rightarrow\) C - IV
D) Cr: The standard electrode potential for \(Cr^{2+}/Cr\) is -0.91 V.
This matches II) -0.91 V in List-II.
\(\rightarrow\) D - II
Step 3: Compile the matches and select the correct option.
The correct pairings are:
A - III
B - I
C - IV
D - II
This combination matches Option (2).
The final answer is \(\boxed{A-III, B-I, C-IV, D-II}\).