Question

\(E^\circ_{M^{2+}/M}\) (in V) is highest for

• A. Fe
• B. Mn
• C. Cr
• D. V

Hint:

For the first transition series, it's useful to remember the approximate \(E^\circ_{M^{2+}/M}\) values and the anomalies. Mn has an unusually low (more negative) value due to the stable \(d^5\) configuration of Mn²⁺. Zn also has a low value due to the stable \(d^{10}\) configuration of Zn²⁺. Cu has a unique positive value (+0.34 V).

Answer 1

The Correct Option is A

Step 1: Understanding Standard Electrode Potential: A high positive value of \( E^\circ_{\text{M}^{3+}|\text{M}^{2+}} \) indicates that the ion \( \text{M}^{3+} \) is easily reduced to \( \text{M}^{2+} \), meaning \( \text{M}^{2+} \) is much more stable than \( \text{M}^{3+} \).
Step 2: Electronic Configurations: - Mn (Z=25): \( \text{Mn}^{2+} \) is \( 3d^5 \) (Half-filled, very stable). \( \text{Mn}^{3+} \) is \( 3d^4 \). Converting \( 3d^4 \to 3d^5 \) is very favorable. Hence, \( E^\circ \) is highly positive (+1.57 V). - Fe (Z=26): \( \text{Fe}^{2+} \) is \( 3d^6 \). \( \text{Fe}^{3+} \) is \( 3d^5 \) (Stable). So reaction \( \text{Fe}^{3+} \to \text{Fe}^{2+} \) is less favorable than Mn (actually \( \text{Fe}^{2+} \to \text{Fe}^{3+} \) is favored). \( E^\circ = +0.77 \text{ V} \). - Cr (Z=24): \( \text{Cr}^{3+} \) (\( 3d^3 \), stable \( t_{2g}^3 \)). \( \text{Cr}^{2+} \) (\( 3d^4 \)). \( E^\circ \) is negative (-0.41 V) because \( \text{Cr}^{3+} \) is more stable. - V (Z=23): \( E^\circ \) is negative (-0.26 V).
Step 3: Conclusion: Among the given options, Manganese (Mn) has the highest positive potential due to the extra stability of the half-filled d-subshell in \( \text{Mn}^{2+} \).