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 the Concept:
The question asks for the highest (most positive or least negative) standard electrode potential (\(E^\circ_{M^{2+}/M}\)) among the given first-row transition metals. The standard electrode potential reflects the tendency of a species to be reduced. A more positive \(E^\circ\) value indicates a greater tendency for the M²⁺ ion to be reduced to the metal M.
Step 2: Analyzing the Trend and Standard Values:
The standard electrode potentials for the M²⁺/M couple in the first transition series generally show a trend of becoming less negative from Ti to Cu, but with some notable exceptions. Let's look at the standard values for the given elements:
- V (Vanadium): \(E^\circ_{V^{2+}/V} = -1.18\) V
- Cr (Chromium): \(E^\circ_{Cr^{2+}/Cr} = -0.91\) V
- Mn (Manganese): \(E^\circ_{Mn^{2+}/Mn} = -1.18\) V
- Fe (Iron): \(E^\circ_{Fe^{2+}/Fe} = -0.44\) V
Step 3: Comparing the Values:
Comparing the given values:
- \(E^\circ(\text{Fe}) = -0.44\) V
- \(E^\circ(\text{Mn}) = -1.18\) V
- \(E^\circ(\text{Cr}) = -0.91\) V
- \(E^\circ(\text{V}) = -1.18\) V
The highest value (least negative) among these is -0.44 V, which corresponds to Iron (Fe).
Step 4: Explaining the Anomalies:
- The \(E^\circ\) value for Mn is more negative than expected from the general trend. This is due to the extra stability of the Mn²⁺ ion, which has a half-filled d-orbital configuration (\(d^5\)). - The value for Fe is less negative (higher) than its neighbors. This can be attributed to a combination of factors including hydration enthalpy and ionization enthalpy.
Final Answer:
Comparing the standard reduction potentials, Iron (Fe) has the highest value (-0.44 V). Therefore, option (A) is correct.