Question

Consider the following reduction processes: \[ \text{Al}^{3+} + 3e^- \longrightarrow \text{Al(s)}, \quad E^\circ = -1.66\,\text{V} \] \[ \text{Fe}^{3+} + e^- \longrightarrow \text{Fe}^{2+}, \quad E^\circ = +0.77\,\text{V} \] \[ \text{Co}^{3+} + e^- \longrightarrow \text{Co}^{2+}, \quad E^\circ = +1.81\,\text{V} \] \[ \text{Cr}^{3+} + 3e^- \longrightarrow \text{Cr(s)}, \quad E^\circ = -0.74\,\text{V} \] The tendency to act as reducing agent decreases in the order:

• A. \( \text{Cr}>\text{Fe}^{2+}>\text{Al}>\text{Co}^{2+} \)
• B. \( \text{Al}>\text{Cr}>\text{Co}^{2+}>\text{Fe}^{2+} \)
• C. \( \text{Al}>\text{Cr}>\text{Fe}^{2+}>\text{Co}^{2+} \)
• D. \( \text{Al}>\text{Fe}^{2+}>\text{Cr}>\text{Co}^{2+} \)

Hint:

More negative standard reduction potential means greater tendency to lose electrons and hence stronger reducing power.

Answer 1

The Correct Option is C

Step 1: Relation between reducing power and electrode potential.
A species acts as a stronger reducing agent if it has a greater tendency to undergo oxidation. This corresponds to a more negative standard reduction potential \(E^\circ\).

Step 2: Compare given standard reduction potentials.
\[ E^\circ(\text{Al}^{3+}/\text{Al}) = -1.66\,\text{V} \] \[ E^\circ(\text{Cr}^{3+}/\text{Cr}) = -0.74\,\text{V} \] \[ E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\,\text{V} \] \[ E^\circ(\text{Co}^{3+}/\text{Co}^{2+}) = +1.81\,\text{V} \] More negative \(E^\circ\) implies stronger reducing agent.

Step 3: Arrange in decreasing order of reducing strength.
\[ \text{Al}>\text{Cr}>\text{Fe}^{2+}>\text{Co}^{2+} \]

Final Answer: \[ \boxed{\text{Al}>\text{Cr}>\text{Fe}^{2+}>\text{Co}^{2+}} \]