Question

The \( E^\circ \) values for \( \text{Cr}^{3+/2+} \), \( \text{Mn}^{3+/2+} \), \( \text{Fe}^{3+/2+} \), and \( \text{Co}^{3+/2+} \) are -0.41, +1.57, +0.77, and +1.97 V respectively. For which of these metals, the change in oxidation state from +2 to +3 is the easiest?

• A. Fe
• B. Mn
• C. Cr
• D. Co

Hint:

When determining which metal is easiest to oxidize, look for the metal with the lowest \( E^\circ \) value, as this indicates the greatest tendency to lose electrons.

Answer 1

The Correct Option is C

Step 1: Understand the meaning of \( E^\circ \).
The \( E^\circ \) value (standard electrode potential) indicates the tendency of a metal to be oxidized. A higher \( E^\circ \) value means that the metal is more easily oxidized (i.e., it more readily loses electrons). Conversely, a lower \( E^\circ \) value indicates a lesser tendency to oxidize.

Step 2: Analyze the oxidation states.
We are asked to find the metal for which the change in oxidation state from +2 to +3 is the easiest. This means we need to focus on the reduction half-reaction, which involves the metal being oxidized from +2 to +3. The metal with the lowest \( E^\circ \) for the \( \text{M}^{3+}/\text{M}^{2+} \) transition will be the easiest to oxidize.

Step 3: Examine the \( E^\circ \) values for the given metals.
The given \( E^\circ \) values are:
- \( \text{Cr}^{3+/2+} \) has \( E^\circ = -0.41 \) V
- \( \text{Mn}^{3+/2+} \) has \( E^\circ = +1.57 \) V
- \( \text{Fe}^{3+/2+} \) has \( E^\circ = +0.77 \) V
- \( \text{Co}^{3+/2+} \) has \( E^\circ = +1.97 \) V

Step 4: Determine the easiest metal to oxidize.
The \( E^\circ \) value for \( \text{Cr}^{3+/2+} \) is the lowest, meaning it is the easiest for chromium to go from the +2 oxidation state to the +3 oxidation state. This is because a lower \( E^\circ \) indicates that the metal is more likely to lose electrons (i.e., oxidize).

Step 5: Conclusion.
Thus, the change in oxidation state from +2 to +3 is easiest for \( \text{Cr} \), corresponding to option (C).