The \(E^0_{M^{3+}/M^{2+}}\) values for Cr, Mn, Fe and Co are –0.41, +1.57, +0.77 and +1.97 V respectively. For which one of these metals the change in oxidation state from +2 to +3 is easiest?
Answer & explanation
Correct answer: option 4
The correct answer is option 4. \(Cr\).
The ease of oxidation from +2 to +3 can be determined by looking at the standard reduction potentials \( E^\circ_{M^{3+}/M^{2+}} \). The more positive (or less negative) the reduction potential, the easier it is for the metal to be oxidized from +2 to +3.
Given:
\( E^\circ_{\text{Cr}^{3+}/\text{Cr}^{2+}} = -0.41 \, \text{V} \)
\( E^\circ_{\text{Mn}^{3+}/\text{Mn}^{2+}} = +1.57 \, \text{V} \)
\( E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77 \, \text{V} \)
\( E^\circ_{\text{Co}^{3+}/\text{Co}^{2+}} = +1.97 \, \text{V} \)
Since these are reduction potentials, a higher (more positive) reduction potential indicates a greater tendency to gain electrons (be reduced). Conversely, a lower (more negative) reduction potential indicates a greater tendency to lose electrons (be oxidized).
To find the easiest metal to oxidize from +2 to +3:
Look for the metal with the least positive (or most negative) reduction potential, as this indicates it is easiest to lose an electron and be oxidized.
Analysis:
Cr: \( E^\circ = -0.41 \, \text{V} \)
Mn: \( E^\circ = +1.57 \, \text{V} \)
Fe: \( E^\circ = +0.77 \, \text{V} \)
Co: \( E^\circ = +1.97 \, \text{V} \)
Conclusion:
The metal with the most negative reduction potential is Cr with \( E^\circ = -0.41 \, \text{V} \). Therefore, for chromium, the change in oxidation state from +2 to +3 is the easiest.