Explanation:
Vanadium has a lower electrode potential for the V^(3+)/V^(2+) couple despite V^(2+) being more stable than V^(3+) because the electrode potential is related to the ease with which a species can either gain or lose electrons.
Here’s why V^(3+) has a higher electrode potential (is more positive) compared to V^(2+):
1. **Definition of Electrode Potential:** Electrode potential is a measure of the tendency of a species to gain electrons and reduce. A more positive electrode potential indicates a greater tendency to gain electrons (be reduced).
2. **Stability vs. Oxidation State:** Stability of an ion (V^(2+) or V^(3+)) relates to its tendency to exist in that oxidation state in a given chemical environment. V^(2+) is indeed more stable than V^(3+) under normal conditions because the +2 oxidation state is favored due to its electronic configuration and the energy considerations.
3. **Electrochemical Considerations:** In electrochemistry, when considering the reduction process (going from V^(3+) to V^(2+)), the species that is easier to reduce (has a greater tendency to gain electrons) will have a less positive (or more negative) electrode potential.
4. **Standard Reduction Potentials:** The standard reduction potential (E°) of the V^(3+)/V^(2+) couple is approximately +0.34 V. This positive value indicates that V^(3+) is easier to reduce (gains electrons more readily) compared to V^(2+). This suggests that V^(2+) has a lower electrode potential (more negative) than V^(3+).
In summary, despite V^(2+) being more stable than V^(3+), the electrode potential of the V^(3+)/V^(2+) couple is positive, indicating that V^(3+) has a higher electrode potential (more positive) than V^(2+). This is because V^(3+) is more readily reduced (gains electrons more easily) compared to V^(2+), which affects the overall electrode potential of the couple.
