- For vanadium, we need to consider the following standard electrode potential values
- The half equations are arranged from the most negative Eθ at the top to the most positive Eθ at the bottom
- The strongest oxidising agent is the reactant from the electrode reaction with the most positive standard electrode potential
- In this case, the most positive value is + 1.00 V, which means that VO2+ is the strongest oxidising agent
- The strongest reducing agent is the product from the electrode reaction with the most negative standard electrode potential
- In this case, the most negative value is – 1.18 V, which means that V is the strongest reducing agent
- For all of the following reactions, we will use zinc as the reducing agent
Reduction of V from +5 to +4 by Zn
- Vanadium is being reduced from an oxidation number of +5 to +4 in half equation 5
- So, the two half equations we need to consider are:
- 2. Zn2+ (aq) + 2e– Zn (s) Eθ = – 0.76 V
- 5. VO2+ (aq) + 2H+ (aq) + e– VO2+ (aq) + H2O (l) Eθ = + 1.00 V
- Half equation 5 has the most positive standard electrode potential, Eθ, value
- Therefore, this is the reduction reaction
- VO2+ (aq) + 2H+ (aq) + e– VO2+ (aq) + H2O (l)
- Half equation 2 has the least positive standard electrode potential, Eθ, value
- Therefore, this is the oxidation reaction and needs to be reversed
- Zn (s) Zn2+ (aq) + 2e–
- We can obtain the overall equation by combining the reduction equation (half equation 5) and the oxidation equation (the reversed half equation 2)
- Remember: When combining half equations, they must have the same number of electrons in both equations
2VO2+ (aq) + 4H+ (aq) + 2e– |
|
2VO2+ (aq) + 2H2O (l) |
Zn (s) |
|
Zn2+ (aq) + 2e– |
- The equal number of electrons on both sides of the equation cancel out to give the overall equation:
2VO2+ (aq) + 4H+ (aq) + Zn (s) 2VO2+ (aq) + Zn2+ (aq) + 2H2O (l)
Reduction of V from +4 to +3 by Zn
- Vanadium is being reduced from an oxidation number of +4 to +3 in half equation 4
- So, the two half equations we need to consider are:
- 2. Zn2+ (aq) + 2e– Zn (s) Eθ = – 0.76 V
- 4. VO2+ (aq) + 2H+ (aq) + e– V3+ (aq) + H2O (l) Eθ = + 0.34 V
- Half equation 4 has the most positive standard electrode potential, Eθ, value
- Therefore, this is the reduction reaction
- VO2+ (aq) + 2H+ (aq) + e– V3+ (aq) + H2O (l)
- Half equation 2 has the least positive standard electrode potential, Eθ, value
- Therefore, this is the oxidation reaction and needs to be reversed
- Zn (s) Zn2+ (aq) + 2e–
- We can obtain the overall equation by combining the reduction equation (half equation 5) and the oxidation equation (the reversed half equation 2)
- Remember: When combining half equations, they must have the same number of electrons in both equations
2VO2+ (aq) + 4H+ (aq) + 2e– |
|
2V3+ (aq) + 2H2O (l) |
Zn (s) |
|
Zn2+ (aq) + 2e– |
- The equal number of electrons on both sides of the equation cancel out to give the overall equation:
2VO2+ (aq) + 4H+ (aq) + Zn (s) 2V3+ (aq) + Zn2+ (aq) + 2H2O (l)
Reduction of V from +3 to +2 by Zn
- Vanadium is being reduced from an oxidation number of +3 to +2 in half equation 3
- So, the two half equations we need to consider are:
- 2. Zn2+ (aq) + 2e– Zn (s) Eθ = – 0.76 V
- 3. V3+ (aq) + e– V2+ (aq) Eθ = – 0.26 V
- Half equation 3 has the most positive / least negative standard electrode potential, Eθ, value
- Therefore, this is the reduction reaction
- V3+ (aq) + e– V2+ (aq)
- Half equation 2 has the least positive standard electrode potential, Eθ, value
- Therefore, this is the oxidation reaction and needs to be reversed
- Zn (s) Zn2+ (aq) + 2e–
- We can obtain the overall equation by combining the reduction equation (half equation 5) and the oxidation equation (the reversed half equation 2)
- Remember: When combining half equations, they must have the same number of electrons in both equations
2V3+ (aq) + 2e– |
|
2V2+ (aq) |
Zn (s) |
|
Zn2+ (aq) + 2e– |
- The equal number of electrons on both sides of the equation cancel out to give the overall equation:
2V3+ (aq) + Zn (s) 2V2+ (aq) + Zn2+ (aq)
Reduction of V from +2 to 0 by Zn
- Vanadium is being reduced from an oxidation number of +3 to +2 in half equation 1
- So, the two half equations we need to consider are:
- 2. Zn2+ (aq) + 2e– Zn (s) Eθ = – 0.76 V
- 1. V2+ (aq) + 2e– V (s) Eθ = – 1.18 V
- Half equation 1 has the most negative standard electrode potential, Eθ, value
- Therefore, this is the oxidation reaction and needs to be reversed
- V (s) V2+ (aq) + 2e–
- Half equation 2 has the most positive / least negative standard electrode potential, Eθ, value
- Therefore, this is the reduction reaction
- Zn2+ (aq) + 2e– Zn (s)
- We can obtain the overall equation by combining the reduction equation (half equation 5) and the oxidation equation (the reversed half equation 2)
- When combining these half equations, there is no need to change them as they have the same number of electrons in both equations
V (s) |
|
V2+ (aq) + 2e– |
Zn2+ (aq) + 2e– |
|
Zn (s) |
- The equal number of electrons on both sides of the equation cancel out to give the overall equation:
V (s) + Zn2+ (aq) V2+ (aq) + Zn (s)
-
- Zn is not electron releasing with respect to V2+
- This means this reaction is not thermodynamically feasible
Predicting oxidation reactions
- The same method can be used to predict whether a given oxidising agent will oxidise a vanadium species to one with a higher oxidation number