Yeah, so that's what I was thinking, but the answer is actually C (I probably should have said that at the start)...
This is what the examiner's report said:
Linking the electrochemical series data
O2(g) + 4H+(aq) + 4e– → 2 H2O(l) E0 = +1.23 V
VO2+(aq) + 2H+(aq) + e– → VO2+(aq) + H2O(l) E0 = +1.00 V
VO2+(aq) + 2H+(aq) + e– → V3+(aq) + H2O(l) E0 = +0.34 V
V3+(aq) + e– → V2+(aq) E0 = –0.26 V
The strongest reductant, V2+(aq), is oxidised to V3+(aq), which is both an oxidant and a reductant.
Subsequently as the [V2+(aq)] decreases, the next strongest reductant, V3+(aq), is oxidised to VO2+(aq) by O2.
If the reaction is allowed to continue, since VO2+(aq) is also both an oxidant and a reductant, VO2+(aq) can be oxidised to VO2+(aq) by O2(g).
So V2+(aq) can be oxidised to V3+(aq), VO2+(aq) and VO2+(aq).
I could not make sense of any of that
(I definitely think it's a weird question )
This answer is a bit odd, but
What this answer is saying is that V2+ is the strongest reductant which means it will be oxidised first. After all the V2+ is oxidised, the next strongest reductant will be oxidised (in this case V3+).
You can determine the strongest reductant by constructing an electrochemical series by listing all the possible half cell reactions from the greatest resultant voltage to the lowest. Species on the right side of the equation are reductants and increase in strength as you move further down the electrochemical series. Species on the left side of the equation are oxidants and increase in strength as you move further up the electrochemical series
I assume there was more information given in a question stem above that applied to the question, hence the 4 half cell equations provided in the answer guide.