oxidation of vanadium(ll) to vanadium(lll) and no further

You can try H2SO4 because

V3+(aq) + e− → V2+(aq) E=−0.26V
VO2+(aq) + 2H+ + e− → V3+(aq) E=+0.34V

so the E of oxidizing agent should be positive and smaller than +0.34V, there are a lot of choices, H2SO4 is chosen because it is common

2H+(aq) + 2e− → H2(g) E=0.00V
HSO4−(aq) + 3H+ + 2e− → 2H2O(l) + SO2(aq) E=+0.16V
SO42−(aq) + 4H+ + 2e− → 2H2O(l) + SO2(aq) E=+0.17V