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vanadium oxidation states calculation help!

i did experiment were i prepared a stock solution of vanadium by dissolving 0.5404g of V2O5 and 2.01g of NaOH in 5 cm3 of water. i then diluted it by transferring it in to a 100 cm3 graduated flask and adding 50cm3 of distilled water and 50cm3 of bench dilute H2SO4. the solution was a yellow/orange colour indicating Vanadium was at its highest oxidation state of 5+.

I worked out the concentration of the stock solution with respect to V2O5. i did n=mass/mr therefore; n=0.5404/181.893 > n=0.00297098 moles. now for concentration i did c=n/v therefore; c=n/(100cm3/1000) > c=0.02970977 mol dm-3. i am told that the molarity with respect to vanadium content is double the molarity based on V2O5. Therefore, i times the concentration by 2. giving me a concentration of 0.0594 mol dm-3. for the stock solution.

now for part A of the experiment i did the following: i pipette 20cm3 of my stock solution in to a 250cm3 conical flask diluted with approximately 55cm3 of water added 3g of zinc and 25cm3 of dilute H2SO4 i let this solution heat until it went from yellow > lavender colour indicating a oxidation state of 2+ for vanadium.

next step was to reoxidise the reduced solution back to its 5+ oxidation state from 2+ by titrating with KMnO4 solution. i added 31.20cm3. to go from lavender > yellow/orange colour.

now the aim for me is to deduce a value for the lower oxidation state of vanadium mathematically. i am stuck on this part. i have tried by working out the moles of KMnO4 and V(v) then getting a ratio. but i think i am doing something wrong as i get a value of 0.5? can someone please help me with this problem. I need to get it done in 4 days. Thanks.

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