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Vanadium question - A-level aqa specimen paper 1

IMG_1882.jpg
Can anyone help me on this question I've managed to work out the moles of NH4VO3 and SO2, but not sure what to do next. I think you need to work out which oxidation state SO2 is oxididised to and V5+ reduced to but not sure.
Original post by Emma23114
IMG_1882.jpg
Can anyone help me on this question I've managed to work out the moles of NH4VO3 and SO2, but not sure what to do next. I think you need to work out which oxidation state SO2 is oxididised to and V5+ reduced to but not sure.


Yes that's the start.

Then you need to see how the moles of each compare. Dont forget that the moles of electrons lost by the sulfur(IV) must be gained by the vanadium, bearing in mind that sulfur(IV) can only get oxidised to sulfur(VI).
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Emma23114
OP
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Original post by charco
Yes that's the start.

Then you need to see how the moles of each compare. Dont forget that the moles of electrons lost by the sulfur(IV) must be gained by the vanadium, bearing in mind that sulfur(IV) can only get oxidised to sulfur(VI).


Thanks I lot I've got an answers now of 4. But I still don't understand why at the end you divide the electron moles by the moles of NH4VO3 to find how many electrons are lost in the reduction of V +5 image.jpg
Original post by Emma23114
Thanks I lot I've got an answers now of 4. But I still don't understand why at the end you divide the electron moles by the moles of NH4VO3 to find how many electrons are lost in the reduction of V +5 image.jpg


mol vanadium = 0.05 x 0.8 = 0.04 mol
oxidation state of vanadium = +5

mol sulfur dioxide = PV/RT = (98 x 0.506)/(8.31 x 293) = 0.02 mol
oxidation state sulfur = +4

When sulfur dioxide gets oxidised it becomes S(VI)
i.e. it releases 2 electrons
Hence 0.02 mol releases 0.04 mol electrons

therefore 0.04 mol electrons go to 0.04 mol vanadium
hence 1 mol electron per mol of vanadium

Vanadium is reduced from +5 to +4
Reply 4
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Emma23114
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Original post by charco
mol vanadium = 0.05 x 0.8 = 0.04 mol
oxidation state of vanadium = +5

mol sulfur dioxide = PV/RT = (98 x 0.506)/(8.31 x 293) = 0.02 mol
oxidation state sulfur = +4

When sulfur dioxide gets oxidised it becomes S(VI)
i.e. it releases 2 electrons
Hence 0.02 mol releases 0.04 mol electrons

therefore 0.04 mol electrons go to 0.04 mol vanadium
hence 1 mol electron per mol of vanadium

Vanadium is reduced from +5 to +4


Thank you! I understand now 😊
Original post by Emma23114
Thank you! I understand now 😊


Please help! I dont get any of what you done; I only understand how you calculated the moles part.

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