Transition metal titration - A-level Chem help!
Hi, I’m doing some chem past paper questions and have come across this one that I just don’t get.
Here it is:
Vanadium can exist in a number of different oxidation states. One compound of vanadium is ammonium vanadate(V) and this contains the ion VO3–. This can be reduced to V2+ in several steps, using zinc metal and aqueous sulphuric acid.
(a) 25.0 cm3 of 0.100 mol dm–3 ammonium vanadate(V) is completely reduced to V2+(aq) using zinc and aqueous sulphuric acid. The resulting solution is titrated with 0.0500 mol dm–3 MnO4–(aq) and 30.0 cm3 is required to oxidise the V2+(aq) back to VO3–(aq).
The half equation for acidified MnO4– acting as an oxidising agent is shown below.
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O
Show that the vanadium has changed oxidation state from +2 to +5 in this
titration.
I get the two marking points: calculate the moles of both VO2+ and MnO4-, but I’ve got absolutely no idea how they get 1 moles MnO4- creating 1.67 V2+. Any help would be greatly appreciated.
Thanks!
Here it is:
Vanadium can exist in a number of different oxidation states. One compound of vanadium is ammonium vanadate(V) and this contains the ion VO3–. This can be reduced to V2+ in several steps, using zinc metal and aqueous sulphuric acid.
(a) 25.0 cm3 of 0.100 mol dm–3 ammonium vanadate(V) is completely reduced to V2+(aq) using zinc and aqueous sulphuric acid. The resulting solution is titrated with 0.0500 mol dm–3 MnO4–(aq) and 30.0 cm3 is required to oxidise the V2+(aq) back to VO3–(aq).
The half equation for acidified MnO4– acting as an oxidising agent is shown below.
MnO4– + 8H+ + 5e– → Mn2+ + 4H2O
Show that the vanadium has changed oxidation state from +2 to +5 in this
titration.
I get the two marking points: calculate the moles of both VO2+ and MnO4-, but I’ve got absolutely no idea how they get 1 moles MnO4- creating 1.67 V2+. Any help would be greatly appreciated.
Thanks!
Original post by lordaxil
For every 1 mole of Mn2+ oxidised, 5e- are released, but 3e- are required to reduce every mole of VO3- to V2+, so the ratio of MnO4- reduced to V2+ created is 5/3 = 1.67 (2 s.f.).
Omg how did I not realise this? Thank you so much this seems obvious now haha
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