The Student Room Group

I don't understand mark schemes

Aqueous vanadium(II) chloride, VCl2(aq), can be oxidised by bubbling gaseous chlorine, Cl2(g), through the solution in the absence of air.
40.0cm3 of 0.100mol dm−3
VCl2 solution was oxidised by 144cm3 of chlorine gas,
at room temperature and pressure (r.t.p.).
The chlorine was reduced to chloride ions, according to the half-equation
Cl2(g) + 2e− →  2Cl−(aq)
[Molar]
(i) Use these data to calculate the final oxidation state of vanadium

how did they deduce the whole number ratio of V(II) to Cl2?
(might be a stupid question)
Reply 1
Original post by DUCKYisBEST
Aqueous vanadium(II) chloride, VCl2(aq), can be oxidised by bubbling gaseous chlorine, Cl2(g), through the solution in the absence of air.
40.0cm3 of 0.100mol dm−3
VCl2 solution was oxidised by 144cm3 of chlorine gas,
at room temperature and pressure (r.t.p.).
The chlorine was reduced to chloride ions, according to the half-equation
Cl2(g) + 2e− →  2Cl−(aq)
[Molar]
(i) Use these data to calculate the final oxidation state of vanadium

how did they deduce the whole number ratio of V(II) to Cl2?
(might be a stupid question)

A vanadium(II) ion has a charge of +2 :smile:
If that’s what you mean?
Reply 2
Original post by bl0bf1sh
A vanadium(II) ion has a charge of +2 :smile:
If that’s what you mean?

yes V does have a +2 charge. but the ratio they stated was 2V : 3Cl2
idk how to work that ratio out
Reply 3
Original post by DUCKYisBEST
yes V does have a +2 charge. but the ratio they stated was 2V : 3Cl2
idk how to work that ratio out

Ah right

It probably has something to do with the “40.0cm3 of 0.100mol dm−3VCl2 solution was oxidised by 144cm3 of chlorine” see what you get when you calculate the moles of each species

Quick Reply

Latest

Trending

Trending