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# can someone help me with this quick question? Watch

1. Hi everyone, i hope you're all doing well. Im a bit stuck on the following question. Its basically to do with the ratio bit. My question is, how is ratio linked with finding the change in oxidation state, as well as even finding the oxidation state.

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.

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[2]
[Total 6 marks]

33. (a) Moles V2+ = 25.0 × 0.100 / 1000 = 0.0025 mols 1
Moles MnO4– = 30.0 × 0.0500 / 1000 = 0.00150 mols 1
1 mole of MnO4- changes its Oxidation State by 5 to change
the Oxidation State of 1.67 moles of V2+ 1
Oxidation State of V2+ changes by 5 / 1.67 = 3

The italic bit is the mark scheme for the answer.
can somebody please also explain to me the underlined bit, i dont get what they mean by it.

thanks
2. anyone please?
3. I think it's saying that if you had 1 mol of MnO4-, then it can take 5 electrons off an equal amount of V2+. But since you have more V2+ than MnO4-, it can't take the full 5 from every V2+.

Since you have 1.67/1 = 1.67 times more V2+ than Mn04-, then each V2+ can only give 1/1.67 = 0.6 times as many electrons as it would if you had equal amount. And 0.6 * 5 is 3.

I hope that makes sense.
4. (Original post by Gordon1985)
I think it's saying that if you had 1 mol of MnO4-, then it can take 5 electrons off an equal amount of V2+. But since you have more V2+ than MnO4-, it can't take the full 5 from every V2+.

Since you have 1.67/1 = 1.67 times more V2+ than Mn04-, then each V2+ can only give 1/1.67 = 0.6 times as many electrons as it would if you had equal amount. And 0.6 * 5 is 3.

I hope that makes sense.
unfortunately i didn't quite understand.
5. (Original post by mabz123)
unfortunately i didn't quite understand.
If you had 1 mol of Mn04- and 1 mol of V2+, then each MnO4- molecule would take 5 electrons from each V2+ molecule, meaning the oxidation states of both molecules change by 5.

But, you have more V2+ than you have Mn04-, so the Mn04- can't take 5 electrons from every V2+ because each MnO4- can only accept 5 electrons.

Don't think in mols, think in individual molecules. If you had 1 MnO4- molecule and 1 V2+ molecule. The Mn04- would take 5 electrons and that would be that. If you had 1 Mn04- molecule and 5 V2+ molecules, than the one Mn04- could only take 1 electron from each V2+. So the oxidation state of Mn04- would change by 5 but the oxidation state of V2+ would only change by (the amount of Mn04- / the amount of V2+)* the number of electrons Mn04- can accept, which is (1/5)*5 = 1. This makes sense as you know each V2+ molecule only gave 1 of it's electrons to the Mn04-.

Basically the oxidation state of V2+ chnages by....

(The amount of Mn04- / the amount of V2+) * the number of electrons Mn04- can accept. This is (1/5)*5 = 1.

Going back to mols, you have 1 Mn04- and 1.67 V2+, and Mn04- can accept 5 electrons. So just doing the same thing as above...

(1/1.67)*5 = 3.

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