Calculate the oxidation state of the vanadium ion?! URGENT
This is the question:
The vanadium in 50.0 cm3 of a 0.800 mol dm–3 solution of NH4VO3 reacts with 506 cm3 of sulfur(IV) oxide gas measured at 20.0 °C and 98.0 kPa.
Use this information to calculate the oxidation state of the vanadium in the solution after the reduction reaction with sulfur(IV) oxide.
Explain your working. The gas constant R=8.3 J K-1 mol-1 .
I've found the moles of V and the moles of SO2, but after that I'm stuck. the answer is +4. can someone please help and explain the steps too please!!
The vanadium in 50.0 cm3 of a 0.800 mol dm–3 solution of NH4VO3 reacts with 506 cm3 of sulfur(IV) oxide gas measured at 20.0 °C and 98.0 kPa.
Use this information to calculate the oxidation state of the vanadium in the solution after the reduction reaction with sulfur(IV) oxide.
Explain your working. The gas constant R=8.3 J K-1 mol-1 .
I've found the moles of V and the moles of SO2, but after that I'm stuck. the answer is +4. can someone please help and explain the steps too please!!
Original post by nfs.
This is the question:
The vanadium in 50.0 cm3 of a 0.800 mol dm–3 solution of NH4VO3 reacts with 506 cm3 of sulfur(IV) oxide gas measured at 20.0 °C and 98.0 kPa.
Use this information to calculate the oxidation state of the vanadium in the solution after the reduction reaction with sulfur(IV) oxide.
Explain your working. The gas constant R=8.3 J K-1 mol-1 .
I've found the moles of V and the moles of SO2, but after that I'm stuck. the answer is +4. can someone please help and explain the steps too please!!
The vanadium in 50.0 cm3 of a 0.800 mol dm–3 solution of NH4VO3 reacts with 506 cm3 of sulfur(IV) oxide gas measured at 20.0 °C and 98.0 kPa.
Use this information to calculate the oxidation state of the vanadium in the solution after the reduction reaction with sulfur(IV) oxide.
Explain your working. The gas constant R=8.3 J K-1 mol-1 .
I've found the moles of V and the moles of SO2, but after that I'm stuck. the answer is +4. can someone please help and explain the steps too please!!
Do you know what sulphur(IV) oxide gets oxidised to?
Original post by charco
Do you know what sulphur(IV) oxide gets oxidised to?
In the mark scheme it says:
When SO2 (sulfur(IV) oxide) acts as a reducing agent, it is oxidised to sulfate(VI) ions so this is a two electron change
But I don't understand how you are meant to know that from the question and how you are meant to know the so2 is being oxidised?
Original post by nfs.
In the mark scheme it says:
When SO2 (sulfur(IV) oxide) acts as a reducing agent, it is oxidised to sulfate(VI) ions so this is a two electron change
But I don't understand how you are meant to know that from the question and how you are meant to know the so2 is being oxidised?
When SO2 (sulfur(IV) oxide) acts as a reducing agent, it is oxidised to sulfate(VI) ions so this is a two electron change
But I don't understand how you are meant to know that from the question and how you are meant to know the so2 is being oxidised?
OK, so you can now see how many electrons are released by 1 mol of sulfur(IV) oxide as it turns to sulfate ions.
Well, the same number of electrons MUST be absorbed by the vanadium atoms.
So if you know the moles of vanadium and the moles of electrons you can work out the moles of electrons per mol of vanadium and hence how many oxidation numbers it changes by ..
Original post by nfs.
In the mark scheme it says:
When SO2 (sulfur(IV) oxide) acts as a reducing agent, it is oxidised to sulfate(VI) ions so this is a two electron change
But I don't understand how you are meant to know that from the question and how you are meant to know the so2 is being oxidised?
When SO2 (sulfur(IV) oxide) acts as a reducing agent, it is oxidised to sulfate(VI) ions so this is a two electron change
But I don't understand how you are meant to know that from the question and how you are meant to know the so2 is being oxidised?
It tells you in the question:
"...oxidation state of the vanadium in the solution after the reduction reaction with sulfur(IV) oxide"
The V is being reduced so the SO2 must be being oxidised. The only possible S species with a higher oxidation state than the +4 in SO2 is +6 in SO4 (2-)...
Original post by MrChemTeacher
It tells you in the question:
"...oxidation state of the vanadium in the solution after the reduction reaction with sulfur(IV) oxide"
The V is being reduced so the SO2 must be being oxidised. The only possible S species with a higher oxidation state than the +4 in SO2 is +6 in SO4 (2-)...
"...oxidation state of the vanadium in the solution after the reduction reaction with sulfur(IV) oxide"
The V is being reduced so the SO2 must be being oxidised. The only possible S species with a higher oxidation state than the +4 in SO2 is +6 in SO4 (2-)...
It says in the mark scheme that that is a two electron change... what does that mean?
Actually i think I get that - it has lost 2 electrons...
Just don't get how on earth they got the last part.
''4.00 × 10–2 mol vanadium has gained 4.08 × 10–2 mol of electrons therefore 1 mol vanadium has gained 4.08 × 10–2 / 4.00 × 10–2 = 1 mol of electrons to the nearest integer, so new oxidation state is 5–1=4.''
Why did they divide it?
(edited 7 years ago)
You work out the moles of SO2, then work out the half equation
O2 2H2O--> SO4(2-) 4H 2e-So the moles of electrons produced will be 2x the moles of SO2 (= 4.00x10-2)If you've got 4.00x10-2 moles of NH4VO3 then it must be the case that each V gains 1 electron so the oxidation state drops by one.
O2 2H2O--> SO4(2-) 4H 2e-So the moles of electrons produced will be 2x the moles of SO2 (= 4.00x10-2)If you've got 4.00x10-2 moles of NH4VO3 then it must be the case that each V gains 1 electron so the oxidation state drops by one.
Original post by charco
Do you know what sulphur(IV) oxide gets oxidised to?
After you find the moles of sulfur dioxide, you multiply that 0.0204 moles into 2 (that's the electrons lost when SO2 gets oxidised to SO42-) to give the total number of moles of electrons i.e. 0.0408 lost by SO2.
Now, 0.04 moles of V in NH4VO3 will gain 0.0408 moles of electrons thanks to the sulfur dioxide
Thus, 1 mole of V in NH4VO3 will gain 0.0408 x 1)/0.04 = 1 mole of electrons.
The original oxidation number of V in NH4NO3 was +5,
new oxidation No will be 5 - 1 (as 1 mole of electrons is gained) = +4.
Quick Reply
Related discussions
- I don't understand mark schemes
- Help with E.cell question
- Please help
- shapes and bonding
- Oxidation states
- Transition Metal Ions
- Chemistry balancing redox equations
- chem mcq help please
- Does vanadium’s 3d orbital fill before 4s?
- question about a half equation
- A Level Chemistry Exam Question-need help!
- Oxidation number format vs ion charge format
- Chemistry OCR A
- Why does Scandium only exist in the 3+ ion
- chemistry question redox reaction
- Etha
- Chemistry (Complex ions and aqua ions)
- A-level Chemistry Study Group 2022-2023
- Redox question; help needed. Oxidation of Thiosulfate
- Learning Transition metal complex ions.
Latest
Trending
