1 ) Vanadium (II) ions can be oxidised quantitively by acidified manganate (VII) to a higher oxidation state. In an experiment, 25.0 cm³ of 0.02 mol/dm³ V²+ was found to react with 15.0cm³ of 0.02 mol/dm³ potassium managanate (VII). acidified with dilute sulphuric acid. What is the new oxidation state of vanadium after oxidation? Ans: +5
2 ) The addition of dilute sulphuric acid to Na3CrO4 produces a solution containing chromium (III) ions and dichromate (VI) ions. Construct a balanced equation for the reaction of Na3CrO4 with dilute sulphuric acid.
ans: 3 Na3CrO4 + 5 H2SO4 → Cr³+ + Cr2O7²- + 9Na+ + 5SO4²- + 5 H2O
3) The metallic ion X^n+ is oxidised to XO3- By MnO4- in acidic medium. If 2.68 X 10^-3 moles of X^n+ require 1.61 X 10^-3 moles of MnO4- for oxidation, what is the value of n? Construct an equation for the reaction.
Originally posted by Destined2REIGN:1 ) Vanadium (II) ions can be oxidised quantitively by acidified manganate (VII) to a higher oxidation state. In an experiment, 25.0 cm³ of 0.02 mol/dm³ V²+ was found to react with 15.0cm³ of 0.02 mol/dm³ potassium managanate (VII). acidified with dilute sulphuric acid. What is the new oxidation state of vanadium after oxidation? Ans: +5
Let the new ovidation state of vanadium be +n
No. of moles of V2+ / No. of moles of MnO4- = 5/3 (do some calculations yourself here)
Hence,
5V2+ + 3MnO4- + xH+ --> 5Vn+ + 3Mn2+ + yH2O
Charges on L.H.S. = Charges on R.H.S.
Can you do it now? Double check by posting.
2 ) The addition of dilute sulphuric acid to Na3CrO4 produces a solution containing chromium (III) ions and dichromate (VI) ions. Construct a balanced equation for the reaction of Na3CrO4 with dilute sulphuric acid.
ans: 3 Na3CrO4 + 5 H2SO4 → Cr³+ + Cr2O7²- + 9Na+ + 5SO4²- + 5 H2OzCrO4 3- + xH+ --> Cr3+ Cr2O7 2- + yH2O
Write this out first before putting back into the original equation, if necessary. Otherwise I don't see the issue of not being able to write out the whole equation entirely.
I know that there are other further parts for this question.
3) The metallic ion X^n+ is oxidised to XO3- By MnO4- in acidic medium. If 2.68 X 10^-3 moles of X^n+ require 1.61 X 10^-3 moles of MnO4- for oxidation, what is the value of n? Construct an equation for the reaction.No. of moles of Xn+ / No. of moles MnO4- = 5/3
So do you agree that for every 3 moles of MnO4-, 15 electrons are transferred away from 5 moles of Xn+ ?
Can you do it now?
Done. What JC are you from, btw?
2) (CrO4)3- ---> Cr3+ reduction (5+ to 3+)
(CrO4)3- ----> (Cr2O7)2- ( 5+ to 6+)
Write out half equations for each step
1. Balance atoms other than O and H
2. Once balanced, balanced excess O by inserting H2O
3. Balance excess H by inserting H+
4. Finally balance charge by adding electrons
(CrO4)3- -----> Cr3+
(CrO4)3- ----> Cr3+ + 4H2O
(CrO4)3- + 8H+ ----> Cr3+ + 4H2O
(CrO4)3- + 8 H+ + 2e ----> Cr3+ + 4H2O
(CrO4)3- -----> (Cr2O7)2-
2(CrO4)3- -----> (Cr2O7)2-
2(CrO4)3- -----> (Cr2O7)2- + H2O
2(CrO4)3- + 2H+ -----> (Cr2O7)2- + H2O
2(CrO4)3- + 2H+ -----> (Cr2O7)2- + H2O + 2e
Combining both halves ( making sure the electrons cancel out)
3(CrO4)3- + 10 H+ -------> Cr3+ + (Cr2O7)2- + 5H2O
Adding Na+ to Cr species to balance charge and SO42- to make H2SO4
6Na3CrO4 + 10H2SO4 ------> Cr2(SO4)3 + 2Na2Cr2O7 + 7Na2SO4 + 10H2O
Originally posted by Garrick_3658:Done. What JC are you from, btw?
No. of moles of Xn+ / No. of moles MnO4- =
5/3
when I put (2.68 X 10^-3) / (1.61 X 10^-3) I get 268/161 not 5/3. I know I'm doing something wrong.
Originally posted by Destined2REIGN:1 ) Vanadium (II) ions can be oxidised quantitively by acidified manganate (VII) to a higher oxidation state. In an experiment, 25.0 cm³ of 0.02 mol/dm³ V²+ was found to react with 15.0cm³ of 0.02 mol/dm³ potassium managanate (VII). acidified with dilute sulphuric acid. What is the new oxidation state of vanadium after oxidation? Ans: +5
2 ) The addition of dilute sulphuric acid to Na3CrO4 produces a solution containing chromium (III) ions and dichromate (VI) ions. Construct a balanced equation for the reaction of Na3CrO4 with dilute sulphuric acid.
ans: 3 Na3CrO4 + 5 H2SO4 → Cr³+ + Cr2O7²- + 9Na+ + 5SO4²- + 5 H2O
3) The metallic ion X^n+ is oxidised to XO3- By MnO4- in acidic medium. If 2.68 X 10^-3 moles of X^n+ require 1.61 X 10^-3 moles of MnO4- for oxidation, what is the value of n? Construct an equation for the reaction.
>>> 3) The metallic ion X^n+ is oxidised to XO3- By MnO4- in acidic medium. If 2.68 X 10^-3 moles of X^n+ require 1.61 X 10^-3 moles of MnO4- for oxidation, what is the value of n? Construct an equation for the reaction. <<<
Solution :
1.61 x 10^-3 mol of MnO4- implies 5 x 1.61 x 10^-3 mol of electrons removed from 2.68 x 10^-3 mol of X^n.
This implies 3 mol of electrons removed per mole of X^n.
O.S. of X in XO3- = +5
Therefore, n = O.S. of X before oxidation = +2
[Reduction] MnO4– + 8H+ + 5e– ---> Mn2+ + 4H2O
[Oxidation] X2+ + 3H2O --> XO3- + 6H+ + 3e-
[Balanced Redox] 3MnO4- + 5X2+ + 3H2O --> 3Mn2+ + 5XO3- + 6H+
Originally posted by Destined2REIGN:No. of moles of Xn+ / No. of moles MnO4- = 5/3
when I put (2.68 X 10^-3) / (1.61 X 10^-3) I get 268/161 not 5/3. I know I'm doing something wrong.
Ratio.
268/161 x 3 gets you somewhere near 5. It's like doing empirical formula for mole concept.
Originally posted by Garrick_3658:Ratio.
268/161 x 3 gets you somewhere near 5. It's like doing empirical formula for mole concept.
didnt know I could do that. Noted with thanks.