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 H2O
zCrO4 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?