25.0 cm3 of a solution of M2O5 of concentration
0.100moldm-3 is reduced by sulphur dioxide to a lower oxidation
state. To re-oxidise M to its original oxidation number, this
required 50.00cm3 of 0.0200moldm-3 potassium manganate (VII) in
acidified medium. To what oxidation number was M reduced to by
sulfur dioxide?
I am really confused by the question. How do I use the mol ratio
to find the oxidation number. I can't solve similar questions due
to this problem too....
Thanks in advance
Write reduction half-equation of MnO4- to Mn2+.
MnO4–
+
8H+
+
5e–
Mn2+
+
4H2O
Find moles of MnO4- required.
50.00cm3 of 0.0200moldm-3 = 1 x
10^-3 mol
Hence find moles of electrons transferred.
1 x 10^-3 x 5 = 5 x 10^-3
mol
Find moles of M2O5 present.
25.0 cm3 of 0.100moldm-3 =
2.5 x 10^-3 mol
Find moles of M atoms present.
2.5 x 10^-3 x 2 = 5 x
10^-3
Hence find moles of electrons transferred per mole of M
atoms.
(5 x 10^-3) / (5 x 10^-3) =
1
Find OS of M in M2O5.
+5
Hence find reduced OS of M.
(+5) + (-1) = +4