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1st qn : qn is flawed becoz the counter-anion is not specified
(so let's assume it's uninegative), and the molarity of sulphite
solution not given (so let's assume it's also 0.1M). 2 mol of M3+
reacts with 1 mol of SO3 2-, which when oxidized to SO4 2-,
releases 2 mol of e-. Hence 2 mol of M3+ accepts 2 mol of e-, ie.
final OS of metal = +2.
2nd qn : each mol of FeC2O4 releases 3 mol of e- upon oxidation
(as each Fe2+ is oxidized to Fe3+, and each C2O4 2- is oxidized to
2 CO2). Each mol of MnO4- accepts 5 mol of e- upon reduction in
acidic pH. Hence each mol of FeC2O4 will react with 3/5 mol of
MnO4-.
Hi ultima,how do you solve these two questions?
'50cm3 of a 0.1 mol dm3 of a solution of a metallic salt reacted
exactly with 25cm3 of sulphite solution.The half equation for the
oxidation of sulphite ion is
SO3^2- + H2O -> SO4^2- + 2H^+ +2e
If the original exidation number of the metal in the salt was 3,
calculate the new oxidation number of the metal'
I only managed to find the no.moles of the salt and I'm
stucked.
'Half-equations
2CO2 + 2e -> C2O4^2-
FE^3+ + e -> FE^2+
MNO4^- + 8H^+ + 5e -> MN^2+ + 4H2O
Calculate the no.moles of MNO4^- ion that will react with 1 mol
of iron(II) ethanedioate;FeC2O4 in acidic solution'
This one,I dont even know where to start since there are no
numbers provided.
