In an experiment 50 cm^3 of a 0.10 mol dm^3 solution of a metallic salt reached exactly with 25cm^3 of 0.10 mol dm^3 aqueous sodium sulphite. The half equation for oxidation of sulphite ion is shown below
SO3^-2 (aq) + H2O(l)---SO4-2(aq) + 2H(aq) + 2e^-
If the original oxidation number of the metal in the salt was +3 what would be the new oxidation number of the metal.
a) +1 b) +2 c) +4 d) +5

Question

In an experiment 50 cm^3 of a 0.10 mol dm^3 solution of a metallic salt reached exactly with 25cm^3 of 0.10 mol dm^3 aqueous sodium sulphite. The half equation for oxidation of sulphite ion is shown below 
SO3^-2 (aq) + H2O(l)--->SO4-2(aq) + 2H(aq) + 2e^-
If the original oxidation number of the metal in the salt was +3 what would be the new oxidation number of the metal.
a) +1 b) +2 c) +4 d) +5

anonymous · Answer

its D

abmon98 · Answer

When I saw the markscheme it appears to be B

anonymous · Answer

SO4^-2  , Sulphur +5 , Oxygen -8 , Sodium +1 = -2

abmon98 · Answer

Okay that's right but there is another part of redox reaction of the unknown metal salt

anonymous · Answer

wait the Salt is Na2SO4 right ?

abmon98 · Answer

No metallic salt http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_w07_qp_1.pdf

abmon98 · Answer

This is the link of the published question

anonymous · Answer

ok checkin it

abmon98 · Answer

Thank you for your help

anonymous · Answer

thank you for what ??? I didn't help you still :(

abmon98 · Answer

You are trying your best to figure out the answer

anonymous · Answer

well if we consider  
Metal(O.N) + Sulphur(O.N) + Oxygen(O.N) = -2
(+3) + (+1) + (-6) = -2

anonymous · Answer

Can't figure out .really sorry

wolfe8 · Answer

@aaronq