Question

0.1 mole of a metal is burnt in air it forms an oxide. The same oxide is then reduced by $4\text{L}$ of $0.05\text{M}$ $S_2{O}_3^{2-}$ (acidic medium) to a +3 oxidation state of the metal. What is the oxidation state of the metal in the oxide?

Answer 1

$M+O_2\to M^{n+}$ (oxide formation)
$M^{n+}\to M^{3+}$ (because of thiosulphate)
The number of equivalents need to be the same in the second redox reacion, hence:
$(0.1\times (n-3){)}_{metal}=(4\times 1\times 0.05{)}_{S_2{O}_3^{2-}}$
$n=5$