Question

1 mole of $N{a_2}{S_2}{O_3}$ loses 10 moles of ${e^ - }$. Determine final oxidation number of S.

Answer 1

Hint: As we have given compound $N{a_2}{S_2}{O_3}$. Now we will find the oxidation number of sulfur in the given compound. After finding the oxidation number we will calculate the as per moles of compound, how many moles of sulfur is required. After this we will calculate per sulfur number of electrons loss.
That number of loss in electrons will be equal to increases in the oxidation number.

Complete answer:
Let oxidation of S in the $N{a_2}{S_2}{O_3}$ is $x$
As we all know that normally oxidation number of Na is +1 and oxidation number of the O is -2
So after balancing the positive and negative charges:
We get,
$
  2 + 2x - 6 = 0 \\
  2x = 4 \\
  x = 2
 $
Now we have calculated that oxidation number of sulfur(S) in $N{a_2}{S_2}{O_3}$ is +2
$\therefore $ We all have seen that for one mole of $N{a_2}{S_2}{O_3}$ compound we should have 2 moles of sulfur(S)
And we have given that from one mole of $N{a_2}{S_2}{O_3}$ compound loses 10 moles of electrons.
So we have 2 moles of sulfur(S) and lose 10 moles of electrons.
Hence, by simple mathematics:
We can say that 1 mole of sulfur will lose 5 moles of electrons.
But we all know that losing electrons from an atom is known as oxidation.
So, if one mole of sulfur(s) loses 5 moles of electrons to form a new compound.
Hence, new compound oxidation of sulfur(s) will increase by 5 units.
So if one moles of $N{a_2}{S_2}{O_3}$ compound loses 10 moles of electrons then finally oxidation of sulfur(S) will be 7(because 5+2, two is initial oxidation number and 5 is after the loss of 5 moles of electrons from that one mole of $N{a_2}{S_2}{O_3}$)

So the final oxidation of $S$ is +7

Note:
Oxidation: oxidation is defined as the losing of electrons from an atom. Or increasing the oxidation number of an atom in the compound is known as the oxidation occurs for that atom.
Reduction: gaining of the electron from an atom is known as the reduction of that atom or decreasing the oxidation number of an atom is known as reduction of that atom.