Question

In an experiment, 50.0 cm3 of a 0.10 mol dm–3 solution of a metallic salt reacted exactly with
25.0 cm3 of 0.10 mol dm–3 aqueous sodium sulphite.
The half-equation for oxidation of sulphite ion is shown below.
SO^-2(_3) (aq) + H2O(I) → SO^2_4(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

Answer 1

Can you calculate moles first?

Answer 2

But before that, convert the units.

Answer 3

No. of moles will be equal to the product of 50.0 cm^3 and 0.10 mol dm^{-3} . Notice that the units are not same, first make the units same and then find the product.

Answer 4

i didnt get your solution :/

Answer 5

As thoughts said above first make the units same.Then find out the number of moles that actually react here.

Answer 6

First, find the no. of moles of the metallic salt and sodium sulphite by the formula
n=c×V

where n= no. of moles
c= concentration in mol/dm^3
V= volume in dm^3

No. of moles of metallic salt= 0.005
No. of moles of sodium sulphite= 0.0025

So, you see that 0.005 moles of metallic salt react with 0.0025 moles of sodium sulphite. You have to think how many moles of sodium sulphite will react with 1 mole of metallic salt?
0.005→0.0025

1→x

x=0.5

Now from the half- equation given, you can see that 1 mole of SO3^-2 gives 2 electrons.
Half a mole will therefore give 1 electron.

This 1 electron will be accepted by the metal in the salt as this is a redox reaction.
So
+3−1=+2