What is the oxidation number for sulfur in sulfuric acid?

Answer 1

To figure this out, we first need to figure out the charge on S2O4.

We know that Na has a +1 charge, so the Na2 portion of the salt will have an overall +2 charge (2 x Na).

Provided that Na2S2O4 is a neutral molecule, we know that the charge of S2O4 must balance Na2, which has a +2 charge. Therefore, the charge on S2O4 is -2.

If the charge on S2O4 is -2, that must mean that there is an excess of 2 electrons in the molecule. We know that Oxygen usually has a charge of -2, and given that there are 4 Oxygen atoms in S2O4, the charge contributed by Oxygen is 4 x (-2) = -8. Since the charge of S2O4 is -2, we know that the total charge contributed by sulfur is +6.

+6 divided by 2 Sulfur atoms gives you the oxidation state (or number) of S: +3

Answer 2

H2SO4... so Hydrogen has a charge of +1

Oxygen has a charge of -2

2 hydrogens(+1), Sulfur is gonna be represented as X, and there are 4 oxygens(-2)... so the equation is 2(1)+x+4(-2)=0 >>>> 2+x-8=0

+8

2+x=8

-2

{ x=6 }

Therefore, oxidation #

of Sulfur is 6

Answer 3

H2SO4 has no overall charge, so the sum of the oxidation numbers must be equal to 0.

O has oxidation number -2. There are 4 of them: 4 * -2 = -8

H has oxidation number +1. There are 2 of them: 2 * 1 = 2

The current total is 2 - 8 = -6

Therefore, the oxidation number of sulfur must be +6

Answer 4

In a sulfuric acid molecule, a sulfur atom is attached to four oxygen atoms through covalent bonds. Oxygen is more electronegative than sulfur. Therefore sulfur has +6 oxidation number in sulfate ion.

Answer 5

+2