What is the change in electrons for sulfur in the following reaction? so 2 ( aq ) + cr 2 o 7 2- ( aq ) -> so 4 2- ( aq ) + cr 3+ ( aq )

Loss of 2 electrons

Explanation:

The reaction given is a chemical reaction. Chemical reactions involves loss and gain of electrons.

Note that the reaction is a redox reaction.

For this reaction, our concern is the sulfur. Let us look at the oxidation numbers of sulfur on the reactant and product side.

Keep in mind the following rules:

1. Elements in uncombined states whose atoms combine to form molecules have an oxidation number of 0.

2. The charge on simple ions is their oxidation number.

3. Algebraic sum of oxidation numbers of atoms in a neutral compound is zero. For radicals, the charge on the ion is the oxidation number.

Now we find the oxidation number of S in SO₂:

S + 2(-2) = 0

S - 4 =0

S = +4

Oxidation number of S in SO₄²⁻

S + 4(-2) = - 2

S - 8 = - 2

S = - 2 + 8

S = +6

Sulfur changed from a state of +4 to +6, this is a loss of 2 electrons.

they it depends on if one believes about radioactivity, they make a geiger counter beep but maybe it was going to beep anyway electronics do wierd things all the time or it was programmed to beep near certain metals.