Question

Balance using oxidation numbers

CrO4^2- + SO3 ^2+ + H+ = CrO2^- + SO4^2- + H2O

Answer 1

To balance the given equation using oxidation numbers, we need to assign oxidation numbers to each element in both the reactants and the products.

Oxidation numbers:
- Oxygen typically has an oxidation number of -2, except when in peroxides where it is -1.
- Hydrogen typically has an oxidation number of +1.
- The sum of oxidation numbers in a compound must equal the charge of the compound.

Let's assign the oxidation numbers:

CrO4^2-:
Oxygen (O) = -2 (since there are four oxygen atoms, the total oxidation number is -8).
Chromium (Cr) = ?

SO3^2+:
Oxygen (O) = -2 (since there are three oxygen atoms, the total oxidation number is -6).
Sulfur (S) = ?

H+:
Hydrogen (H) = +1.

CrO2^-:
Oxygen (O) = -2 (since there are two oxygen atoms, the total oxidation number is -4).
Chromium (Cr) = ?

SO4^2-:
Oxygen (O) = -2 (since there are four oxygen atoms, the total oxidation number is -8).
Sulfur (S) = ?

H2O:
Hydrogen (H) = +1.
Oxygen (O) = -2.

Now, let's determine the oxidation numbers for the unknown elements:

For CrO4^2-:
-8 + 2 = -6 (since the overall charge is 2-)
Therefore, Chromium (Cr) has an oxidation number of +6.

For SO3^2+:
-6 + 2 = -4 (since the overall charge is 2+)
Therefore, Sulfur (S) has an oxidation number of +4.

For CrO2^-:
-4 + 2 = -2 (since the overall charge is 1-)
Therefore, Chromium (Cr) has an oxidation number of +4.

For SO4^2-:
-8 + 2 = -6 (since the overall charge is 2-)
Therefore, Sulfur (S) has an oxidation number of +6.

Now, we can write the balanced equation using oxidation numbers:

2CrO4^2- + 3SO3^2+ + 16H+ = 2CrO2^- + 3SO4^2- + 8H2O