Verified answer

CrO3 2- + SO3 2- => Cr(OH)3 + SO4 2-

Balance the Oxidation number

CrO4 2- = oxidation of Cr is

-2 = 4(-2) + x

x = +6

Cr(OH)3 = oxidation of Cr is

0 = 3(-1) +x

x = +3

SO3 2- = oxidation of S

-2 = 3(-2) + x

x = +4

SO4 2- = oxidation of S

-2 = 4(-2) + x

x = +6

Do the half equations of each are:

3e + CrO4 2- => Cr(OH)3

Cr goes from an oxidation state of +6 to +3 so you add 3e to the reactant side and Cr is being oxidized

and

SO3 2- => SO4 2- +2e

S goes from an oxidation state of +4 to +6 so you add 2e to the product side and S is being reduced.

That will leave you with...

2[3e + CrO4 2- => Cr(OH)3]

3[SO3 2- => SO4 2- +2e]

Multiply the whole equation so it'll have the same number of electron to cancel out.

6e + 2CrO4 2- => 2Cr(OH)3

3SO3 2- => 3SO4 2- + 6e

The whole equation will be

6e + 2CrO4 2- + 3SO3 2- => 2Cr(OH)3 + 3SO4 2- + 6e

6e is a spectator so you cross it off, leaving you with

2CrO4 2- + 3SO3 2- => 2Cr(OH)3 + 3SO4 2-

Now you check the charges on the reactants and products

-4 + -6 => 0 + -6

-10 => -6

Since the equation is basic, you'll add OH- to balance the overall charge... so

-10 => -6 + 4OH-

Now the reaction is:

2CrO4 2- + 3SO3 2- => 2Cr(OH)3 + 3SO4 2- + 4OH-

Now the charges are balanced with 10- on both sides.

Then, balance either the Hydrogen or the Oxygen.

There are 0 Hydrogen on the reactants and 10 Hydrogens on the product so add 5H2O to the reactants

5H2O + 2CrO4 2- + 3SO3 2- => 2Cr(OH)3 + 3SO4 2- + 4OH-

Lastly, check the overall charges and the hydrogen/oxygen to make sure all are balanced.

10 hydrogens => 10 hydrogens

10 - => 10-

Done. The equation is balanced.

5H2O + 2CrO4 2- + 3SO3 2- => 2Cr(OH)3 + 3SO4 2- + 4OH-