**Balancing the Equation using the Oxidation Number Method**

To balance the equation Cr2O7 + SO32- → Cr3+ + SO42-, we can use the oxidation number method. This method involves assigning oxidation numbers to each element in the equation and then adjusting the coefficients to ensure that the total increase in oxidation numbers equals the total decrease in oxidation numbers.

**Assigning Oxidation Numbers**

1. Start by assigning oxidation numbers to each element in the equation. The oxidation number is a positive or negative value that represents the charge an atom would have if electrons were completely transferred.

- Cr2O7: Since oxygen usually has an oxidation number of -2, and there are a total of 7 oxygen atoms in Cr2O7, the total oxidation number contributed by oxygen is -14. Therefore, the oxidation number of Cr must be +6 to balance the equation (2 x +6 + (-14) = 0).

- SO32-: The oxidation number of oxygen is -2, so the total oxidation number contributed by oxygen is -6. Since the overall charge on SO32- is -2, the sum of the oxidation numbers must equal -2. Therefore, the oxidation number of S must be +4 to balance the equation (x + (-6) = -2).

- Cr3+: The oxidation number of Cr3+ is +3, as indicated by the charge.

- SO42-: The oxidation number of oxygen is -2, so the total oxidation number contributed by oxygen is -8. Since the overall charge on SO42- is -2, the sum of the oxidation numbers must equal -2. Therefore, the oxidation number of S must be +6 to balance the equation (x + (-8) = -2).

**Balancing the Equation**

Now that we have assigned oxidation numbers, we can balance the equation by adjusting the coefficients in front of each compound.

1. Balance the Cr atoms first. Since there are 2 Cr atoms on the left side and 1 Cr atom on the right side, we need to multiply Cr3+ by 2 to balance the equation: Cr2O7 + SO32- → 2Cr3+ + SO42-.

2. Next, balance the S atoms. Since there is 1 S atom on both sides of the equation, the sulfur atoms are already balanced.

3. Balance the O atoms. On the left side, there are 7 O atoms in Cr2O7 and 3 O atoms in SO32-, for a total of 10 O atoms. On the right side, there are 4 O atoms in SO42-. To balance the O atoms, we need to add 6 H2O molecules to the left side of the equation: Cr2O7 + SO32- + 6H2O → 2Cr3+ + SO42- + 6H2O.

4. Finally, balance the H atoms. On the left side, there are 12 H atoms in the 6 H2O molecules. To balance the H atoms, we need to add 12 H+ ions to the right side of the equation: Cr2O7 + SO32- + 6H2O → 2Cr3+ + SO42- + 6H2O + 12H+.

The balanced equation is: Cr2O7 + SO