Calculate the oxidation number of Sulphur in S₂O₈²⁻.

To determine the oxidation number of sulfur in the sulfate ion, S₂O₈²⁻, we can follow a systematic approach. The overall charge of the ion is -2, and we need to account for the oxidation states of both sulfur and oxygen to find the value for sulfur.

Understanding Oxidation States

Oxidation states are a way to keep track of electrons in chemical compounds. The oxidation number of an element in a compound can be thought of as the hypothetical charge that atom would have if all bonds were ionic. For oxygen, the common oxidation state is -2.

Step-by-Step Calculation

Let’s break down the calculation for S₂O₈²⁻:

• First, identify the oxidation state of oxygen. In most compounds, oxygen has an oxidation state of -2.
• In S₂O₈²⁻, there are 8 oxygen atoms. Therefore, the total contribution from oxygen is:
• 8 atoms × -2 = -16
• Let the oxidation state of sulfur be represented as x. Since there are 2 sulfur atoms, their total contribution will be:
• 2x
• Now, we can set up the equation based on the overall charge of the ion:
• 2x + (-16) = -2

Solving the Equation

Now, let’s solve for x:

• 2x - 16 = -2
• 2x = -2 + 16
• 2x = 14
• x = 14 / 2
• x = 7

Final Result

Thus, the oxidation number of sulfur in S₂O₈²⁻ is +7. This indicates that sulfur is in a high oxidation state, which is characteristic of its role in various oxidation-reduction reactions.

Real-World Implications

Understanding the oxidation state of sulfur is crucial in fields such as environmental science and chemistry, particularly when studying compounds like sulfuric acid or in processes like the sulfur cycle. It helps in predicting the behavior of sulfur in reactions and its interactions with other elements.