To determine the oxidation number of sulfur in \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) (copper(II) sulfate pentahydrate), we first need to identify the oxidation state of sulfur in the sulfate ion \( \text{SO}_4^{2-} \).
In the sulfate ion \( \text{SO}_4^{2-} \):
- The oxidation number of oxygen is generally \(-2\).
- There are four oxygen atoms, contributing a total of \(4 \times (-2) = -8\).
- The sulfate ion has an overall charge of \(-2\).
Let \( x \) represent the oxidation number of sulfur. The equation for the sulfate ion can be set up as follows:
\[ x + (-8) = -2 \]
Solving for \( x \):
\[ x - 8 = -2 \] \[ x = -2 + 8 \] \[ x = +6 \]
Therefore, the oxidation number of sulfur in \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \) is \( +6 \).
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