A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.
8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻
Let's consider the following unbalanced redox reaction.
I⁻ + SO₄²⁻ → I₂ + H₂S
- The oxidation number of I goes from -1 (I⁻) to 0 (I₂) so it is oxidized.
- The oxidation number of S goes from +6 (SO₄²⁻) to -2 (H₂S) so it is reduced.
The corresponding half-reactions are:
I⁻ → I₂
SO₄²⁻ → H₂S
We will perform the mass balance adding OH⁻ and H₂O where appropriate.
2 I⁻ → I₂
6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻
Then, we will perform the charge balance adding electrons where appropriate.
2 I⁻ → I₂ + 2 e⁻
8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻
Finally, we will multiply the first half-reaction by 4 and the second by 1, and add them.
4 × (2 I⁻ → I₂ + 2 e⁻)
1 × (8 e⁻ + 6 H₂O + SO₄²⁻ → H₂S + 10 OH⁻)
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8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻
A sample of crude sodium iodide was analyzed by the following balanced reaction. The oxidation number of S in SO₄²⁻ is +6.
8 I⁻ + 6 H₂O + SO₄²⁻ → 4 I₂ + H₂S + 10 OH⁻
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