Question

A sample of crude sodium iodide was analysed by the following rection.

Answer 1

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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