Question

What is the oxidation half-reaction, reduction half-reaction and balanced equation?H2SO4(aq) + KI(s) rightwards arrow K2SO4(aq) + I2(g) + SO2(g) + H2O(l)

Answer 1

Step-by-step explanation:

Given the reaction:

__ H₂SO₄ (aq) + __ KI (s) ----> __ K₂SO4 (aq) + __ I₂ (g) + __ SO₂ (g) + __ H₂O (l)

First we have to balance it.

__H₂SO₄ (aq) + __ KI (s) ----> __ K₂SO₄ (aq) + __ I₂ (g) + __ SO₂ (g) + __ H₂O (l)

H: 2 H: 2

I: 1 I: 2

K: 1 K: 2

S: 1 S: 2

O: 4 O: 7

We have two atoms of H on both sides, but 1 atom of I and K on the left and 2 atoms of K and I on the right. If we change the coefficient for KI on the left we will balance both elements.

__H₂SO₄ (aq) + 2 KI (s) ----> __ K₂SO₄ (aq) + __ I₂ (g) + __ SO₂ (g) + __ H₂O (l)

H: 2 H: 2

I: 2 I: 2

K: 2 K: 2

S: 1 S: 2

O: 4 O: 7

Now we have two atoms of S on the right and just 1 atom of S on the left. We can change the coefficient for H₂SO₄ and write a 2 there.

2 H₂SO₄ (aq) + 2 KI (s) ----> __ K₂SO₄ (aq) + __ I₂ (g) + __ SO₂ (g) + __ H₂O (l)

H: 4 H: 2

I: 2 I: 2

K: 2 K: 2

S: 2 S: 2

O: 8 O: 7

We balanced the S but unbalanced the H. We have to change the coefficient for H₂O on the right and write a 2 there.

2 H₂SO₄ (aq) + 2 KI (s) ----> __ K₂SO₄ (aq) + __ I₂ (g) + __ SO₂ (g) + 2 H₂O (l)

H: 4 H: 4

I: 2 I: 2

K: 2 K: 2

S: 2 S: 2

O: 8 O: 8

The equation is balanced now. The balanced equation is:

2 H₂SO₄ (aq) + 2 KI (s) ----> K₂SO₄ (aq) + I₂ (g) + SO₂ (g) + 2 H₂O (l)

To find both half reactions we can split these compounds into its ions.

4 H⁺ + 2 SO₄²⁻ + 2 K⁺ + 2 I⁻ ---> 2 K⁺ + SO₄²⁻ + I₂ + SO₂ + 2 H₂O

Let's see which are the species that are changing their oxidations states.

Iodide goes from -1 in I⁻ to 0 in I₂. Its oxidation state is increasing, so it is being oxidized.

Sulfur goes from +6 in SO₄²⁻ to +4 in SO₂. Its oxidation state is decreasing so it is being reduced.

2 I⁻ ---> I₂ oxidation half-reaction

SO₄²⁻ ---> SO₂ reduction half-reaction

Finally we have to balance these equation. To balance the iodine equation we only have to add electrons. To balance the sulfur half-reaction (since it is taking place in an acidic medium) we have to add molecules of water to balance the oxygen atoms, H+ to balance the hydrogen atoms and finally electrons to balance the charges.

2 I⁻ ---> I₂ + 2 e⁻ oxidation half-reaction

SO₄²⁻ ---> SO₂

SO₄²⁻ ---> SO₂ + 2 H₂O

SO₄²⁻ + 4 H⁺ ---> SO₂ + 2 H₂O

SO₄²⁻ + 4 H⁺ + 2 e⁻ ---> SO₂ + 2 H₂O reduction half-reaction

Answer:

2 H₂SO₄ (aq) + 2 KI (s) ----> K₂SO₄ (aq) + I₂ (g) + SO₂ (g) + 2 H₂O (l)

2 I⁻ ---> I₂ + 2 e⁻ oxidation half-reaction

SO₄²⁻ + 4 H⁺ + 2 e⁻ ---> SO₂ + 2 H₂O reduction half-reaction