Answer to Question #270868 in Chemistry for kweni

Question

1. Balance in acidic medium. Show step by step process.

Iron (solid) reacts with hydrochloric acid (liquid) to form aqueous iron(II) chloride and gaseous hydrogen.

2. Balance in acidic medium. Show step by step process.

𝑁𝑂2− + 𝐴𝑙(𝑠) → 𝑁𝐻4+ (𝑎𝑞)

(𝑎𝑞)

+ 𝐴𝑙𝑂2−

(𝑎𝑞)

3. Balance the following reaction in basic medium. Show step by step process.

𝑆 𝑂2− + 𝑁𝑖(𝑂𝐻) → 𝑆𝑂2− + 𝑁𝑖𝑂

2 8(𝑎𝑞) 2(𝑠) 4(𝑎𝑞) 2(𝑠)

Answer 1

Solution:

(1):

Unbalanced chemical equation (acidic medium):

Fe(s) + HCl(aq) → FeCl₂(aq) + H₂(g)

Step 1: Separate the half-reactions:

Fe(s) → Fe²⁺(aq)

H⁺(aq) → H₂(g)

Step 2: Balance elements:

Fe(s) → Fe²⁺(aq)

2H⁺(aq) → H₂(g)

Step 3: Balance the charge of each equation with electrons.

The iron reaction has no charge on the left side and (2+) on the right side. To balance, add 2 electrons (each with a charge of -1) to the right side:

Fe(s) → Fe²⁺(aq) + 2e⁻

For the other reaction, there is (2×1+) = (2+) charge on the left and no charge on the right. So add 2 electrons to the left side:

2H⁺(aq) + 2e⁻ → H₂(g)

Step 4: Scale the reactions so that they have an equal amount of electrons.

In this case, it is already done.

Step 5: Add the reactions and cancel the electrons:

Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g)

Balanced chemical equation:

Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)

(2):

Unbalanced chemical equation (acidic medium):

NO₂⁻(aq) + Al(s) → NH₄⁺(aq) + AlO₂⁻(aq)

Step 1: Separate the half-reactions:

Al(s) → AlO₂⁻(aq)

NO₂⁻(aq) → NH₄⁺(aq)

Step 2: Balance elements other than O and H.

In this case, it is already done.

Step 3: Add H₂O to balance oxygen.

The aluminium reaction needs to be balanced by adding 2H₂O molecules:

Al(s) + 2H₂O(l) → AlO₂⁻(aq)

The other reaction also needs to be balanced by adding 2H₂O molecules:

NO₂⁻(aq) → NH₄⁺(aq) + 2H₂O(l)

Step 4: Balance hydrogen by adding protons (H⁺).

4H⁺ need to be added to the right side of the aluminium reaction to balance the 4 hydrogens (2 per water molecule × 2H₂O molecules):

Al(s) + 2H₂O(l) → AlO₂⁻(aq) + 4H⁺(aq)

4H⁺ need to be added to the left side of the other reaction:

NO₂⁻(aq) + 4H⁺(aq) → NH₄⁺(aq) + 2H₂O(l)

Step 5: Balance the charge of each equation with electrons.

The aluminium reaction has no charge on the left side and (1−) + (4+) = (3+) on the right side. To balance, add 3 electrons (each with a charge of -1) to the right side:

Al(s) + 2H₂O(l) → AlO₂⁻(aq) + 4H⁺(aq) + 3e⁻

For the other reaction, there is (1-) charge on the left and (1+) on the right. So add 2 electrons to the left side:

NO₂⁻(aq) + 4H⁺(aq) + 2e⁻ → NH₄⁺(aq) + 2H₂O(l)

Step 6: Scale the reactions so that they have an equal amount of electrons.

Al(s) + 2H₂O(l) → AlO₂⁻(aq) + 4H⁺(aq) + 3e⁻. It should be multiplied by 2. This gives:

2Al(s) + 4H₂O(l) → 2AlO₂⁻(aq) + 8H⁺(aq) + 6e⁻

NO₂⁻(aq) + 4H⁺(aq) + 2e⁻ → NH₄⁺(aq) + 2H₂O(l). It should be multiplied by 3. This gives:

3NO₂⁻(aq) + 12H⁺(aq) + 6e⁻ → 3NH₄⁺(aq) + 6H₂O(l)

Step 7: Add the reactions and cancel out common terms/electrons:

2Al(s) + 4H₂O(l) + 3NO₂⁻(aq) + 12H⁺(aq) + 6e⁻ → 3NH₄⁺(aq) + 6H₂O(l) + 2AlO₂⁻(aq) + 8H⁺(aq) + 6e⁻

2Al(s) + 3NO₂⁻(aq) + 4H⁺(aq) → 3NH₄⁺(aq) + 2H₂O(l) + 2AlO₂⁻(aq)

Balanced chemical equation:

2Al(s) + 3NO₂⁻(aq) + 4H⁺(aq) → 3NH₄⁺(aq) + 2H₂O(l) + 2AlO₂⁻(aq)

(3):

Unbalanced chemical equation (basic medium):

S₂O₈²⁻(aq) + Ni(OH)₂(s) → SO₄²⁻(aq) + NiO₂(s)

Step 1: Separate the half-reactions:

Ni(OH)₂(s) → NiO₂(s)

S₂O₈²⁻(aq) → SO₄²⁻(aq)

Step 2: Balance elements other than O and H.

Ni(OH)₂(s) → NiO₂(s)

S₂O₈²⁻(aq) → 2SO₄²⁻(aq)

Step 3: Balance hydrogen by adding protons (H⁺).

2H⁺ need to be added to the right side of the nickel reaction to balance the 2 hydrogens:

Ni(OH)₂(s) → NiO₂(s) + 2H⁺(aq)

S₂O₈²⁻(aq) → 2SO₄²⁻(aq)

Step 4: Balance the charge of each equation with electrons.

The nickel reaction has no charge on the left side and (2+) on the right side. To balance, add 2 electrons (each with a charge of -1) to the right side:

Ni(OH)₂(s) → NiO₂(s) + 2H⁺(aq) + 2e⁻

For the other reaction, there is (2-) charge on the left and (2×2−) = (4−) on the right. So add 2 electrons to the left side:

S₂O₈²⁻(aq) + 2e⁻ → 2SO₄²⁻(aq)

Step 5: Scale the reactions so that they have an equal amount of electrons.

In this case, it is already done.

Step 6: Add the reactions and cancel out common terms/electrons:

Ni(OH)₂(s) + S₂O₈²⁻(aq) + 2e⁻ → 2SO₄²⁻(aq) + NiO₂(s) + 2H⁺(aq) + 2e⁻

Ni(OH)₂(s) + S₂O₈²⁻(aq) → 2SO₄²⁻(aq) + NiO₂(s) + 2H⁺(aq)

Step 7: Add OH^- to balance H⁺.

There are 2 net protons (H⁺) in this equation, so add 2OH⁻ ions to each side.

Ni(OH)₂(s) + S₂O₈²⁻(aq) + 2OH⁻(aq) → 2SO₄²⁻(aq) + NiO₂(s) + 2H⁺(aq) + 2OH⁻(aq)

Step 8: Combine OH⁻ ions and H⁺ ions that are present on the same side to form water and cancel common terms:

Ni(OH)₂(s) + S₂O₈²⁻(aq) + 2OH⁻(aq) → 2SO₄²⁻(aq) + NiO₂(s) + 2H₂O(l)

Balanced chemical equation:

Ni(OH)₂(s) + S₂O₈²⁻(aq) + 2OH⁻(aq) → 2SO₄²⁻(aq) + NiO₂(s) + 2H₂O(l)