Answer:
Part A: (1, 1, 4, 1, 1, 1)
Part B: (2, 6, 4, 2, 3, 8)
Explanation:
Redox reactions can be balanced using the half-reaction method. It has the following steps:
- We write both half-reactions (reduction and oxidation)
- We balance the masses using H⁺ and H₂O in acidic media or OH⁻ and H₂O in basic media.
- We add electrons to balance electrically the half-reaction
- We multiply the half-reaction by numbers to make sure the number of electrons gained and lost are the same.
- We add both half-reactions and take the numbers to the general equation.
<em>Acidic solution</em>
SO₄²⁻(aq) + Sn²⁺(aq) + X ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + Y
1.
Reduction: SO₄²⁻ ⇒ SO₃²⁻
Oxidation: Sn²⁺ ⇒ Sn⁴⁺
2.
2 H⁺ + SO₄²⁻ ⇒ SO₃²⁻ + H₂O
Sn²⁺ ⇒ Sn⁴⁺
3.
2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O
Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻
4.
1 x [2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O]
1 x [Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻]
5.
2 H⁺ + SO₄²⁻ + 2 e⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺ + 2 e⁻
2 H⁺ + SO₄²⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺
Taking this to the general equation:
SO₄²⁻(aq) + Sn²⁺(aq) + 2 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)
Since H⁺ are spectator ions, they are not balanced automatically through this method and we have to balance them manually. In this case, we need to add 2 more H⁺ to the left.
SO₄²⁻(aq) + Sn²⁺(aq) + 4 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)
<em>Basic solution</em>
MnO₄⁻(aq) + F⁻(aq) + X ⇄ MnO₂(s) + F₂(aq) + Y
1.
Reduction: MnO₄⁻ ⇒ MnO₂
Oxidation: F⁻ ⇒ F₂
2.
2 H₂O + MnO₄⁻ ⇒ MnO₂ + 4 OH⁻
2 F⁻ ⇒ F₂
3.
2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻
2 F⁻ ⇒ F₂ + 2 e⁻
4.
2 × (2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻)
3 × (2 F⁻ ⇒ F₂ + 2 e⁻)
5.
4 H₂O + 2 MnO₄⁻ + 6 e⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂ + 6 e⁻
4 H₂O + 2 MnO₄⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂
Taking this to the general equation:
2 MnO₄⁻(aq) + 6 F⁻(aq) + 4 H₂O ⇄ 2 MnO₂(s) + 3 F₂(aq) + 8 OH⁻
This equation is balanced.
