Question

Complete and balance the following redox equation. What is the coefficient of H2O when the equation is balanced using the set of smallest whole-number coefficients?

MnO₄⁻ + SO₃²⁻ → Mn²⁺ + SO₄²⁻ (acidic solution)

Answer 1

The coefficient of H2O when the equation MnO4⁻ + SO3²⁻ → Mn²⁺ + SO4²⁻ is balanced in acidic solution is 3.

Step-by-step explanation:

The redox reaction in question is between permanganate ion (MnO₄⁻) and sulfite ion (SO₃²⁻) in an acidic solution which forms manganese ion (Mn²⁺) and sulfate ion (SO₄²⁻). To balance this redox equation, we need to balance the changes in oxidation states of the substances involved and also balance the atoms and charges in the reaction.

First, the oxidation half-reaction: MnO₄⁻ → Mn²⁺

Mn is reduced from +7 to +2, a change of 5 electrons:

5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O

Next, the reduction half-reaction: SO₃²⁻ → SO₄²⁻

S is oxidized from +4 to +6, a change of 2 electrons:

SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻

Now, we need to balance the electron transfer by multiplying the oxidation half-reaction by 2 and the reduction half-reaction by 5 so that the number of electrons lost equals the number of electrons gained:

2(SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻)

5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O

Now add the half-reactions together and simplify:

2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ → 2Mn²⁺ + 5SO₄²⁻ + 3H₂O

Therefore, the coefficient of H₂O when the equation is balanced using the set of smallest whole-number coefficients is 3.