Question

Методом полуреакций подберите стехиометрические коэффициенты в приведенном уравнении окислительно-восстановительной реакции. Укажите, какие сопряженные окислительно-восстановительные пары принимают участие в указанном превращении. H2SO4(конц) + FeO -> Fe2(SO4)3 + SO2 ↑ + H2O

Answer 1

4H2SO4(конц) + 2FeO -> Fe2(SO4)3 + SO2 ↑ + 4H2O
SO₄²⁻ + 4H⁺ + 2e  = SO₂ + 2H₂O   1
FeO + 2H⁺ - 1e  =  Fe³⁺ + H₂O      2
SO₄²⁻ + 8H⁺ + 2FeO = SO₂ + Fe³⁺ + 4H₂O

SO₄²⁻/FeO  ⇒ SO₂/Fe³⁺

Answer 2

Balancing the Redox Reaction using Half-Reaction Method

To balance the given redox reaction using the half-reaction method, we need to separate the reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction.

Let's start by identifying the oxidation and reduction half-reactions:

Oxidation Half-Reaction: In the given reaction, iron (Fe) is being oxidized from a lower oxidation state (+2) to a higher oxidation state (+3). Therefore, the oxidation half-reaction can be written as follows:

FeO → Fe^3+

Reduction Half-Reaction: In the given reaction, sulfuric acid (H2SO4) is being reduced. The sulfur in sulfuric acid is going from an oxidation state of +6 to +4. Therefore, the reduction half-reaction can be written as follows:

H2SO4 → SO2 + H2O

Now, let's balance each half-reaction individually:

Balancing the Oxidation Half-Reaction: To balance the oxidation half-reaction, we need to balance the number of atoms on each side of the equation and the charges. Since there is only one iron (Fe) atom on each side, the equation is already balanced in terms of atoms. However, the charge is not balanced. To balance the charge, we need to add three electrons (e-) to the left side of the equation:

FeO → Fe^3+ + 3e-

Balancing the Reduction Half-Reaction: To balance the reduction half-reaction, we need to balance the number of atoms on each side of the equation and the charges. The number of sulfur (S) atoms is already balanced, but the number of hydrogen (H) and oxygen (O) atoms is not balanced. To balance the atoms, we need to add two water (H2O) molecules to the right side of the equation:

H2SO4 → SO2 + 2H2O

Now, let's balance the charges. The charge on the left side of the equation is +2 (from the hydrogen atoms), and the charge on the right side is 0 (since SO2 is neutral). To balance the charges, we need to add two electrons (2e-) to the left side of the equation:

H2SO4 + 2e- → SO2 + 2H2O

Balanced Redox Reaction:

Now that we have balanced the oxidation and reduction half-reactions, we can combine them to form the balanced redox reaction:

FeO + H2SO4 → Fe^3+ + 3e- + SO2 + 2H2O

To simplify the equation, we can cancel out the electrons on both sides:

FeO + H2SO4 → Fe^3+ + SO2 + 2H2O

Conjugate Redox Pairs:

In the given redox reaction, the following conjugate redox pairs are involved:

1. FeO/Fe^3+: Iron (II) oxide (FeO) is oxidized to iron (III) ion (Fe^3+). 2. H2SO4/SO2: Sulfuric acid (H2SO4) is reduced to sulfur dioxide (SO2).

These pairs involve the transfer of electrons, where one species is oxidized (loses electrons) and the other is reduced (gains electrons).

I hope this helps! Let me know if you have any further questions.