What happens to a reducing agent during a redox reaction?
2 answers:
The reducing agent is oxidized during a redox reaction
The oxidation-reduction reaction or abbreviated as Redox is a chemical reaction in which there is a change in oxidation number
3 basic theories explain this Redox concept:
- 1. Binding/release of oxygen
The oxidation reaction is the binding of a substance with oxygen. (O₂)
For example:
2SO₂ + O₂ ----> 2SO₃
The reduction reaction is the release of oxygen from a substance.
For example:
2CuO → 2Cu + O₂
- 2. Electron release / binding reaction
Oxidation is an electron release event
Example:
2F ---> 2Fe³⁺ + 6e⁻
The reduction is an electron capture event
Example:
3O₂ + 6e⁻ ---> 3O²⁻
- 3. The reaction of addition/reduction of oxidation number
Oxidation is an increase/increase in oxidation number, while reduction is a decrease/decrease in oxidation number.
In this reaction, it is also known
Reducing agents are substances that experience oxidation
The oxidizer is a substance that is reduced
Let's look at the answer choices
- The reducing agent is reduced.
This statement is wrong because the reducing agent undergoes oxidation
- The reducing agent is oxidized.
This statement is true
- The reducing agent remains unchanged.
The statement is wrong because the oxidation number of the oxidized compound will increase
- The reducing agent gains electrons.
The statement is wrong because the reducing agent releases electrons
So the correct statement is B The reducing agent is oxidized
chemical reactions are an oxidation-reduction reaction
the reduction process
an element's oxidation
an element's oxidation number increase
Keywords: oxidation, reduction, The oxidation number
You might be interested in
The rate law equation for Ozone reaction
r=k[O][O₂]
<h3>Further explanation</h3>Given
Reaction of Ozone :.
O(g) + O2(g) → O3(g)
Required
the rate law equation
Solution
The rate law is a chemical equation that shows the relationship between reaction rate and the concentration / pressure of the reactants
For reaction
aA + bB ⇒ C + D
The rate law can be formulated:
where
r = reaction rate, M / s
k = constant, mol¹⁻⁽ᵃ⁺ᵇ⁾. L⁽ᵃ⁺ᵇ⁾⁻¹. S⁻¹
a = reaction order to A
b = reaction order to B
[A] = [B] = concentration of substances
So for Ozone reaction, the rate law (first orde for both O and O₂) :