Identify the oxidizing agent and the reducing agent for 4Li(s) + O_2 (g) to 2Li_2O(s).
2 answers:
The Li element is the reducing agent and the O element is the oxidizing agent
<h3>Further explanation</h3>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 in oxidation number.
In the redox reaction, it is also known
Reducing agents are substances that experience oxidation
The oxidizing agent is a substance that is reduced
The formula for determining Oxidation Numbers in general:
- 1. Single element atomic oxidation number = 0. Examples of Ar, Mg, Cu, Fe, N₂, O₂, etc. = 0
Group IA (Li, Na, K, Rb, Cs, and Fr): +1
Group IIA (Be, Mg, Ca, Sr and Ba): +2
H in compound = +1, except metal hydride compounds (Hydrogen which binds to IA or IIA groups) oxidation number H= -1, for example, LiH, MgH₂, etc.
- 2. Oxidation number O in compound = -2, except OF2 = + 2 and in peroxide (Na₂O₂, BaO₂) = -1 and superoxide, for example KO₂ = -1/2.
- 3 The oxidation number in an uncharged compound = 0,
Total oxidation number in ion = ion charge, Example NO₃⁻ = -1
Redox reactions are reactions that are accompanied by changes in oxidation numbers, so what must be examined is whether there are elements that experience changes in oxidation numbers in the reaction
Let's look at the reaction
4Li (s) + O₂ (g) → 2 Li₂O (s)
Let see the change in the oxidation number of each element
Li on the left = 0 (single element)
O₂ on the left = 0 (single element)
For Li₂O = 0 then (oxidation number O = -2, see rule no 2 above), so
2. Li + O = 0
2. Li-2 = 0
2Li = 2
Li = + 1
Means that the element Li has increased oxidation number from 0 to +1 so that it experiences an oxidation reaction and acts as a reducing agent
While O2 has decreased the oxidation number from 0 to -2, so it has a reduction reaction and acts as an oxidizer
<h3>Learn more</h3>an oxidation-reduction reaction
a reducing agent
element is reduced
Keywords: oxidation-reduction, an oxidizing agent, a reducing agent
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<h2>Hello!</h2>
The answer is:
The percent yield of the reaction is 32.45%
<h2>Why?</h2>To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.
We are given that:
Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.
So, calculating we have:
Hence, we have that the percent yield of the reaction is 32.45%.
Have a nice day!