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3 years ago

Identify the oxidizing agent and the reducing agent for 4Li(s) + O_2 (g) to 2Li_2O(s).

Chemistry

2 answers:

ki77a [65]3 years ago

8 0

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

brainly.com/question/2973661

a reducing agent

brainly.com/question/2890416

element is reduced

brainly.com/question/4924694

Keywords: oxidation-reduction, an oxidizing agent, a reducing agent

wolverine [178]3 years ago

7 0

Answer: Li is the reducing agentg and O is the oxidizing agent.

Explanation:

1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.

2) The given reaction is:

4Li(s) + O₂ (g) → 2 Li₂O(s)

3) Determine the oxidation states of each atom:

Li(s): oxidation state = 0 (since it is alone)

O₂ (g): oxidation state = 0 (since it is alone)

Li in Li₂O (s) +1

O in Li₂O -2

That because 2× (+1) - 2 = 0.

4) Determine the changes:

Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.

O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.

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$n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

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Any gas at standard temperature and pressure (STP) has a volume of 22.4 liters per mole or 22.4 L/mol. We can create a proportion with this value.

$\frac {22.4 \ L \ Kr}{1 \ mol \ Kr}$

Multiply both sides of the equation by 6.8 moles of krypton.

$6.8 \ mol \ Kr *\frac {22.4 \ L \ Kr}{1 \ mol \ Kr}$

The units of moles of krypton will cancel.

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The denominator of 1 can be ignored, so this becomes a simple multiplication problem.

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If we round to the nearest whole number, the 3 in the tenths place tells us to leave the 2 in the ones place.

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