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Kamila [148]
3 years ago
11

Which statements are true regarding redox reactions?

Chemistry
1 answer:
Maslowich3 years ago
5 0

Answer : The correct statements are,

(a) A molecule that has gained H atoms is said to be reduced.

(b) Oxidizing agents accept electrons.

(d) If a molecule accepts electrons, it has been reduced.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases.

The loss of electrons is also known as oxidation.

The gain of oxygen is also known as oxidation.

The loss of hydrogen is also known as oxidation.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases.

The gain of electrons is also known as reduction.

The loss of oxygen is also known as reduction.

The gain of hydrogen is also known as reduction.

Reducing agent : It is defined as the agent which helps the other substance to reduce and itself gets oxidized. Thus, it will undergo oxidation reaction.

Reducing agents loose electrons.

Oxidizing agent : It is defined as the agent which helps the other substance to oxidize and itself gets reduced. Thus, it will undergo reduction reaction.

Oxidizing agents accept electrons.

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Substitute natural gas (SNG) is a gaseous mixture containing CH4(g) that can be used as a fuel. One reaction for the production
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ΔH° of the reaction is -747.54kJ

Explanation:

Based on gas law, it is possible to find the ΔH of a reaction using ΔH of half reactions.

Using the reactions:

<em>(1) </em>C(graphite) + 1/2O₂(g) → CO(g) ΔH° = -110.5 kJ

<em>(2) </em>CO(g) + 1/2O₂(g) → CO₂(g) ΔH° = -283.0 kJ

<em>(3) </em>H₂(g) + 1/2O₂(g) → H₂O(l) ΔH° = -285.8 kJ

<em>(4) </em>C(graphite) + 2H₂(g) → CH₄(g) ΔH° = -74.81 kJ

<em>(5) </em>CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH° = -890.3 kJ

The sum of 4×(4) + (5) gives:

4C(graphite) + 8H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -74.81 kJ ×4 - 890.3 kJ = -1189.54kJ

Now, this reaction - 4×(1) gives:

4CO(g) + 8H₂(g) → CO₂(g) + 2H₂O(l) + 3CH₄(g)

ΔH° = -1189.54kJ - 4×-110.5 = <em>-747.54kJ</em>

<em></em>

Thus <em>ΔH° of the reaction is -747.54kJ</em>

3 0
3 years ago
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