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Phantasy [73]
2 years ago
6

Why must every redox reaction involve an oxidizing agent and a reducing agent?

Chemistry
1 answer:
vagabundo [1.1K]2 years ago
4 0

A reduced element (which gains electrons) and an oxidized element are required for redox reactions (gives electrons). It is not a redox reaction if we lack both of them (an element can not receive electrons if no element gives electrons and vice versa).

A reduced half and an oxidized half, which always occur together, make up redox processes. While the oxidized half experiences electron loss and an increase in oxidation number, the reduced half obtains electrons and the oxidation number declines. The mnemonic devices OIL RIG, which stand for "oxidation is loss" and "reduction is gain," are simple ways to memorize this. In a redox process, the total number of electrons stays constant. In the reduction half reaction, another species absorbs those that were released in the oxidation half reaction.

In a redox reaction, two species exchange electrons, and they are given unique names:

  • The ion or molecule that accepts electrons is called the oxidizing agent - by accepting electrons it oxidizes other species.
  • The ion or molecule that donates electrons is called the reducing agent - by giving electrons it reduces the other species.

Hence, what is oxidized is the reducing agent and what is reduced is the oxidizing agent.

<h3>What is the purpose of oxidizing agents and reducing agents?</h3>

By reducing other compounds and shedding electrons, a reducing agent raises its oxidation state. An oxidizing agent gets electrons by oxidizing other compounds; as a result, its oxidation state lowers.

<h3>What is a redox reaction?</h3>

Oxidation-reduction (or "redox") reactions are chemical processes in which electrons are exchanged between two substances. An oxidation-reduction reaction is any chemical process in which a molecule, atom, or ion alters the number of electrons it has, hence increasing or decreasing its oxidation state.

Learn more about redox reaction:  brainly.com/question/13293425

#SPJ4

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What would be the freezing point of a solution that has a molality of 1.468 m which was prepared by dissolving biphenyl (C12H10)
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Answer:

Freezing point solution = 70.131 °C

Explanation:

Step 1: Data given

Molality = 1.468 molal

A solution is created by dissolving biphenyl (C12H10) into naphthalene

Biphenyl is a non-electrolyte

Freezing point of naphthalene = 80.26 °C

Step 2: Calculate the freezing point depression

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = TO BE DETERMINED

⇒with i = the van't Hoff factor of biphenyl = 1

⇒with Kf = the freezing point depression constant of naphthalene = 6.90 °C/m

⇒with m = the molality = 1.468 molal

ΔT = 1 * 6.90 °C/m * 1.468 °C

ΔT = 10.13 °C

Step 3: Calculate the freezing point of the solution

ΔT = 10.13 °C

Freezing point solution = freezing point naphthalene - 10.13 °C

Freezing point solution = 80.26 °C - 10.129 °C

Freezing point solution = 70.131 °C

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