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Chlorine \(\left(\mathrm{Cl}_{2}\right)\) is a strong germicide used to disinfect drinking water and to kill microbes in swimming pools. If the product is \(\mathrm{Cl}^{-}\), was the elemental chlorine oxidized or reduced?

Short Answer

Expert verified
The elemental chlorine was reduced.

Step by step solution

01

- Define Oxidation and Reduction

Oxidation is the loss of electrons, while reduction is the gain of electrons. To determine if chlorine was oxidized or reduced, we need to compare the oxidation states of chlorine in \(\text{Cl}_2\) and \(\text{Cl}^-\).
02

- Determine the Oxidation State of \(\text{Cl}_2\)

Elemental chlorine \(\text{Cl}_2\) is in its diatomic form, meaning each chlorine atom has an oxidation state of 0, as it is in its elemental form.
03

- Determine the Oxidation State of \(\text{Cl}^-\)

In the \(\text{Cl}^-\) ion, chlorine has an oxidation state of -1, since it has gained an electron to form the ion.
04

- Compare Oxidation States

Chlorine goes from an oxidation state of 0 in \(\text{Cl}_2\) to an oxidation state of -1 in \(\text{Cl}^-\). Since the oxidation state decreases, this indicates that chlorine has gained electrons.
05

- Identify the Process

Because chlorine \(\text{Cl}_2\) gains electrons to form \(\text{Cl}^-\), it has undergone reduction. Therefore, elemental chlorine was reduced in the process.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation States
Understanding oxidation states is essential in chemistry, as they help us track electron movement in reactions. An oxidation state is a number that represents the total number of electrons an atom gains, loses, or shares when forming chemical bonds.

For example, in elemental chlorine \( \text{Cl}_{2} \), each chlorine atom has an oxidation state of 0 because they share their electrons equally. However, in the chloride ion \( \text{Cl}^{-} \), chlorine has an oxidation state of -1 because it has gained an electron.
Electron Transfer
Electron transfer is a crucial concept in oxidation and reduction reactions (redox reactions). It refers to the movement of electrons from one atom or molecule to another.

In the context of chlorine disinfection, elemental chlorine (\( \text{Cl}_{2} \)) undergoes a redox reaction when it interacts with microbes. Each chlorine atom in \( \text{Cl}_{2} \) has an oxidation state of 0, but when it gains an electron to become \( \text{Cl}^{-} \), its oxidation state changes to -1. This gain of an electron is referred to as reduction.

Oxidation is the opposite process, where an atom loses electrons and increases its oxidation state.
Chlorine Disinfection
Chlorine disinfection is a widely used method for purifying water. When chlorine gas \( \text{Cl}_{2} \) is added to water, it reacts to form hypochlorous acid (\( \text{HOCl} \)) and hydrochloric acid (\( \text{HCl} \)). These substances are effective in killing bacteria, viruses, and other microbes.

The chemistry behind this disinfection process involves redox reactions. Elemental chlorine (\( \text{Cl}_{2} \)) undergoes reduction when it gains electrons to form chloride ions (\( \text{Cl}^{-} \)). This electron gain reduces the chlorine and helps in the disinfection process.

The effectiveness of chlorine as a disinfectant depends on various factors, including the concentration of chlorine, the presence of organic matter, and the pH of the water.

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Most popular questions from this chapter

Calculate the molar mass for each of the following compounds: a. \(\mathrm{FeSO}_{4}\) (iron supplement) b. \(\mathrm{Al}_{2} \mathrm{O}_{3}\) (absorbent and abrasive) c. \(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{NO}_{3} \mathrm{~S}\) (saccharin) d. \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) (rubbing alcohol) e. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) (baking powder) f. \(\mathrm{Zn}\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)_{2}\) (zinc dietary supplement)

When linoleic acid, an unsaturated fatty acid, reacts with hydrogen, it forms a saturated fatty acid. Is linoleic acid oxidized or reduced in the hydrogenation reaction? $$ \mathrm{C}_{18} \mathrm{H}_{32} \mathrm{O}_{2}+2 \mathrm{H}_{2} \longrightarrow \mathrm{C}_{18} \mathrm{H}_{36} \mathrm{O}_{2} $$

Write all the mole-mole factors for each of the following equations: a. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) b. \(4 \mathrm{P}(s)+5 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{P}_{2} \mathrm{O}_{5}(s)\)

When lead(II) sulfide ore burns in oxygen, the products are solid lead(II) oxide and sulfur dioxide gas. a. Write the balanced equation for the reaction. b. How many grams of oxygen are required to react with \(0.125\) mole of lead(II) sulfide? c. How many grams of sulfur dioxide can be produced when \(65.0 \mathrm{~g}\) of lead(II) sulfide reacts? d. How many grams of lead(II) sulfide are used to produce \(128 \mathrm{~g}\) of lead(II) oxide?

In each of the following reactions, identify the reactant that is oxidized and the reactant that is reduced: a. \(2 \mathrm{Li}(s)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{LiF}(s)\) b. \(\mathrm{Cl}_{2}(g)+2 \mathrm{KI}(a q) \longrightarrow 2 \mathrm{KCl}(a q)+\mathrm{I}_{2}(s)\) c. \(2 \mathrm{Al}(s)+3 \mathrm{Sn}^{2+}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+3 \operatorname{Sn}(s)\) d. \(\mathrm{Fe}(s)+\mathrm{CuSO}_{4}(a q) \longrightarrow \mathrm{FeSO}_{4}(a q)+\mathrm{Cu}(s)\)

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