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Is it possible to have a reaction in which oxidation occurs and reduction does not? Explain.

Short Answer

Expert verified
No, a reaction with oxidation must also involve reduction due to electron transfer.

Step by step solution

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01

Understand Oxidation

Oxidation is a chemical process in which a substance loses electrons. This results in an increase in the oxidation state of the substance. For example, when a metal such as iron oxidizes, it loses electrons to form iron ions.
02

Understand Reduction

Reduction is the chemical process where a substance gains electrons, resulting in a decrease of its oxidation state. In essence, for a substance to be reduced, it must take the electrons that another substance loses.
03

Explore Oxidation-Reduction Relationship

In any chemical reaction involving redox (reduction-oxidation) processes, one reactant loses electrons (oxidation) while another gains those same electrons (reduction). This means that oxidation and reduction always happen simultaneously in a reaction, which is why they are collectively referred to as redox reactions.
04

Conclude Possibility of Independent Oxidation

Since oxidation involves losing electrons, those electrons must be gained by another reactant. Therefore, reduction is a necessary and simultaneous part of any oxidation reaction. Thus, it is impossible to have a chemical reaction where oxidation occurs without reduction also taking place.

Key Concepts

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

Oxidation
In the world of chemistry, oxidation refers to the process where a substance loses electrons. This concept might sound complex at first, but it's essentially what happens when a metal turns rusty or when an apple browns. When a substance undergoes oxidation, its oxidation state, which measures the degree of oxidation of an atom in a compound, increases.
This rise indicates that the substance has lost electrons. For example, consider a metal like iron. When iron oxidizes, it loses electrons and transforms into iron ions, contributing to that familiar orange rust.
Understanding oxidation is crucial because it sets the stage for complementary processes in reactions, establishing a balance of electron exchange in the universe of chemistry.
Reduction
Reduction is the process in which a substance gains electrons. Think of it as the counterpart or the other half of oxidation. When a substance undergoes reduction, its oxidation state decreases because the gain of electrons lowers the positive charge of the ion or atom.
In simpler terms, the substance being reduced literally becomes 'reduced' in charge by gaining those electrons. A practical example can be seen in the reaction of hydrogen with another element like oxygen, where hydrogen atoms gain electrons to form water, showcasing reduction.
This process is necessary as it balances the loss of electrons during oxidation. Without reduction, the electrons lost in oxidation would have nowhere to go, disrupting electron balance in the reaction.
Electron Transfer
Electron transfer is the heart of redox reactions. It involves the movement of electrons from one reactant, undergoing oxidation, to another, undergoing reduction. Imagine electrons as tiny energy balls being passed from one atom to another.
This transfer is what keeps the principles of conservation in chemical reactions intact, as electrons are neither created nor destroyed, only moved. The simultaneous nature of oxidation and reduction, achieved through electron transfer, means that every electron "lost" by one reactant is "gained" by another.
Without this essential transfer, redox reactions can't occur, which confirms the impossibility of having an oxidation without a simultaneous reduction.
Oxidation State
Oxidation state is a number assigned to an element in a chemical compound. It represents the number of electrons lost or gained by an atom of that element in the compound. When we talk about changes in oxidation, it's reflected in these oxidation states.
Each element in the periodic table has specific tendencies whether to gain or lose electrons, determining its typical oxidation state. For example, hydrogen usually has an oxidation state of +1. This shows a consistency in how it normally donates its single electron in chemical reactions.
Tracking oxidation states is like keeping score in chemical reactions, helping chemists determine which species are oxidized and reduced. It provides an organized way to predict and understand chemical behavior in reactions.

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Most popular questions from this chapter

Calculate the molarity of each of the following solutions: (a) \(6.57 \mathrm{~g}\) of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in \(1.50 \times 10^{2} \mathrm{~mL}\) of solution, (b) \(10.4 \mathrm{~g}\) of calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) in \(2.20 \times 10^{2} \mathrm{~mL}\) of solution, \((\mathrm{c}) 7.82 \mathrm{~g}\) of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) in \(85.2 \mathrm{~mL}\) of benzene solution.

Draw molecular models to represent the following acidbase reactions: (a) \(\mathrm{OH}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{NH}_{4}^{+}+\mathrm{NH}_{2}^{-} \longrightarrow 2 \mathrm{NH}_{3}\) Identify the Bronsted acid and base in each case.

A \(325-\mathrm{mL}\) sample of solution contains \(25.3 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\) (a) Calculate the molar concentration of \(\mathrm{Cl}^{-}\) in this solution. (b) How many grams of \(\mathrm{Cl}^{-}\) are in \(0.100 \mathrm{~L}\) of this solution?

For each of the following pairs of combinations, indicate which one will produce the greater mass of solid product: a) \(105.5 \mathrm{~mL} 1.508 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(250.0 \mathrm{~mL}\) \(1.2075 \mathrm{M} \mathrm{KCl}\) or \(138.5 \mathrm{~mL} 1.469 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{~mL} 2.115 \mathrm{M} \mathrm{KCl}\) b) \(32.25 \mathrm{~mL} 0.9475 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(92.75 \mathrm{~mL} 0.7750 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) or \(52.50 \mathrm{~mL} 0.6810 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(39.50 \mathrm{~mL} 1.555 \mathrm{M}\) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) c) \(29.75 \mathrm{~mL} 1.575 \mathrm{M} \mathrm{AgNO}_{3}\) and \(25.00 \mathrm{~mL} 2.010 \mathrm{M}\) \(\mathrm{BaCl}_{2}\) or \(52.80 \mathrm{~mL} 2.010 \mathrm{M} \mathrm{AgNO}_{3}\) and \(73.50 \mathrm{~mL} 0.7500 \mathrm{M}\) \(\mathrm{BaCl}_{2}\)

You are given two colorless solutions, one containing \(\mathrm{NaCl}\) and the other sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) .\) Suggest a chemical and a physical test that would allow you to distinguish between these two solutions.

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