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Chapter 4: Problem 6

Why is it that when something gains electrons, it is said to be reduced? What is being reduced?

Short Answer

Expert verified
When something gains electrons, it is said to be reduced because its oxidation number decreases, signifying a reduction in its charge. In a redox reaction, one species undergoes reduction (electron gain) while another undergoes oxidation (electron loss). The term "reduction" comes from the Latin word "reducere," meaning "to bring back," as the process involves bringing back the charge to a lower, more negative value. The change in oxidation numbers helps track electron transfer, highlighting its importance in chemical reactions.

Step by step solution

01

Definition of Reduction

In chemistry, reduction refers to the process in which a chemical species gains electrons. This process is usually accompanied by a decrease in the oxidation number. The term "reduction" is derived from the Latin word "reducere," which means "to bring back." In the context of chemical reactions, reduction occurs when a molecule, ion, or atom undergoes a change in its oxidation state by gaining electrons.
02

Understanding Oxidation Numbers

Oxidation numbers are a way to keep track of electrons and represent the apparent charge of an atom in a compound. In a chemical reaction, oxidation numbers help to determine whether an atom has gained or lost electrons. For example, in a reaction between hydrogen (H) and chlorine (Cl) to produce hydrochloric acid (HCl), the oxidation number of hydrogen increases from 0 to +1, and the oxidation number of chlorine decreases from 0 to -1. So, hydrogen undergoes oxidation (loses electrons) while chlorine undergoes reduction (gains electrons).
03

Explaining Electron Gain and Reduction

When a species gains electrons during a reaction, its oxidation number decreases, resulting in a reduction of its charge. This electron gain occurs because the species is more electronegative than the other species in the reaction, which means it has a greater tendency to attract electrons.
04

Reduction in Redox Reactions

Reduction is one half of a redox (reduction-oxidation) reaction. A redox reaction consists of two parts: one species that undergoes reduction (gains electrons) and another species that undergoes oxidation (loses electrons). These two processes always occur simultaneously in a redox reaction because, as one species gains electrons, another species must lose electrons to balance the charges.
05

What is Being Reduced?

When something gains electrons and is said to be reduced, it is the oxidation number of the species that is being reduced. The reduced oxidation number means that the species has gained electrons, resulting in a decrease in its charge or "bringing back" its charge to a lower, more negative value. This process highlights the importance of electron transfer in chemical reactions and how changes in oxidation numbers signify the gain or loss of electrons.

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Most popular questions from this chapter

Citric acid, which can be obtained from lemon juice, has the molecular formula \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} .\) A \(0.250-\mathrm{g}\) sample of citric acid dissolved in \(25.0 \mathrm{~mL}\) of water requires \(37.2 \mathrm{~mL}\) of \(0.105 \mathrm{M} \mathrm{NaOH}\) for complete neutralization. What number of acidic hydrogens per molecule does citric acid have?

A solution is prepared by dissolving \(10.8 \mathrm{~g}\) ammonium sulfate in enough water to make \(100.0 \mathrm{~mL}\) of stock solution. A \(10.00\) mL sample of this stock solution is added to \(50.00 \mathrm{~mL}\) of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

Suppose \(50.0 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{CoCl}_{2}\) solution is added to \(25.0 \mathrm{~mL}\) of \(0.350 \mathrm{M} \mathrm{NiCl}_{2}\) solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing. Assume that the volumes are additive.

What volume of each of the following bases will react completely with \(25.00 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\) ? a. \(0.100 \mathrm{M} \mathrm{NaOH}\) b. \(0.0500 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) c. \(0.250 \mathrm{M} \mathrm{KOH}\)

The vanadium in a sample of ore is converted to \(\mathrm{VO}^{2+}\). The VO \(^{2+}\) ion is subsequently titrated with \(\mathrm{MnO}_{4}^{-}\) in acidic solution to form \(\mathrm{V}(\mathrm{OH})_{4}{ }^{+}\) and manganese(II) ion. The unbalanced titration reaction is \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$ \mathrm{V}(\mathrm{OH})_{4}^{+}(a q)+\mathrm{Mn}^{2+}(a q)+\mathrm{H}^{+}(a q) $$ To titrate the solution, \(26.45 \mathrm{~mL}\) of \(0.02250 \mathrm{M} \mathrm{MnO}_{4}^{-}\) was required. If the mass percent of vanadium in the ore was \(58.1 \%\), what was the mass of the ore sample? Hint: Balance the titration reaction by the oxidation states method.
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