Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Where in the periodic table are the most easily reduced elements found? The most easily oxidized?

Short Answer

Expert verified
Most easily reduced elements are in the top right (halogens), and most easily oxidized are in the bottom left (alkali metals).

Step by step solution

01

Understanding Reduction and Oxidation

Reduction refers to the gain of electrons while oxidation refers to the loss of electrons. To understand where these processes are most likely to occur in the periodic table, we must consider the electron affinity and ionization energy of elements.
02

Identify Elements with High Electron Affinity

Elements with high electron affinity easily gain electrons. These elements are most easily reduced. In the periodic table, the elements with the highest electron affinity are found in the top right, excluding the noble gases, primarily in the halogens group (Group 17), such as fluorine and chlorine.
03

Identify Elements with Low Ionization Energy

Elements with low ionization energy easily lose electrons and are readily oxidized. These elements are found in the bottom left of the periodic table, especially the alkali metals (Group 1), such as lithium, sodium, and potassium.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Reduction and Oxidation
Reduction and oxidation, often abbreviated as redox reactions, are essential chemical processes. Reduction involves the gain of electrons, resulting in a decrease in oxidation state. Oxidation is the opposite process, where an atom or molecule loses electrons, leading to an increase in oxidation state.
In the context of the periodic table, different groups of elements exhibit distinct tendencies to undergo these processes. Elements on the right side of the periodic table, specifically in Groups 16 and 17, have a high ability to gain electrons, making them easily reducible. Conversely, metals on the left, particularly in Group 1 known as alkali metals, lose electrons readily, thus they are easily oxidizable."
  • Reduction: Gain of electrons; decrease in oxidation state.
  • Oxidation: Loss of electrons; increase in oxidation state.

Understanding the placement of these elements can help predict their chemical behavior in reactions.
Electron Affinity
Electron affinity refers to the amount of energy released when an electron is added to a neutral atom in the gaseous state. Elements with high electron affinities release more energy upon gaining an electron and often form negative ions easily.
In the periodic table, electron affinity generally increases across a period from left to right and decreases down a group. The halogens, found in Group 17, possess some of the highest electron affinities because they need only one more electron to reach a stable noble gas configuration.
  • High electron affinity means a greater tendency to gain electrons and be reduced.
  • Halogens like fluorine and chlorine are excellent examples of elements with high electron affinities.
Understanding electron affinity helps in predicting which elements are likely to gain electrons and be reduced.
Ionization Energy
Ionization energy is the energy required to remove an electron from an isolated atom in the gaseous phase. Elements with low ionization energies lose their outer electrons easily, making them naturally prone to oxidation.
Across the periodic table, ionization energy generally increases from left to right across a period and decreases down a group.
This trend is why alkali metals, found in Group 1, are known for their low ionization energy. They can easily lose one electron to achieve a stable electronic configuration, which accounts for their high reactivity and tendency to undergo oxidation.
  • Low ionization energy indicates that an element loses electrons easily.
  • Alkali metals are prime examples due to their single outer electron.
Comprehending ionization energy makes it clearer which elements will tend to be oxidized.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), is a toxic substance found in spinach leaves. What is the molarity of a solution made by dissolving \(12.0 \mathrm{~g}\) of oxalic acid in enough water to give \(400.0 \mathrm{~mL}\) of solution? How many milliliters of \(0.100 \mathrm{M}\) KOH would you need to titrate \(25.0 \mathrm{~mL}\) of the oxalic acid solution according to the following equation? $$ \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+2 \mathrm{KOH}(a q) \longrightarrow \mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$

Compound \(\mathrm{X}\) contains only the elements \(\mathrm{C}_{2} \mathrm{H}, \mathrm{O}\), and \(\mathrm{S} . \mathrm{A}\) \(5.00 \mathrm{~g}\) sample undergoes complete combustion to give \(4.83 \mathrm{~g}\) of \(\mathrm{CO}_{2}, 1.48 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\), and a certain amount of \(\mathrm{SO}_{2}\) that is further oxidized to \(\mathrm{SO}_{3}\) and dissolved in water to form sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) On titration of the \(\mathrm{H}_{2} \mathrm{SO}_{4}, 109.8 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) is needed for complete reaction. (Both \(\mathrm{H}\) atoms in sulfuric acid are acidic and react with \(\mathrm{NaOH}\).) (a) What is the empirical formula of \(\mathrm{X} ?\) (b) When \(5.00 \mathrm{~g}\) of \(\mathrm{X}\) is titrated with \(\mathrm{NaOH}\), it is found that \(\mathrm{X}\) has two acidic hydrogens that react with \(\mathrm{NaOH}\) and that \(54.9 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) is required to completely neutralize the sample. What is the molecular formula of \(\mathrm{X} ?\)

Predict whether a precipitation reaction will occur when aqueous solutions of the following substances are mixed. For those that form a precipitate, write the net ionic reaction. (a) \(\mathrm{MnCl}_{2}+\mathrm{Na}_{2} \mathrm{~S}\) (b) \(\mathrm{HNO}_{3}+\mathrm{CuSO}_{4}\) (c) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{Na}_{3} \mathrm{PO}_{4}\) (d) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{KOH}\)

A bottle of \(12.0 \mathrm{M}\) hydrochloric acid has only \(35.7 \mathrm{~mL}\) left in it. What will the HCl concentration be if the solution is diluted to \(250.0 \mathrm{~mL} ?\)

Assume that you have an aqueous solution of an unknown salt. Treatment of the solution with dilute \(\mathrm{BaCl}_{2}, \mathrm{AgNO}_{3}\), and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) produces no precipitate. Which of the following anions might the solution contain? (a) \(\mathrm{Cl}\) (b) \(\mathrm{NO}_{3}^{-}\) (c) \(\mathrm{OH}^{-}\) (d) \(\mathrm{SO}_{4}{ }^{2-}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free