Why are oxidation and reduction said to be complementary processes?

Short Answer

Expert verified
Oxidation and reduction are complementary because one cannot occur without the other; when one substance loses electrons, another must gain them.

Step by step solution

01

- Understand Oxidation

Oxidation is a chemical process where an atom or molecule loses electrons. When a substance undergoes oxidation, its oxidation state increases.
02

- Understand Reduction

Reduction is a chemical process where an atom or molecule gains electrons. When a substance undergoes reduction, its oxidation state decreases.
03

- Complementary Nature

Oxidation and reduction always occur together because the electrons lost in oxidation are gained by another species in reduction. Without one, the other cannot happen.
04

- Example

Consider the reaction: \[ \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \] In this reaction, Zn is oxidized as it loses electrons ( \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\) ). Cu is reduced as it gains electrons ( \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) ).

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.

chemical reactions
Chemical reactions are the processes where substances interact to form new substances. They involve breaking old bonds and forming new ones. Chemical reactions can be represented by balanced equations, showcasing the conversion of reactants to products. In the context of oxidation and reduction, these reactions always involve electron transfer. Think of chemical reactions like making a cake: ingredients (reactants) combine and interact to form a cake (products). Sometimes, you need to follow specific steps to get the desired result.
electron transfer
Electron transfer is the movement of electrons from one atom or molecule to another. This movement is crucial in oxidation and reduction processes. When a substance loses electrons, it undergoes oxidation. When another substance gains those electrons, it undergoes reduction. Imagine electrons as little balls being passed around. If one friend (atom) hands over a ball (electron) to another friend, the first friend is oxidized, and the second friend is reduced.
  • Oxidation: Losing electrons
  • Reduction: Gaining electrons
Without electron transfer, oxidation and reduction reactions wouldn't occur.
oxidation state
The oxidation state, also known as the oxidation number, is a value that helps us keep track of electrons in atoms during chemical reactions. It's especially useful for redox reactions (reactions involving both oxidation and reduction). When an atom's oxidation state increases, it means the atom has lost electrons (oxidation). When an atom's oxidation state decreases, it means the atom has gained electrons (reduction).
Think of oxidation states as scores in a game. If your score goes up, you've lost some tokens (electrons). If your score goes down, you've gained some tokens (electrons). Keeping track of these scores helps us understand and predict the behavior of substances in a chemical reaction.

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

The following observations were made concerning metals \(A, B, C\), and \(D\). (a) When a strip of metal \(\mathrm{A}\) is placed in a solution of \(\mathrm{B}^{2+}\) aons, no reaction is observed. (b) Similarly, \(A\) in a solution containing \(\mathrm{C}^{+}\)ions produces no reaction. (c) When a strip of metal \(D\) is placed in a solut?on of " \(\mathrm{C}^{+}\)àns, black metallic C deposits on the surface of D, and the solution tests posatively for \(\mathrm{D}^{2+}\) tons. (d) When a piece of metallic B is placed in a solution of \(D^{2+}\) ions, metallic \(D\) appears on the surface of \(\mathrm{B}\) and \(\mathrm{B}^{2+}\) jons are found in the solution. Arrange the ions \(\mathrm{A}^{4}, \mathrm{~B}^{2+}, \mathrm{C}^{4}\), and \(\mathrm{D}^{24}\) in order of their ability to attract clectrons. List them in order of increasing ability.

(a) Cu is oxidized; Sn is reduced (b) anode: \(\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 e^{-}\) cathode: \(\mathrm{Sn}^{2+}(a q q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{St}(s)\)

What mass of \(\mathrm{KMnO}_{4}\) is needed to react with \(100, \mathrm{~mL}, \mathrm{H}_{2} \mathrm{O}_{2}\) solution? ( \(a=1,031 \mathrm{~g} / \mathrm{mL}, 9.0^{3} \% \mathrm{H}_{2} \mathrm{O}_{2}\) by mass) $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}+& \mathrm{KMnO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \\ & \mathrm{O}_{2}+\mathrm{MnSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O} \quad \text { (acídic solution) } \end{aligned} $$

The chemical reactions taking place during discharge in a lead storage battery are $$ \begin{gathered} \mathrm{Pb}+\mathrm{SO}^{-} \longrightarrow \mathrm{PbSO}_{4} \\ \mathrm{PbO}_{2}+\mathrm{SO}_{4}^{2}+4 \mathrm{H}^{+} \longrightarrow \mathrm{PbSO}_{4}+2 \mathrm{H}_{4} \mathrm{O} \end{gathered} $$ (a) Complete each half-reaction by supplying electrons. (b) Which reaction is oxidation, and which is reduction? (c) Which reaction occurs at the anode of the battery?

Why is a porous barrier or a salt bridge necessary in some voltaxe cells?

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