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Chapter 17: Problem 3

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 ( \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) ).

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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.

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Most popular questions from this chapter

Determine whether the following oxidation-reduction reactions ate balanced correctly, If they are not, provide the correct balanced reaction. (a) unbalanced: \(\mathrm{MnO}_{2}(s)+\mathrm{Al}(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{Al} \mathrm{O}_{3}(s)\) balanced: \(\mathrm{MnO}_{2}(s)+2 A 1(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)\) (b) unbalanced: \(\mathrm{Cu}(s)+\mathrm{Ag}^{4}(a q) \longrightarrow \mathrm{Cu}^{24}(a q)+A \mathrm{~g}(s)\) balanced: \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)\) (c) unbalanced: $$ \mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(d)+\mathrm{Mn}^{2+}(a q) $$ (icidic solution) balanced: $$ \begin{aligned} 16 \mathrm{H}^{+}(a q)+10 \mathrm{Br}^{-}(a q)+& 2 \mathrm{MnO}_{4}^{-}(a q) \longrightarrow \\ & 5 \mathrm{Br}(l)+2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ (d) unhalanced: $$ \begin{aligned} &\mathrm{MnO}_{4}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s) \quad \text { (basic solution) } \\ &\text { balanced: } \\ &8 \mathrm{H}^{-1}(a q)+\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \\ &\qquad \mathrm{S}(s)+\mathrm{MnS}(s)+4 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$

How many grams of zinc are reguired to reduce Fe3+ when \(25.0 \mathrm{~mL}\) of \(1.2 \mathrm{M} \mathrm{FeCl}\) are reacted? $$ \mathrm{Zn}+\mathrm{Fe}^{3+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}

Balance these ionic redox equations using the ion-electron method. These reactions occur in acidic solution. (a) \(\mathrm{Zn}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{NH}_{4}^{+}\) (b) \(\mathrm{NO}_{3}^{-}+\mathrm{S} \longrightarrow \mathrm{NO}_{2}+\mathrm{SO}_{4}^{-}\) (c) \(\mathrm{PH}_{3}+\mathrm{I}_{2} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{2}+\mathrm{I}^{-}\) (d) \(\mathrm{Cu}+\mathrm{NO}_{4}^{-} \longrightarrow \mathrm{Cu}^{24}+\mathrm{NO}\) (c) \(\mathrm{ClO}_{4}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}\)

In one type of alkaline cell used to power devices such as portable radios, \(\mathrm{Hg}^{20}\) ions are reduced to metallac mercury when the cell is bcing discharged. Does this reduction occur at the anode or the cathode? Explain.

Most explosive reactions are complex redox reactions with multiple oxidations and reductions. The reaction of gunpowder is shown below. Determine the element(s) oxidized and reduced. $$ \begin{aligned} &4 \mathrm{KNO}_{s}(s)+7 \mathrm{C}(s)+\mathrm{S}(s) \longrightarrow \\ &3 \mathrm{CO}_{2}(g)+3 \mathrm{CO}(g)+2 \mathrm{~N}_{3}(g)+\mathrm{K}_{2} \mathrm{CO}_{s}(s)+\mathrm{K}_{3} \mathrm{~S}(s) \end{aligned} $$
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