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Chapter 4: Problem 11

Why must oxidation and reduction occur together in a reaction?

Short Answer

Expert verified
Oxidation and reduction occur together because electrons lost by one species are gained by another, forming a redox reaction.

Step by step solution

01

Understanding Oxidation

Oxidation refers to the process by which an atom, ion, or molecule loses electrons. This can be remembered as "OIL" from "Oxidation Is Loss." For example, in the reaction of magnesium with oxygen, magnesium loses electrons to form Mg虏鈦� ions.
02

Understanding Reduction

Reduction is the process by which an atom, ion, or molecule gains electrons. This can be remembered as "RIG" from "Reduction Is Gain." In the magnesium and oxygen reaction, oxygen gains electrons to form O虏鈦� ions.
03

Explaining the Simultaneity

Since electrons cannot exist freely in most chemical environments, when one species loses electrons (oxidation), another must gain those electrons (reduction). This electron transfer ensures the conservation of charge and maintains the balance of energy in the system.
04

The Redox Pair

Because oxidation and reduction processes are linked through electron transfer, they occur simultaneously. Together they form a "redox" reaction, where the term 'redox' is a shorthand for reduction-oxidation. In essence, you can't have one without the other.

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Key Concepts

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

Oxidation
Oxidation is an essential chemical process where an atom, ion, or molecule loses electrons. You might hear the phrase "Oxidation Is Loss" or "OIL" as it describes this very concept. When an element undergoes oxidation, it results in an increase in its oxidation state. Let's take magnesium as an example. In the presence of oxygen, magnesium atoms lose two electrons, transforming into magnesium ions (\( \text{Mg}^{2+} \)).
  • Oxidized substances lose electrons.
  • Oxidation is often associated with adding oxygen or removing hydrogen.
  • This process is crucial in a variety of chemical reactions, from rusting of metals to metabolism in living organisms.
Understanding oxidation helps us grasp why it always pairs with reduction; as oxidation progresses by losing electrons, these electrons must be received or captured by another participant in the reaction.
Reduction
Reduction is the counterpart to oxidation and involves the gain of electrons by an atom, ion, or molecule. "Reduction Is Gain" or "RIG" neatly sums up this process. When an element experiences reduction, its oxidation state decreases. Consider the same magnesium reaction; while magnesium loses electrons, oxygen gains them, forming oxide ions (\( \text{O}^{2-} \)).
  • Reduced substances gain electrons.
  • Reduction typically involves removing oxygen or adding hydrogen.
  • This process is equally important in both natural phenomena and industrial applications.
Reduction cannot happen alone; as it gains electrons, it relies on another process (oxidation) to supply these electrons. This interdependency is a hallmark of redox reactions.
Electron Transfer
The concept of electron transfer is pivotal in understanding redox reactions. During such reactions, electrons shift from one species to another, facilitating both oxidation and reduction. Since electrons are not usually stable as free entities in chemical environments, this transfer is necessary.
  • Electrons move from oxidized to reduced species.
  • This process is a natural means to achieve chemical stability and energy balance.
  • Electron transfer is the basis for processes like energy production in cells and corrosion in metals.
In essence, electron transfer acts as the bridge that links oxidation and reduction processes, ensuring that both can occur simultaneously. This linkage underscores the existence of redox pairs.
Conservation of Charge
The conservation of charge is a fundamental principle in chemistry ensuring that during any chemical reaction, the total charge before and after the reaction remains the same. In redox reactions, as electrons are neither created nor destroyed but merely transferred, charge conservation is crucial.
  • Charge is balanced by equal numbers of electrons being lost and gained.
  • This principle keeps the universality of charge balance intact across all reactions.
  • It ensures stability, making sure no imbalances occur in reaction dynamics.
When oxidation and reduction occur as a redox pair, the electrons given up in oxidation must match those consumed in reduction, maintaining a balanced and neutral charge state. This balance is vital for the consistency and feasibility of chemical reactions.

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Most popular questions from this chapter

How many milliliters of \(0.250 \mathrm{M} \mathrm{KMnO}_{4}\) are needed to react with \(3.55 \mathrm{~g}\) of iron(II) sulfate, \(\mathrm{FeSO}_{4}\) ? The reaction is as follows: $$ \begin{array}{r} 10 \mathrm{FeSO}_{4}(a q)+2 \mathrm{KMnO}_{4}(a q)+8 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\ 5 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+2 \mathrm{MnSO}_{4}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q)+ \\ 8 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$

A 3.75-g sample of iron ore is transformed to a solution of iron(II) sulfate, \(\mathrm{FeSO}_{4}\), and this solution is titrated with \(0.150 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (potassium dichromate). If it requires \(43.7 \mathrm{~mL}\) of potassium dichromate solution to titrate the iron(II) sulfate solution, what is the percentage of iron in the ore? The reaction is $$ \begin{array}{r} 6 \mathrm{FeSO}_{4}(a q)+\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(a q)+7 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\ 3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l)+ \\ \mathrm{K}_{2} \mathrm{SO}_{4}(a q) \end{array} $$

How many milliliters of \(0.250 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) (sulfuric acid) are required to react with \(8.20 \mathrm{~g}\) of sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3}\), according to the following equation? $$ \begin{aligned} \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaHCO}_{3}(a q) \longrightarrow & \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \end{aligned} $$

Potassium hydrogen phthalate (abbreviated as KHP) has the molecular formula \(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}\) and a molar mass of \(204.22 \mathrm{~g} / \mathrm{mol}\). KHP has one acidic hydrogen. A solid sample of KHP is dissolved in \(50 \mathrm{~mL}\) of water and titrated to the equivalence point with \(22.90 \mathrm{~mL}\) of a \(0.5010 \mathrm{M}\) NaOH solution. How many grams of KHP were used in the titration?

Mercury(II) nitrate is treated with hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S},\) forming a precipitate and a solution. Write the molecular equation and the net ionic equation for the reaction. An acid is formed; is it strong or weak? Name each of the products. If \(65.65 \mathrm{~g}\) of mercury(II) nitrate and \(4.26 \mathrm{~g}\) of hydrogen sulfide are mixed in \(395.0 \mathrm{~g}\) of water to form \(54.16 \mathrm{~g}\) of precipitate, what is the mass of the solution after the reaction?
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