! ∠ EDUCATION ∠ SCIENCE ∠ CHEMISTRY ∠ REDOX REACTIONS: OXIDATION AND REDUCTION
REDOX REACTIONS: OXIDATION AND REDUCTION Redox reactions — reactions in which there’s a simultaneous transfer of electrons from one chemical species to another — are really composed of two different reactions: oxidation (a loss of electrons) and reduction (a gain of electrons).
The electrons that are lost in the oxidation reaction are the same electrons that are gained in the reduction reaction. These two reactions are commonly called half-reactions; the overall reaction is called a redox (reduction/oxidation) reaction.
OXIDATION There are three definitions you can use for oxidation: The loss of electrons The gain of oxygen The loss of hydrogen
LOSS OF ELECTRONS
One way to define oxidation is with the reaction in which a chemical substance loses electrons in going from reactant to product. For example, when sodium metal reacts with chlorine gas to form sodium chloride (NaCl), the sodium metal loses an electron, which is then gained by chlorine. The following equation shows sodium losing the electron:
When it loses the electron, chemists say that the sodium metal has been oxidized to the sodium cation. (A cation is an ion with a positive charge due to the loss of electrons.)
Reactions of this type are quite common in electrochemical reactions, reactions that produce or use electricity.
GAIN OF OXYGEN Sometimes, in certain oxidation reactions, it’s obvious that oxygen has been gained in going from reactant to product. Reactions where the gain of oxygen is more obvious than the gain of electrons include combustion reactions (burning) and the rusting of iron. Here are two examples. Burning of coal:
Rusting of iron:
In these cases, chemists say that the carbon and the iron metal have been oxidized to carbon dioxide and rust, respectively.
LOSS OF HYDROGEN
In other reactions, oxidation can best be seen as the loss of hydrogen. Methyl alcohol (wood alcohol) can be oxidized to formaldehyde:
In going from methanol to formaldehyde, the compound went from having four hydrogen atoms to having two hydrogen atoms.
REDUCTION Like oxidation, there are three definitions you can use to describe reduction: The gain of electrons The loss of oxygen The gain of hydrogen
GAIN OF ELECTRONS Reduction is often seen as the gain of electrons. In the process of electroplating silver onto a teapot, for example, the silver cation is reduced to silver metal by the gain of an electron. The following equation shows the silver cation gaining the electron:
When it gains the electron, chemists say that the silver cation has been reduced to silver metal.
LOSS OF OXYGEN In other reactions, it’s easier to see reduction as the loss of oxygen in going from reactant to product. For example, iron ore (primarily rust) is reduced to iron metal in a blast furnace by a reaction with carbon monoxide:
The iron has lost oxygen, so chemists say that the iron ion has been reduced to iron metal.
GAIN OF HYDROGEN In certain cases, a reduction can also be described as the gain of hydrogen atoms in going from reactant to product. For example, carbon monoxide and hydrogen gas can be reduced to methyl alcohol:
In this reduction process, the CO has gained the hydrogen atoms.
ONE’S LOSS IS THE OTHER’S GAIN Neither oxidation nor reduction can take place without the other. When those electrons are lost, something has to gain them. Consider, for example, the net-ionic equation (the equation showing just the chemical substances that are changed during a reaction) for a reaction with zinc metal and an aqueous copper(II) sulfate solution:
This overall reaction is really composed of two half-reactions, shown below. Oxidation half-reaction — the loss of electrons:
Reduction half-reaction — the gain of electrons:
Zinc loses two electrons; the copper(II) cation gains those same two electrons. Zn is being oxidized. But without that copper cation (the oxidizing agent) present, nothing will happen. It’s a necessary agent for the oxidation process to proceed. The oxidizing agent accepts the electrons from the chemical species that is being oxidized. The copper(II) cation is reduced as it gains electrons. The species that furnishes the electrons is called the reducing agent. In this case, the reducing agent is zinc metal.
The oxidizing agent is the species that’s being reduced, and the reducing agent is the species that’s being oxidized. Both the oxidizing and reducing agents are on the left (reactant) side of the redox equation.
! ∠ EDUCATION ∠ SCIENCE ∠ CHEMISTRY ∠ DRAWING LEWIS DOT STRUCTURES FOR CHEMISTRY
DRAWING LEWIS DOT STRUCTURES FOR CHEMISTRY RELATED BOOK
Chemistry Workbook For Dummies, 2nd Edition
By Peter J. Mikulecky, Chris Hren Part of Chemistry Workbook For Dummies Cheat Sheet
In chemistry, drawing Lewis dot structures can be challenging, but they provide a wealth of information about the molecules they represent. Remember that Lewis dot structures are drawn for covalent (molecular) compounds that share electrons. Follow these simple steps to correctly draw a Lewis dot structure:
1
Add up the total number of valence electrons found in the entire compound. Don’t forget to include any positive or negative charges when determining this.
2
Draw the simple structure (skeleton structure) of the compound by connecting
3
Add electrons to all the noncentral atoms.
everything with single bonds only.
Note that most atoms want eight total electrons, so atoms that have only one bond will each need to receive six electrons. However, hydrogen wants only two electrons; each bond counts as two shared electrons, so don’t add any electrons to hydrogen.
4
Put any unused electrons on the central atom. If all atoms (except hydrogen) now have eight electrons, you’re done.
5
If one or more atoms do not have eight electrons, you must form double or triple bonds between them. Keep in mind that each bond counts for two shared electrons.
6
If all atoms now have eight electrons around them, you’re done. If you have valence electrons left over, add them to the central atom, even if it violates the octet rule.
! ∠ EDUCATION ∠ SCIENCE ∠ CHEMISTRY ∠ HOW TO CALCULATE ENDOTHERMIC AND EXOTHERMIC REACTIONS
HOW TO CALCULATE ENDOTHERMIC AND EXOTHERMIC REACTIONS RELATED BOOK
Chemistry Workbook For Dummies, 2nd Edition
By Peter J. Mikulecky, Chris Hren By calculating the enthalpy change in a chemical reaction, you can determine whether the reaction is endothermic or exothermic. Chemical reactions transform both matter and energy. Though chemical equations usually list only the matter components of a reaction, you can also consider heat energy as a reactant or product. When chemists are interested in heat flow during a reaction (and when the reaction is run at constant pressure), they may list an enthalpy change
to the right of the reaction equation. At constant pressure, heat flow equals enthalpy change:
If the enthalpy change listed for a reaction is negative, then that reaction releases heat as it proceeds — the reaction is exothermic (exo- = out). If the enthalpy change listed for the reaction is positive, then that reaction absorbs heat as it proceeds — the reaction is endothermic (endo- = in). In other words, exothermic reactions release heat as a product, and endothermic reactions consume heat as a reactant.
The sign of the
tells you the direction of heat flow, but what about the magnitude? The coefficients of a chemical reaction represent molar equivalents, so the value listed for the
refers to the enthalpy change for one mole equivalent of the reaction. Here’s an example:
This reaction equation describes the combustion of methane, a reaction you might expect to release heat. The enthalpy change listed for the reaction confirms this expectation: For each mole of methane that combusts, 802 kJ of heat is released. The reaction is highly exothermic. Based on the stoichiometry of the equation, you can also say that 802 kJ of heat is released for every 2 mol of water produced.
So reaction enthalpy changes (or reaction “heats”) are a useful way to measure or predict chemical change. But they’re just as useful in dealing with physical changes, like freezing and melting, evaporating and condensing, and others. For example, water (like most substances) absorbs heat as it melts (or fuses) and as it evaporates. Here are the molar enthalpies for such changes: Molar enthalpy of fusion:
Molar enthalpy of vaporization:
The same sorts of rules apply to enthalpy changes listed for chemical changes and physical changes. Here’s a summary of the rules that apply to both: The heat absorbed or released by a process is proportional to the moles of substance that undergo that process. For example, 2 mol of combusting methane release twice as much heat as 1 mol of combusting methane. Running a process in reverse produces heat flow of the same magnitude but of opposite sign as running the forward process. For example, freezing 1 mol of water releases the same amount of heat that is absorbed when 1 mol of water melts. Try an example: here is a balanced chemical equation for the oxidation of hydrogen gas to form liquid water, along with the corresponding enthalpy change:
How much electrical energy must be expended to perform electrolysis of 3.76 mol of liquid water, converting that water into hydrogen gas and oxygen gas? First, recognize that the given enthalpy change is for the reverse of the electrolysis reaction, so you must reverse its sign from –572 kJ to 572 kJ. Second, recall that heats of reaction are proportional to the amount of substance reacting (2 mol of H2O in this case), so the calculation is
! ∠ EDUCATION ∠ SCIENCE ∠ CHEMISTRY ∠ HOW TO CALCULATE THE EMPIRICAL FORMULA OF A COMPOUND
HOW TO CALCULATE THE EMPIRICAL FORMULA OF A COMPOUND RELATED BOOK
Chemistry Workbook For Dummies, 2nd Edition
By Peter J. Mikulecky, Chris Hren If you don’t know the empirical formula of a compound, you can analyze samples of the unknown compound to identify the percent composition. From there, you calculate the ratios of different types of atoms in the compound. You express these ratios as the empirical formula. ADVERTISING inRead invented by Teads ADVERTISING
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An empirical formula represents the lowest whole-number ratio of elements in a compound.
Here’s how to find an empirical formula when given percent composition:
1
Assume that you have 100 g of the unknown compound. The beauty of this little trick is that you conveniently gift yourself with the same number of grams of each elemental component as its contribution to the percent composition. For example, if you assume that you have 100 g of a compound composed of 60.3% magnesium and 39.7% oxygen, you know that you have 60.3 g of magnesium and 39.7 g of oxygen. (The only time you don’t do this is if the problem specifically gives you the masses of each element present in the unknown compound.)
2
Convert the masses from Step 1 into moles using the molar mass.
3
Determine which element has the smallest mole value. Then divide all the mole values you calculated in Step 2 by this smallest value. This division yields the mole ratios of the elements of the compound.
4
If any of your mole ratios aren’t whole numbers, multiply all numbers by the smallest possible factor that produces whole-number mole ratios for all the elements. For example, if you have 1 nitrogen atom for every 0.5 oxygen atoms in a compound, the empirical formula is not N1O0.5. Such a formula casually suggests that an oxygen atom has been split, something that would create a small-scale nuclear explosion. Though
impressive sounding, this scenario is almost certainly false. Far more likely is that the atoms of nitrogen and oxygen are combining in a 1 : 0.5 ratio but do so in a larger but equivalent ratio of 2 : 1. The empirical formula is thus N2O. Because the original percent composition data is typically experimental, expect to see a bit of error in the numbers. For example, 2.03 is probably within experimental error of 2, 2.99 is probably 3, and so on.
5
Write the empirical formula by attaching these whole-number mole ratios as subscripts to the chemical symbol of each element. Order the elements according to the general rules for naming ionic and molecular compounds.
Here’s an example: What is the empirical formula of a substance that is 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass?
For the sake of simplicity, assume that you have a total of 100 g of this mystery compound. Therefore, you have 40.0 g of carbon, 6.7 g of hydrogen, and 53.3 g of oxygen. Convert each of these masses to moles by using the gram atomic masses of C, H, and O:
Notice that the carbon and oxygen mole numbers are the same, so you know the ratio of these two elements is 1:1 within the compound. Next, divide all the mole numbers by the smallest among them, which is 3.33. This division yields
The compound has the empirical formula CH2O. The actual number of atoms within each particle of the compound is some multiple of the numbers expressed in this formula.