11.3: Oxidation States: Electron Bookkeeping
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Redox reactions are all about electrons being transferred from one substance to another, so it would be useful if we had a system for keeping track of what gains and what loses electrons, and how many electrons are involved. We do - our record-keeping system is called Oxidation Numbers. You may remember electronegativity from the previous section. When determining the oxidation number on an individual atom, the shared electrons are assigned to the most electronegative atom.
Electronegativity
The ability of a bonded atom to attract shared electrons towards itself
When two atoms of different elements are bonded together by a covalent bond (sharing electrons), the electrons are generally not shared equally between the two atoms due to differences in their electronegativity. Think of this as a tug-of-war for electrons. Sometimes both atoms pull with equal strength on shared electrons; other times there is a clearly stronger player who will pull the electrons closer to itself.
Consider the bond between a hydrogen atom (with one valence electron) and an oxygen atom (with its six valence electrons):
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Because oxygen has a higher electronegativity than hydrogen, the shared electrons are closer to the oxygen atom than to the hydrogen atom. This is not the total transfer of electrons that would create an ion, but partial charges do form - the hydrogen end of the bond is partially positive (+1) because it has partially lost one electron, and the oxygen end of the H–O is partially negative (-1) because it has partially gained the one electron from hydrogen:
Our molecule is incomplete however, because there is a lone electron around oxygen. Let's add one more hydrogen to complete our water molecule:
We see that each of the two hydrogen has "lost" one electron to oxygen. Oxygen has "gained" two electrons - one from each hydrogen. Again, these are not true ions but it is useful to think of them in the same way.
Charges given to atoms in a molecule in this way are called oxidation numbers. We can use oxidation numbers to keep track of where electrons are in a molecule, and how they move during a reaction. In our water example, hydrogen is assigned an oxidation number of +1 because each individual hydrogen has "lost" one electron. Oxygen has an oxidation number of +2 because the single oxygen atom has "gained" a total of two electrons, one from each hydrogen.
Here is another molecule involving hydrogen and oxygen - hydrogen peroxide, H2O2:
In hydrogen peroxide, each hydrogen still has an oxidation number of +1 because each hydrogen "gives up" a single electron to oxygen. Oxygen, however, now has an oxidation number of -1 because each oxygen gains just one electron from its neighboring hydrogen. The electrons between the two identical oxygen atoms are shared equally, so there is no partial charge resulting from that bond.
Oxidation Numbers
A positive or negative number assigned to an atom in a molecule or ion that reflects a partial gain or loss of electrons
Knowing the oxidation number of each individual element in a molecule will be a key step in our understanding of redox reactions. Fortunately it will not usually involve drawing electron dot diagrams. Instead there are a series of rules that we can use to determine oxidation numbers. Here are the main rules:
1. | The oxidation number of a pure element (by itself, and not an ion) is zero. |
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2. |
The oxidation number of a monatomic ion (by itself or as part of an ionic compound) is equal to its charge. Alkali metals - elements in the first column of the periodic table - will always have an oxidation number of +1; Alkali metals (column 2) are almost always +2 |
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3. | The oxidation number of hydrogen is almost always +1 when it is in a compound. |
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4. |
The oxidation number of oxygen is almost always -2 when it is in a compound The exceptions:
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The sum of the oxidation numbers in a compound is zero. To determine the oxidation number of Mn in Mn2O7 we must work backwards:
-2 × 7 atoms = -14 total Since the sum of oxidation numbers must be zero, the total oxidation number of Mn must be +14 to cancel out oxygen's -14, but since there are 2 Mn atoms, each individual atom will have an oxidation number of +7:
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The sum of the oxidation numbers in a polyatomic ion is equal to the charge on that ion. Again, work backwards to determine the oxidation number of any non-oxygen or non-hydrogen atom. To determine the oxidation number of Cr in Cr2O72- :
-2 × 7 = -14
+12 + (-14) = -2
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It is important to note that oxidation number always refers to each individual atom in the compound, not to the total for that element. For example in H2O - the total positive "charge" for both hydrogen atoms will be +2 (which balances with the -2 from oxygen), but EACH HYDROGEN has an oxidation number of +1.
Example \(\PageIndex{1}\): Assigning Oxidation Numbers
Assign oxidation numbers to the atoms in each substance.
- Br2
- SiO2
- Ba(NO3)2
Solution
- Br2 is the elemental form of bromine. Therefore, by rule 1, each atom has an oxidation number of 0.
- By rule 3, oxygen is normally assigned an oxidation number of −2. For the sum of the oxidation numbers to equal the charge on the species (which is zero), the silicon atom is assigned an oxidation number of +4.
- The compound barium nitrate can be separated into two parts: the Ba2+ ion and the nitrate ion. Considering these separately, the Ba2+ ion has an oxidation number of +2 by rule 2. Now consider the NO3− ion. Oxygen is assigned an oxidation number of −2, and there are three oxygens. According to rule 4, the sum of the oxidation number on all atoms must equal the charge on the species, so we have the simple algebraic equation
x + 3(−2) = −1
where x is the oxidation number of the nitrogen atom and −1 represents the charge on the species. Evaluating,
x + (−6) = −1x = +5
Thus, the oxidation number on the N atom in the nitrate ion is +5.
Exercise \(\PageIndex{1}\): Oxidation States in Phosphoric Acids
Assign oxidation numbers to the atoms in H3PO4.
- Answer
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H = +1, O = −2, P = +5
All redox reactions occur with a simultaneous change in the oxidation numbers of some atoms. At least two elements must change their oxidation numbers. When an oxidation number of an atom is increased in the course of a redox reaction, that atom is being oxidized. When an oxidation number of an atom is decreased in the course of a redox reaction, that atom is being reduced. Oxidation and reduction are thus also defined in terms of increasing or decreasing oxidation numbers, respectively.
Example \(\PageIndex{2}\): Oxidation States in Sodium Chloride
Identify what is being oxidized and reduced in this redox equation:
\[\ce{2Na + Cl_2 → 2NaCl} \nonumber\]
Solution
Consider the reactants. Because both reactants are the elemental forms of their atoms, the Na and Cl atoms as reactants have oxidation numbers of 0. In the ionic product, the sodium ions have an oxidation number of +1, while the chloride ions have an oxidation number of −1:
We note that the sodium is increasing its oxidation number from 0 to +1, so it is being oxidized; chlorine is decreasing its oxidation number from 0 to −1, so it is being reduced:
Because oxidation numbers are changing, this is a redox reaction. Note that the total number of electrons lost by the sodium (two, one lost from each atom) is gained by the chlorine atoms (two, one gained for each atom).
Exercise \(\PageIndex{2}\)
Identify what is being oxidized and reduced in this redox equation.
\[\ce{C + O2 → CO2} \nonumber\]
- Answer
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C is being oxidized from 0 to +4; O is being reduced from 0 to −2.
In this introduction to oxidation-reduction reactions, we are using straightforward examples. However, oxidation reactions can become quite complex; the following equation represents a redox reaction:
To demonstrate that this is a redox reaction, the oxidation numbers of the species being oxidized and reduced are listed; can you determine what is being oxidized and what is being reduced? This is also an example of a net ionic reaction; spectator ions that do not change oxidation numbers are not displayed in the equation.