4.3: Oxidation Numbers and Redox Reactions
- Page ID
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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 may involve proton transfers, electron transfers, and other bond-breaking and bond-making processes. As a result, the equations involved are can be more difficult to interpret and balance than those describing acid-base reactions. To recognize redox reactions, we must be able to identify when a species is oxidized or reduced; to do this we assign assign oxidation numbers to each atom in a reaction. Then, we compare the oxidation states of eac atom on the reactants side and to the atoms on the products side. When changes occur, we know a readox reaction has taken place.
In the reaction above, the nitrogen in the NO3– ion is assigned an oxidation number of +5, and each oxygen is assigned an oxidation number of –2. This assignment corresponds to the nitrogen having lost five valence electrons to the more electronegative oxygen atoms. The nitrogen in the NO2 ion has an oxidation number of + 4, which corresponds to the loss of four valence electrons to the more electronegative oxygen atoms. The nitrogen atom may be thought of as having one valence electron for itself, that is, one more electron than it had in NO3–. Through this reaction, nitrogen gains one electron in going from NO3– to NO2, and we say nitrogen is reduced by the reaction.
In the same reaction, the oxidation number of copper increased from 0 to +2; we say that copper is oxidized by the reaction and lost two electrons. The nitrogen was reduced by electrons donated by copper, and so copper was the reducing agent. Copper was oxidized because its electrons were accepted by an oxidizing agent, nitrogen (or nitrate ion).
A word of caution: Although they are useful and necessary for recognizing redox reactions, oxidation numbers are a highly artificial device. The nitrogen atom in NO3– does not really have a +5 charge which can be reduced to +4 in NO2. Instead, there are covalent bonds and electron-pair sharing between nitrogen and oxygen in both species, and nitrogen has certainly not lost its valence electrons entirely to oxygen. Even though this may (and indeed should) make you suspicious of the validity of oxidation numbers, they are undoubtedly a useful tool for spotting electron-transfer processes. So long as they are used for that purpose only, and not taken to mean that atoms in covalent species actually have the large charges oxidation numbers often imply, their use is quite valid.
The general rules for oxidation numbers are seen below, taken from the following page in the Analytical Chemistry Core Textbook: Oxidation States
Rules for Assigning Oxidation Numbers
- The oxidation state of an uncombined element is zero. This applies regardless of the structure of the element: Xe, Cl2, S8, and large structures of carbon or silicon each have an oxidation state of zero. (Since atoms of the same element always form pure covalent bonds, they share electrons equally, neither losing nor gaining, e.g., Cl2.)
- The sum of the oxidation states of all the atoms or ions in a neutral compound is zero. (When an electron is lost by one atom (+1 contribution to oxidation number), the same electron must be gained by another atom (–1 contribution to oxidation number).)
- The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
- The more electronegative element in a substance is assigned a negative oxidation state. The less electronegative element is assigned a positive oxidation state. Remember that electronegativity is greatest at the top-right of the periodic table and decreases toward the bottom-left.
- Some elements almost always have the same oxidation states in their compounds:
| Element | Usual oxidation state | Exceptions |
| Group 1 metals | Always +1 | |
| Group 2 metals | Always +2 | |
| Oxygen | Usually -2 | Peroxides and F2O (see below) |
| Hydrogen | Usually +1 | Metal hydrides (-1) (see below) |
| Fluorine | Always -1 | |
| Chlorine | usually -1 | Compounds with O or F (see below) |
Exceptions:
Hydrogen in the metal hydrides: Metal hydrides include compounds like sodium hydride, NaH. Here the hydrogen exists as a hydride ion, H-. The oxidation state of a simple ion like hydride is equal to the charge on the ion—in this case, -1.
Alternatively, the sum of the oxidation states in a neutral compound is zero. Because Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0).
Oxygen in peroxides: Peroxides include hydrogen peroxide, H2O2. This is an electrically neutral compound, so the sum of the oxidation states of the hydrogen and oxygen must be zero.
Because each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.
Oxygen in F2O: The deviation here stems from the fact that oxygen is less electronegative than fluorine; the fluorine takes priority with an oxidation state of -1. Because the compound is neutral, the oxygen has an oxidation state of +2.
Chlorine in compounds with fluorine or oxygen: Because chlorine adopts such a wide variety of oxidation states in these compounds, it is safer to simply remember that its oxidation state is not -1, and work the correct state out using fluorine or oxygen as a reference.
What is the oxidation state of chromium in Cr2+?
Solution
For a simple ion such as this, the oxidation state equals the charge on the ion: +2 (by convention, the + sign is always included to avoid confusion)
What is the oxidation state of chromium in CrCl3?
This is a neutral compound, so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1 (no fluorine or oxygen atoms are present). Let n equal the oxidation state of chromium:
n + 3(-1) = 0
n = +3
The oxidation state of chromium is +3.
What is the oxidation state of chromium in Cr(H2O)63+?
Solution
This is an ion and so the sum of the oxidation states is equal to the charge on the ion. There is a short-cut for working out oxidation states in complex ions like this where the metal atom is surrounded by electrically neutral molecules like water or ammonia.
The sum of the oxidation states in the attached neutral molecule must be zero. That means that you can ignore them when you do the sum. This would be essentially the same as an unattached chromium ion, Cr3+. The oxidation state is +3.
What is the oxidation state of chromium in the dichromate ion, Cr2O72-?
The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don't forget that there are 2 chromium atoms present.
2n + 7(-2) = -2
n = +6
What is the oxidation state of copper in CuSO4?
Solution
Unfortunately, it isn't always possible to work out oxidation states by a simple use of the rules above. The problem in this case is that the compound contains two elements (the copper and the sulfur) with variable oxidation states.
In cases like these, some chemical intuition is useful. Here are two ways of approaching this problem:
- Recognize CuSO4 as an ionic compound containing a copper ion and a sulfate ion, SO42-. To form an electrically neutral compound, the copper must be present as a Cu2+ ion. The oxidation state is therefore +2.
- Recognize the formula as being copper(II) sulfate (the (II) designation indicates that copper is in a +2 oxidation state, as discussed below).
Problems
Assign oxidation states to each atom in the reaction. Determine which species are the oxidizing and reducing agents.
\(\ce{2MnO4^{–} + 5SO2 + 6H2O -> 5SO4^{2–} + 2Mn^{2+} + 4H3O^{+}}\)
- Answer
-
The only atoms which change are Mn, from +7 to +2, a reduction, and S, from +4 to +6, an oxidation. The reaction is a redox process. SO2 has been oxidized by MnO4–, and so MnO4–is the oxidizing agent. MnO4– has been reduced by SO2, and so SO2 is the reducing agent.
Assign oxidation states to each atom in the reaction. Determine which species are the oxidizing and reducing agents.
\(\ce{NH4^+ + PO4^{3–} -> NH3 + PO4^{2–}}\)
- Answer
-
Oxidation numbers show that no redox has occurred. This is an acid-base reaction because a proton, but no electrons, has been transferred.
Assign oxidation states to each atom in the reaction. Determine which species are the oxidizing and reducing agents.
\(\ce{HClO + H2S -> H3O^+ + Cl^{–} + S}\)
- Answer
-
H2S has been oxidized, losing two electrons to form elemental S. Since H2S donates electrons, it is the reducing agent. HClO accepts these electrons and is reduced to Cl–. Since it accepts electrons, HClO is the oxidizing agent.
Determine the Oxidation States of each element in the following reactions:
- \(\ce{Fe(s) + O2(g) -> Fe2O3(g)}\)
- \(\ce{Fe^{2+}(aq)}\)
- \(\ce{Ag(s) + H2S -> Ag2S(g) + H2(g)}\)
- Answer
-
- \(\ce{Fe}\) and \(\ce{O2}\) are free elements; therefore, they each have an oxidation state of 0 according to Rule #1. The product has a total oxidation state equal to 0, and following Rule #6, \(\ce{O}\) has an oxidation state of -2, which means \(\ce{Fe}\) has an oxidation state of +3. \
- The oxidation state of \(\ce{Fe}\) ions just corresponds to its charge since it is a single element species; therefore, the oxidation state is +2.
- \(\ce{Ag}\) has an oxidation state of 0, \(\ce{H}\) has an oxidation state of +1 according to Rule #5, \(\ce{H2}\) has an oxidation state of 0, \(\ce{S}\) has an oxidation state of -2 according to Rule #7, and hence \(\ce{Ag}\) in \(\ce{Ag2S}\) has an oxidation state of +1.
Determine the oxidation states of the phosphorus atom bold element in each of the following species:
- \(\ce{Na3PO3}\)
- \(\ce{H2PO4^{-}}\)
- Answer
-
- The oxidation numbers of \(\ce{Na}\) and \(\ce{O}\) are +1 and -2. Because sodium phosphite is neutral species, the sum of the oxidation numbers must be zero. Letting x be the oxidation number of phosphorus, 0= 3(+1) + x + 3(-2). x=oxidation number of P= +3.
- Hydrogen and oxygen have oxidation numbers of +1 and -2. The ion has a charge of -1, so the sum of the oxidation numbers must be -1. Letting y be the oxidation number of phosphorus, -1= y + 2(+1) +4(-2), y= oxidation number of P= +5.
Determine which element is oxidized and which element is reduced in the following reactions (be sure to include the oxidation state of each):
- \(\ce{Zn + 2H^{+} -> Zn^{2+} + H2}\)
- \(\ce{2Al + 3Cu^{2+} -> 2Al^{3+} +3Cu}\)
- \(\ce{CO3^{2-} + 2H^{+} -> CO2 + H2O}\)
- Answer
-
- Zn is oxidized (Oxidation number: 0 → +2); H+ is reduced (Oxidation number: +1 → 0)
- Al is oxidized (Oxidation number: 0 → +3); Cu2+ is reduced (+2 → 0)
- This is not a redox reaction because each element has the same oxidation number in both reactants and products: O= -2, H= +1, C= +4.
Determine what is the oxidizing and reducing agents in the following reaction.
\[\ce{Zn + 2H^{+} -> Zn^{2+} + H2} \nonumber \]
- Answer
-
The oxidation state of \(\ce{H}\) changes from +1 to 0, and the oxidation state of \(\ce{Zn}\) changes from 0 to +2. Hence, \(\ce{Zn}\) is oxidized and acts as the reducing agent. \(\ce{H^{+}}\) ion is reduced and acts as the oxidizing agent.