Oxidation Reduction Reactions

Oxidation/Reduction Reactions

In these reactions some species is oxidized (gives away electrons) and some other species is reduced (accepts electrons). A first-grade method of remembering which is which is LEO says GER. LEO stands for "loose electrons oxidized" and GER stands for "gains electrons reduced".

To recognize oxidation-reduction reactions, we have to look for changes in the oxidation numbers of the elements. Thus we have to be able to determine the oxidation numbers of the elements. So we have some rules to learn:

  1. Atoms in free elements have the oxidation number zero. For diatomics, each element has the oxidation number of zero.
  2. Ions of single atoms involved in ionic compounds have the oxidation number of the charge: Cl- has the oxidation number of -1, Al+3 has the oxidation number of +3, etc. Since we know the charges of the ions from the periodic table so we will know the oxidation numbers of the ions in ionic compounds.
  3. There are two special cases involving H and O when in a compound: the oxidation number of H is +1 and the oxidation number of O is -2. There are a few exceptions:
    a.) in metal H compounds the oxidation number of H is -1.
    Example: NaH (oxidation number of Na is +1, oxidation number of H is -1.
    b.) in peroxides the oxidation number of O is -1.
    Example: H2O2(the oxidation number of H is +1
    and the oxidation number of O is -1.
  4. the sum of the oxidation numbers in neutral compounds must be zero. The sum of the oxidation numbers in polyatomic ions must equal the charge of the polyatomic ion. Example of a neutral compound: Li2O, O has oxidation number of -2 and Li has oxidation number of +1.
  5. Using Rule 2 we know that the oxidation states of ions in compounds according to the columns of the periodic table are IA -> +1, IIA -> +2, IIIA -> +3, VIA -> -2, VIIA -> -1 (except for exceptions: Tl can have +3 and +1; Cl, Br, and I are usually -1 but can have a number of other OxN; S is often -2 but can have +6, +4, +2; Se is usually -2 but can be +6 or +4; Te is usually -2 but can be +6 or +4). In compounds and polyatomic ions we then use the OxN of the ones that we know to calculate the OxN of the ones we don't know.

    So let's look at some examples:

    H3PO4 is neutral and we know that the OxN of H is +1 and the OxN of O is -2 which gives the charges of 3*(+1) = +3 and 4*(-2) = -8 so that the total charge contributed by H and O is -5. Therefore the OxN of P has to be +5 in order for the compound to be neutral.

    MnO4-1 has an overall charge of -1. We know the OxN of O is -2 which gives a total of 4*(-2) = -8, so Mn must have an OxN of +7.

    S2O3-2 has an overall charge of -2. We know that the OxN of O is -2 for a total charge of 3*(-2) = -6 so the oxidation number of S must be +2.

    HNO3 is neutral. The OxN of O is -2 which gives a total of -6, H is +1 which gives a total of +1, so the sum of H and O is -5 and thus the OxN of N must be +5.

    Now we are ready to recognize some Ox-Red reactions (often called Redox reactions). Look at

    ZnO + C arrow CO + Zn and see which atom is oxidized and which is reduced.

    Write the OxN below each atom to get:

             Zn O + C arrow C O + Zn
            +2 -2   0     +2 -2  0
    
    So you see that C looses two electrons in the reaction and Zn gains two electrons. Therefore C is oxidized and Zn is reduced. Whatever is oxidized is called the reducing agent and whatever is reduced is called the oxidizing agent. Except in rare instances, metals acting alone are reducing agents and so are oxidized. You see the reverse of the above reaction has Zn acting alone and you see that in that reverse reaction Zn is oxidized and hence is the reducing agent.

             S + O2 arrow S O2
             0   0     +4 -2
    
    So S is oxidized (0 to +4) and oxygen is reduced (0 to -2).

           3Zn S  +  8 H N O3 arrow Zn S O4 + 8 N O + 4H2O
            +2 -2     +1+5-2      +2 +6-2    +2-2   +1-2
    
    Here we see that N is reduced and S is oxidized.

    helpHere is a tutoring site on determining oxidation states (or charges).

    Practice, practice, practice.

    Balancing Redox reactions

    It is best to break the process into the following steps:

    1. First recognize as an oxidation-reduction reaction by determining which species are reduced and which are oxidized.
    2. Break the overall reaction into an oxidation reaction and a reduction reaction (called half reactions)
    3. Use necessary stoichiometric coefficients to obtain mass balance of each reaction
    4. Achieve charge balance of each reaction by adding electrons appropriately
    5. Multiply both sides of each reaction by the appropriate number to achieve charge balance for each reaction
    6. Add together the two half reactions
    7. Check the resulting equation for mass and charge balance

    Example:

    Step 1
    Cr+2(aq) + I2 arrow Cr+3(aq) + I-1(aq)
    +2         0      +3        -1
    

    We see that Cr is oxidized and I is reduced so this is a Redox reaction.

    Step 2

    Reduction: I2 arrow I-1 Oxidation: Cr+2 arrow Cr+3

    Step 3 Mass balance

    Reduction: I2 arrow 2I-1 Oxidation: Cr+2 arrow Cr+3

    Step 4 Charge balance

    Reduction: I2 +2e- arrow 2I-1 Oxidation: Cr+2 arrow Cr+3 + e-

    Step 5 Each half reaction must have same number of electrons

    Reduction: I2 +2e- arrow 2I-1 Oxidation: 2Cr+2 arrow 2Cr+3 + 2e-

    Step 6 Add the half reactions

    I2 +2e- arrow 2I-1 Cr+2 arrow Cr+3 + e- I2 + 2 Cr2+ arrow 2 Cr3+ + 2I-1

    Step 7 Check result for mass and charge balance

    Both balance so we are through.

    You might think that was very easy! Well, it gets more complicated if we are in acid or basic solutions.

    Acid Solutions

    To demonstrate, let's work an example in an acid solution.

    Cu(s) + NO3-(aq) arrow NO2(g) + Cu2+(aq)

    Step 1
    

    We see that copper is oxidized and nitrogen is reduced so this is a Redox reaction.

    Step 2 Set up half reactions

    reduction: NO3- arrow NO2 oxidation: Cu arrow Cu2+

    Step 3 Balance by mass (now it becomes different)

    We see that the first reaction needs balancing for oxygen. We balance oxygen atoms in acidic solution by adding H2O to the side deficient in oxygen:

    reduction: NO3- arrow NO2 + H2O oxidation: Cu arrow Cu2+

    Now we balance the hydrogens by adding H+'s to the side deficient in H's (we have the H+'s because this is an acidic solution.)

    reduction: NO3- + 2 H+ arrow NO2 + H2O oxidation: Cu arrow Cu2+

    Step 4 Balance the charges

    reduction: NO3- + 2 H+ + e- arrow NO2 + H2O oxidation: Cu arrow Cu2+ + 2e-

    Step 5 Get each half-reaction to have the same number of electrons

    reduction: 2NO3- + 4 H+ + 2e- arrow 2NO2 + 2H2O oxidation: Cu arrow Cu2+ + 2e-

    Step 6 Add the half reactions

    reduction: 2NO3- + 4 H+ + 2e- arrow 2NO2 + 2H2O oxidation: Cu arrow Cu2+ + 2e-

    2NO3- + 4 H+ + Cu arrow Cu2+ + 2NO2 + 2H2O

    Step 7 Check it out

    All checks out!

    So in acid solution, balance the oxygens with waters and then balance the hydrogens with H+'s.

    helpHere is a tutoring site on balancing Redox reactions in acid solution

    How about in Basic Solution? Good question.

    Step 1.  Recognize as Redox
    

    Cr(s) + ClO4-(aq) arrow Cr(OH)3(s) + ClO3-(aq)

    We see that chromium is oxidized and chlorine is reduced so it is a Redox reaction.

    Step 2 Set up half reactions

    reduction: ClO4- arrow ClO3- oxidation: Cr arrow Cr(OH)3

    Step 3. Mass Balance (Cl and Cr ok)

    For every oxygen atom needed, add 2 OH- to the side needing oxygen and add 1 H2O to the other side of the equation since 2 OH- arrow O-2 + H2O unless you have a half reaction like the oxidation reaction above where a OH- ion has appeared. Since this is a basic solution, the oxidation reaction can be mass balanced by adding 3 hydroxide ions to the reactant side. So we have

    reduction: ClO4- + H2 arrow ClO3- + 2 OH- oxidation: Cr + 3 OH- arrow Cr(OH)3

    Step 4. Balance half reactions for charge

    reduction: ClO4- + H2 +2 e-arrow ClO3- + 2 OH- oxidation: Cr + 3 OH- arrow Cr(OH)3 + 3 e-

    Step 5. Get same number of electrons in each reaction

    reduction: 3ClO4- + 3H2 + 6 e-arrow 3ClO3>- + 6 OH- oxidation: 2Cr + 6 OH- arrow2 Cr(OH)3 + 3 e-

    Step 6. Add both half reactions

    reduction: 3ClO4- + 3H2 + 6 e-arrow 3ClO3>- + 6 OH- oxidation: 2Cr + 6 OH- arrow2 Cr(OH)3 + 3 e-

    3 ClO4- + 3 H2O + 2 Cr arrow 2 Cr(OH)3 + 3 ClO3-

    Step 7. Check it out

    It is ok so we are through.

    helpHere is a tutoring site on balancing Redox reactions in basic solution.

    Practice, practice, practice.

    Now take a practice quiz to help you understand if you understand the basic concepts.
    You must use your real name when it asks for a name.
    The test will only submit when you have answers all of the questions correctly.
    If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests.


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    Web Author: Dr. Leon L. Combs
    Copyright 1999 by Dr. Leon L. Combs - ALL RIGHTS RESERVED