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:

Atoms in free elements have the oxidation number zero. For diatomics, each element has the oxidation number of zero.
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.

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.

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.

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 CO + Zn and see which atom is oxidized and which is reduced.

Write the OxN below each atom to get:

         Zn O + C  C O + Zn
        +2 -2   0     +2 -2  0

reducing agent

oxidizing agent

         S + O₂  S O₂
         0   0     +4 -2

       3Zn S  +  8 H N O₃  Zn S O₄ + 8 N O + 4H₂O
        +2 -2     +1+5-2      +2 +6-2    +2-2   +1-2

Here 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:

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

Example:

Step 1
Cr⁺²(aq) + I₂  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  I^-1
Oxidation: Cr⁺²  Cr⁺³

Step 3 Mass balance

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

Step 4 Charge balance

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

Step 5 Each half reaction must have same number of electrons

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

Step 6 Add the half reactions

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

Step 7 Check result for mass and charge balance

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) 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^- NO₂ oxidation: Cu 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^- NO₂ + H₂O oxidation: Cu 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⁺ NO₂ + H₂O oxidation: Cu Cu²⁺

Step 4 Balance the charges

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

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

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

Step 6 Add the half reactions

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

2NO₃^- + 4 H⁺ + Cu Cu²⁺ + 2NO₂ + 2H₂O

Step 7 Check it out

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

Here 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)  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^-  ClO₃^-
oxidation: Cr  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^-  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  ClO₃^- + 2 OH^-
oxidation: Cr + 3 OH^-  Cr(OH)₃

Step 4. Balance half reactions for charge

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

Step 5. Get same number of electrons in each reaction

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

Step 6. Add both half reactions

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

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

Step 7. Check it out

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

Practice, practice, practice.

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