Redox Reactions

Redox reactions are concerned with the transfer of electrons between species. Like acid-base reactions, redox reactions are a matched set -- you don't have an oxidation reaction without a reduction reaction happening at the same time. Oxidation refers to the loss of while reduction refers to the gain of electrons. Each reaction by itself is called a "half-reaction", simply because we need two (2) half-reactions to form a whole reaction. To indicate redox reactions, chemists typically write out the electrons explicitly:

Cu (s) ----> Cu²⁺ + 2 e^-

This half-reaction says that we have solid copper (with no charge) being oxidized (losing electrons) to form a copper ion with a plus-2 charge. Notice that, like the stoichiometry notation, we have a "balance" between both sides of the reaction. We have one (1) copper atom on both sides, and the charges balance as well. The symbol "e-" represent s a free electron with a negative charge that can now go out and reduce some other species, such as in the half-reaction:

2 Ag⁺ (aq) + 2 e^- ------> 2 Ag (s)

Here, a silver ion (silver with a positive charge) is being reduced through the addition of two (2) electrons to form solid silver. The abbreviations "aq" and "s" mean aqueous or solid, respectively. We can now combine the two (2) half-reactions to form a redox equation:

As a summary, here are the steps to follow to balance a redox equation in acidic medium (add the starred step in a basic medium) :

Divide the equation into an oxidation half-reaction and a reduction half-reaction
Balance these
- Balance the elements other than H and O
- Balance the O by adding H₂O
- Balance the H by adding H⁺
- Balance the charge by adding e^-
Multiply each half-reaction by an integer such that the number of e^- lost in one equals the number gained in the other
Combine the half-reactions and cancel
**Add OH^- to each side until all H⁺ is gone and then cancel again**

In considering redox reactions, you must have some sense of the oxidation number (ON) of the compound. The oxidation number is defined as the effective charge on an atom in a compound, calculated according to a prescribed set of rules. An increase in oxidation number corresponds to oxidation, and a decrease to reduction.The rules are shown below. Go through them in the order given until you have an oxidation number assigned.

For atoms in their elemental form, the oxidation number is 0.
For ions, the oxidation number is equal to their charge
For hydrogen, the number is usually +1 but in some cases it is -1
For oxygen, the number is usually -2
The sum of the oxidation number (ONs) of all the atoms in the molecule or ion is equal to its total charge.

Practice Redox Problem: balance the following redox reaction in acidic solution:

S(s) + NO₃^-(aq) --> SO₂(g) + NO(g)

Solution:

The equation is: S + NO₃^- ---> SO₂ + NO

Oxidation half-reaction:

S ---> SO₂
S + 2H₂O ---> SO₂ +4H⁺ + 4e^-

Reduction half-reaction:

NO₃^- ---> NO
NO₃^_ + 4H⁺ + 3e^- ---> NO+2H₂O

To have the same number of electrons in both half-reactions, multiply the oxidation reaction by 3 and multiply the reduction reaction by 4. After this, add the two half-reactions together and then cancel:

4NO₃^- +3S +16H⁺ +6H₂O ---> 3SO₂ + 16H⁺ + 4NO +8H₂O

4NO₃^- + 3S +4H⁺ ---> 3SO₂ + 4NO + 2H₂

Example 2: Balance the following in acid medium:

H₂O₂ + [MnO₄]^-1 --> Mn⁺² + O₂

Let's start with the hydrogen peroxide half-equation. What we know is:

The oxygen is already balanced. What about the hydrogen?

All you are allowed to add to this equation are water, hydrogen ions and electrons.

Add two hydrogen ions to the right-hand side.

Now all you need to do is balance the charges. You would have to add 2 electrons to the right-hand side to make the overall charge on both sides zero.

Now for the other half-equation:

The manganese balances, but you need four oxygens on the right-hand side. These can only come from water - that's the only oxygen-containing thing you are allowed to write into one of these equations in acid conditions.

By doing this, we've introduced some hydrogens. To balance these, you will need 8 hydrogen ions on the left-hand side.

Now that all the atoms are balanced, all you need to do is balance the charges. At the moment there are a net 7+ charges on the left-hand side (1- and 8+), but only 2+ on the right. Add 5 electrons to the left-hand side to reduce the 7+ to 2+.

This is the typical sort of half-equation which you will have to be able to work out. The sequence is usually:

Balance the atoms apart from oxygen and hydrogen.

Balance the oxygens by adding water molecules.

Balance the hydrogens by adding hydrogen ions.

Balance the charges by adding electrons.

Combining the half-reactions to make the ionic equation for the reaction

The two half-equations we've produced are:

You have to multiply the equations so that the same number of electrons are involved in both. In this case, everything would work out well if you transferred 10 electrons.

But this time, you haven't quite finished. During the checking of the balancing, you should notice that there are hydrogen ions on both sides of the equation:

You can simplify this down by subtracting 10 hydrogen ions from both sides to leave the final version of the ionic equation - but don't forget to check the balancing of the atoms and charges!

You will often find that hydrogen ions or water molecules appear on both sides of the ionic equation in complicated cases built up in this way. Always check, and then simplify where possible.

Example 3: Balance the following reaction in aqueous acid solution: MnO₄^- + H₂C₂O₄ --> Mn²⁺ + CO₂
Divide the reaction into two half-reactions. Any element (except hydrogen and oxygen, which you can get from water as needed in aqueous solutions) must appear on both sides of the same half-reaction.

MnO₄^- --> Mn²⁺;
H₂C₂O₄ --> CO₂
Balance all elements except hydrogen and oxygen by proper multiplication of compounds or ions.

MnO₄^- --> Mn²⁺;
H₂C₂O₄ --> 2CO₂
Balance oxygen by adding water, H₂O, as necessary.

MnO₄^- --> Mn²⁺ + 4H₂O;
H₂C₂O₄ --> 2CO₂
Balance hydrogen by adding hydrogen ions, H⁺, as necessary.

8H⁺ + MnO₄^- --> Mn²⁺ + 4H₂O;
H₂C₂O₄ --> 2CO₂ + 2H⁺
Balance charge by adding electrons, e^-, as necessary.

8H⁺ + MnO₄^- + 5e^- --> Mn²⁺ + 4H₂O;
H₂C₂O₄ --> 2CO₂ + 2H⁺ + 2e^-
The two half-reactions (electrode reactions) are balanced separately at this point, and now must be combined. To do so, multiply each equation by the number of electrons appearing in the other.

16H⁺ + 2MnO₄^- + 10e^- --> 2Mn²⁺ + 8H₂O;
5H₂C₂O₄ --> 10CO₂ + 10H⁺ + 10e^-
To complete the combination, add the two equations together, then cancel out anything appearing on both sides. The number of electrons cancelled in this step is the number of electrons transferred in the redox reaction.
16H⁺ + 2MnO₄^- + 10e^- + 5H₂C₂O₄ --> 2Mn²⁺ + 8H₂O + 10CO₂ + 10H⁺ + 10e^- or

6H⁺ + 2MnO₄^- + 5H₂C₂O₄ --> 2Mn²⁺ + 8H₂O + 10CO₂
At this point the redox reaction is balanced; no electrons should be left in the reaction. It is also desirable to check that the equation can no longer be simplified by dividing it through by a simple integer such as 2 or 3. No species should appear on both sides of the equation.

If you are required to balance an equation in basic solution rather than acidic solution, proceed in the manner above. After balancing the reaction as if it took place in acidic solution, realize that protons are not readily available in basic solutions. So remove any protons by:
adding to both sides of the equation a certain number of hydroxide ions, OH^-; that number equals the number of protons, H⁺, appearing on the only side that has any.

6H⁺ + 2MnO₄^- + 5H₂C₂O₄ + 6OH^- --> 2Mn²⁺ + 8H₂O + 10CO₂ + 6OH^-
combining, when they appear on the same side of the equation, hydrogen ions (protons) and hydroxide ions to form water.

6H₂O + 2MnO₄^- + 5H₂C₂O₄ --> 2Mn²⁺ + 8H₂O + 10CO₂ + 6OH^-
removing any water which appears on both sides of the final equation.

2MnO₄^- + 5H₂C₂O₄ --> 2Mn²⁺ + 2H₂O + 10CO₂ + 6OH^-
This procedure, balancing aqueous redox equations first as if they were in acidic solution and then converting formally to a basic solution, will always give a formally correct balanced equation for a redox reaction in basic aqueous solutions.

Example 4: Balance the following in Alkaline Medium

MnO₄^-   +    I^-   -------->   MnO₂     +    I₂   (basic solution)
Step 1. Break into half-reactions:
MnO₄^-    ------->    MnO₂
          I^-     ------->     I₂
Step 2. Balance atoms other than H and O
MnO₄^-    ------->    MnO₂
      2 I^-     ------->     I₂
Step 3. Balance O by adding H₂O
MnO₄^-    ------->   MnO₂   +   2 H₂O
      2 I^-     ------->     I₂
Step 4. Balance H by adding H⁺
4 H⁺ + MnO₄^-    ------->    MnO₂    +   2 H₂O
                  2 I^-     ------->     I₂
Step 5. Balance charge by adding electron(s)
3 e^- + 4 H⁺ + MnO₄^-    ------->   MnO₂       + 2 H₂O
                               2 I^-     ------->     I₂   +    2 e^-
Step 6. Electrons lost = electrons gained
(3 e^- + 4 H⁺ + MnO₄^-    ------->    MnO₂       + 2 H₂O) x 2
(                             2 I^-     ------->     I₂   +    2 e^- ) x 3
Gives:
6 e^- + 8 H⁺ + 2 MnO₄^-    ------->   2 MnO₂       + 4 H₂O
          6 I^-      ------->     3 I₂   +   6 e^-
Step 7. Cancel like terms and add half reactions
8 H⁺ + 2 MnO₄^-   +   6 I^-       ------->   2 MnO₂       + 4 H₂O + 3 I₂
Step 8. Since it is a basic solution, add OH^- for each H⁺ (add OH^- to both sides of eqn)
8 OH^-   +    8 H⁺ + 2 MnO₄^-   +   6 I^-       ------->   2 MnO₂    + 4 H₂O + 3 I₂   +   8 OH^-
9. Combine H⁺ with OH^- to form H₂O
8 OH^-   +    8 H⁺ + 2 MnO₄^-   +   6 I^-       ------->   2 MnO₂   + 4 H₂O +   3 I₂   +   8 OH^-
Gives:
      8 H₂O   +    2 MnO₄^-   +   6 I^-       ------->   2 MnO₂   +   4 H₂O + 3 I₂   +   8 OH^-
10. Cancel like terms:
   4 H₂O   +    2 MnO₄^-   +   6 I^-       ------->   2 MnO₂    + 3 I₂   +   8 OH^-

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