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Chemistry LibreTexts

Redox

Half Reactions

Before one can balance an overall redox equation, one has to be able to balance two half-equations, one for oxidation (electron loss) and one for reduction (electron gain). Collectively, oxidation and reduction are known as redox, or an electron transfer reaction. After balancing the two half-equations one can determine the total net reaction.

Each equation is balanced by adjusting coefficients and adding H2O, H+, and e- in this order:

  1. Balance the number of atoms of each element. 
  2. Balance the number of electrons transferred.
  3. Balance the total charge on reactants and products

(Note: If #1 and #2 are done correctly, #3 will follow. Thus, it serves as a means of checking your work).

Balancing Oxidation-Reduction Reactions

To solve redox reactions accurately, you must first understand how to balance chemical equations. Though this process is more difficult than normal balancing it is a required step in the process of redox reactions. One of the most accepted methods of balancing a redox reaction is known as the half-equation method, however it can become more complex when involving basic or acidic solutions. In this module, a brief introduction to this different method will be explored. For an in depth explanation see: Balancing Oxidation-Reduction Reactions.

Half-Equation Method

The half-equation method (for neutral reactions) involves three basic steps which are as follows:

  • Write and balance the half reactions.
  • Adjust coefficients in both equations so that the same number of electrons appears in each half.
  • Add together both halves, canceling out electrons, to obtain the overall equation.

Balancing in Basic and Acidic Solution

Balancing in acidic solution is similar to balancing in neutral solutions however, instead of three steps to follow, there are six. These rules are:

  • Write and balance the half reactions.
  • Balance oxygen, O, by adding with H2O
  • Balance hydrogen, H, by adding H+ (acidic)
  • Balance charge by adding electrons (you should be adding the same number of electrons as H+ ions)
  • Multiply both half reactions by some integer to cancel out electrons
  • Add the half reactions together and cancel out what appears on both sides

 

Example

Balance the redox reaction in acidic solution: 

\[ MnO_4^- + I^- \rightarrow Mn^{2+} + I_2 (s) \]

Solution

  • Write and balance the half reactions:
           MnO4- + I- --> Mn2+ + I2(s)
    O.S:  +7 -2       -1          +2          0        (Mn is reduced and I- is oxidized)

    Oxidation Rx:2I-(aq) --> I2(s) + 2e-
    Reduction Rx: MnO4- + 5e- --> Mn2+
  • Balance oxygen, O, by adding H2O
    Oxidation Rx: 2I-(aq) --> I2(s) + 2e-
    Reduction Rx: MnO4- + 5e- --> Mn2+ + 4H2O
  • Balance hydrogen, H, by adding H+
    Oxidation Rx: 2I-(aq) --> I2(s) + 2e-
    Reduction Rx: MnO4+ 5e- + 8H--> Mn2+ + 4H2O
  • Balance charge by adding electrons
    Oxidation Rx: 2I-(aq) --> I2(s) + 2e-
    Reduction Rx: MnO4- + 5e- + 8H+ --> Mn2+ + 4H2O
  • Multiply both half reactions by some integer to cancel out electrons
    (Oxidation Rx: 2I-(aq) --> I2(s) + 2e-) * 5
    (Reduction Rx: MnO4- + 5e- + 8H--> Mn2+ + 4H2O) *2
    Oxidation Rx: 10I-(aq) --> 5I2(s) + 10e-
    Reduction Rx: 2MnO4- + 10e- + 16H+ --> 2Mn2+ + 8H2O
  • Add the half reactions together and cancel out what appears on both sides:
    10I-(aq) + 2MnO4-(aq) + 16H+(aq) --> 2Mn2+(aq) + 5I2(s) + 8H2O(l)
    (Note: Don't forget the states of matter! Generally, anything with a charge is (aq) and H2O is (l))

Balancing in basic solution follows balancing in acidic solutions in three steps:

  • Balance the reaction in acidic solution
  • Add the same amount of OH- ions as H+ ions to both sides of the equation. On one side, the OH- and H+ will react to form water (H2O) in a 1:1 ratio. 
  • Cancel out water molecules appearing on both sides
Example

Balance the above redox reaction in basic solution

Solution

  • Balance the reaction in acidic solution
    10I-(aq) + 2MnO4-(aq) + 16H+(aq) --> 2Mn2+(aq) + 5I2(s) + 8H2O(l)
  • Add the same amount of OH- ions as H+ ions to both sides of the equation.
    10I-(aq) + 2MnO4-(aq) + 16H+(aq) + 16OH--> 2Mn2+(aq) + 5I2(s) + 8H2O(l) + 16OH-
    On one side, the OH- and H+ will react to form water (H2O) in a 1:1 ratio.
    10I-(aq) + 2MnO4-(aq) + 16H2O --> 2Mn2+(aq) + 5I2(s) + 8H2O(l) 16OH-
  • Cancel out the water molecules appearing on both sides
    10I-(aq) + 2MnO4-(aq) + 8H2O(l) --> 2Mn2+(aq) + 5I2(s) 16OH-(aq)
Example

Balance the following half-equation:

MnO4- → Mn2+

Solution

  1. Because there is one atom of Mn on both sides, no adjustment is required.
  2. Because manganese is reduced from an oxidation number of +7 to +2, five electrons must be added to the left (MnO4- + 5e- → Mn2+)
  3. There is a total charge of -6 on the left versus +2 on the right. To balance, add eight H+ to the left to give a charge of +2 on both sides. (MnO4- + 8H+ 5e- → Mn2+)
  4. To balance the eight H+ ions on the left, add four H2O molecules to the right. MnO4- + 8H+ + 5e- Mn2+ + 4H2O
  5. Note that there are the same number of oxygen atoms, four, on both sides, as there should be. The equation shown in green is the correctly balanced reduction half-equation.

 

True or False

  1. The oxidation state of an individual atom is always +1
  2. The oxidation agent has to be reduced
  3. Combustion reactions always involve oxygen
  4. Electrons and Hydrogen are almost never on the same side

Solutions

  1. False: is always 0
  2. True
  3. True
  4. False: electrons and hydrogen are almost always on the same side