Redox Reactions

Oxidation-reduction reactions (abbreviated as redox reactions) involve the transferring of electrons from one atom to another, changing the oxidation state of atoms as they chemically react. 

• The reducing agent (electron donor) gets oxidized, and the oxidizing agent (electron acceptor) gets reduced. Electrons travel from reducer → oxidizer. 

Introduction to Oxidation States 

Oxidation State: Hypothetical numbers labelled to elements as a guideline for predicting electron transfer patterns in a redox reaction. 

Charge Number: The actual charge of an atom or ion. 

TIP: Usually, representative elements (s and p block, especially groups 1, 2, and 17) have the same oxidation state as their typical ionic charge. 

Oxidation Number Rules: 

• Electrically-Neutral Compounds: Oxidation states add up to 0. 

• Ions: Oxidation states add up to their charge. 

• Hydrogen: Carries an oxidation number of +1. 

  • When bonding to less electronegative elements, it carries an oxidation number of -1. 

• Fluorine: Carries an oxidation number of -1. 

• Oxygen: Carries an oxidation number of -2. 

  • Oxygen carries an oxidation number of -1 in peroxides (O₂²⁻)

  • When bonded to fluorine, oxygen carries an oxidation number of +2. 

• Alkali metals carry an oxidation number of +1 in compounds. 

• Alkaline earth metals carry an oxidation number of +2 in compounds. 

Assigning Oxidation Numbers to Compounds

• Example #1: ClO₂ (Chlorine Dioxide)

1. The oxidation state of O is -2, and the molecule is neutral in charge. Let the oxidation state of Cl be represented as x. 

2. By adding up the oxidation states, we solve for x from: x + 2(−2) = 0 

3. x = 4, so the oxidation state of Cl is +4. 

• Example #2: NH₃ (Nitrogen Trihydride)

  1. The oxidation state of H is +1, and the molecule is neutral in charge. Let the oxidation state of N be represented as x. 

  2. By adding up the oxidation states, we solve for x from: x + 3(+1) = 0 

  3. x = -3, so the oxidation state for N is -3. 

• Example #3: KMnO₄ (Potassium Permanganate)

  1. The oxidation state of O is -2, the oxidation state of K is +1, and the compound is neutral in charge. Let the oxidation state of Mn be represented as x. 

  2. By adding up the oxidation states, we solve for x from: (+1) + x + 4(−2) = 0 

  3. x = 7, so the oxidation state for Mn is +7. 

Identifying Oxidation and Reduction 

TIP: When an atom is oxidized, its oxidation state increases (gains a more positive value). When an atom is reduced, its oxidation state decreases (gains a more negative value). 

• Example #1: 2Na + F₂ → 2NaF

  1. The oxidation state of Na atoms in 2Na is 0 since it is elemental Na.

  2. The oxidation state of F atoms in F₂ is 0 since it is elemental F.

  3. In 2NaF, the oxidation state of F atoms are -1.

     1. To balance the overall value of 2NaF to 0, the oxidation state of Na atoms in 2NaF must be +1.

  4. Na goes from 0 to +1: oxidized. 

  5. F goes from 0 to -1: reduced. 

• Example #2: Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

  1. The oxidation state of Cu atoms in Cu is 0 since it is elemental Cu.

  2. The oxidation state of Ag atoms in 2AgNO₃ is +1. 

  3. The oxidation state of NO₃ stays as -1 throughout the equation. 

  4. The oxidation state of Cu atoms in Cu(NO₃)₂ is +2, since it is bonded to two NO₃ in this neutral compound. 

  5. The oxidation state of Ag atoms in 2Ag are 0 since it is elemental Ag. 

  6. Cu goes from 0 to +2: oxidized. 

  7. Ag goes from +1 to 0: reduced. 

• Example #3: 2H₂O₂ → 2H₂O + O₂

  1. The oxidation state of H atoms in 2H₂O₂ is +1. 

  2. The oxidation state of O atoms in 2H₂O₂ is -1, since it is in peroxide form. 

  3. In 2H₂O, the oxidation state of H atoms are +1. 

     1. To balance the overall value of 2H₂O to 0, the oxidation state of O atoms in 2H₂O must be -2. 

  4. The oxidation state of O atoms in O₂ is 0, since it is elemental O. 

  5. O goes from -1 to 0: oxidized. 

  6. O goes from -1 to -2: reduced.