Oxidation and Reduction Reactions

Oxidation & Reduction

Development of Scientific Understanding

• Outline the development of scientific understanding of oxidation and reduction reactions. Include:
  • Gain and loss of electrons
  • Oxidizing agent
  • Reducing agent
• Determine the oxidation numbers for atoms in compounds and ions.

Historical Definitions

• Oxidation: The combining of oxygen with other elements.
  • Example: Burning a substance involves adding $O_2$ from the air to the elements being burned.
• Reduction: The removal of oxygen from a compound.
  • Implies a decrease (or reduction) in the mass of the material being reduced.

Redox Reaction Example

• Example:
  $Mg(s) + O_2(g) \rightarrow 2 MgO(s)$
• Ionic form: $2 Mg^0 + O_2^0 \rightarrow 2 (Mg^{2+} O^{2-})$
  • Mg is oxidized to MgO.
  • Mg changes from a 0 to a 2+ charge.
  • $O_2$ changes from a 0 to a 2- charge.

Redox Reactions

• "Redox" reaction: electrons are transferred (lost or gained) between reactants.
• Two chemical reactions (oxidation & reduction) occur simultaneously.
• The electrons lost by the oxidized atom must be gained by the reduced one.
• Example: Electrons are transferred from hydrogen to fluorine.

Oxidation Reaction

• Oxidation: The process by which electrons are removed from an atom/ion.
  • The charge of the atom/ion increases (becomes more positive).
• Example:
  $2 Mg^0 \rightarrow Mg^{2+} + 2 e^{-}$
• Notes:
  • When an atom goes from a 0 charge to a positive charge, electrons have been removed.
  • When an atom or ion becomes more positively charged, oxidation has taken place.

Oxidation Reaction Examples

• Example 1: A rusty chain (a slow oxidation reaction)
  • Iron slowly turns to rust:
    $2 Fe(s) + 3 O<em>2(g) \rightarrow 2 Fe</em>2O_3(s)$
  • Ionic form:
    $2 Fe^0 + 3 O<em>2^0 \rightarrow 2 (Fe^{3+}</em>2 O^{2-}_3)$
  • Iron oxidizes (gains oxygen) to form iron(III) oxide; a charge of 0 to a charge of 3+; three electrons were lost/removed; has three more protons than electrons, resulting in a charge of 3+.
• Example 2: A bonfire, a fast oxidation reaction.
  • Carbon burns in air:
    $C(s) + O<em>2(g) \rightarrow CO</em>2(g)$
  • Ionic form:
    $C^0 + O<em>2^0 \rightarrow C^{4+} O^{2-}</em>2$
  • Carbon oxidizes (gains oxygen) to form carbon dioxide; a charge of 0 to a charge of 4+; four electrons were lost/removed; has four more protons than electrons, resulting in a charge of 4+.

LEO Acronym

• LEO: Loss of Electrons is Oxidation!

Reduction Reaction

• Reduction: The process by which electrons are added to an atom or ion.
  • The charge of the atom or ion decreases (becomes more negative).
• Example:
  $2 Mg^0 + O_2^0 \rightarrow 2 (Mg^{2+}O^{2-})$
• Isolating the oxygen:
  $O_2^0 + 4e^- \rightarrow 2 O^{2-}$
• Notes:
  • When an atom goes from a 0 charge to a negative charge, electrons are gained.
  • When an atom/ion becomes more negatively charged, reduction has taken place.
  • The number of electrons lost must equal the number of electrons gained.
  • The charge must be conserved on both sides of the equation.

Reduction Reaction Example

• Example 3: Iron ore is converted to iron in steelmaking to build bridges.
  • Iron ore is reduced:
    $2 Fe<em>2O</em>3(s) + 3 C(s) \rightarrow 4 Fe(s) + 3 CO_2(g)$
  • Ionic form:
    $2 (Fe^{3+}<em>2O^{2-}</em>3) + 3 C^0 \rightarrow 4 Fe^0 + 3 (C^{4+}O^{2-}_2)$
  • Iron went from a charge of 3+ to a charge of 0; three electrons were gained; has an equal number of protons and electrons, resulting in a charge of 0.

GER Acronym

• GER: Gaining Electrons is Reduction!
• Putting together the two acronyms: LEO the lion says GER!

Assigning Oxidation Numbers

• Oxidation number: A positive or negative number assigned to an atom.
  • Values assigned to individual atoms within molecules/ions for the purpose of tracking ownership of electrons.
• During oxidation and reduction, a charge can become more "+" or more "–".

Notes on Oxidation Numbers:

• Ionic charge is written as 2+.
• An oxidation number is written as +2.
• A change in the oxidation number of an atom or ion indicates a redox reaction has occurred.
• Oxidation numbers are also used to help balance the redox reaction.

Oxidation Number Rules

• Rule 1: The oxidation number of any elemental atom is 0.
  • Ex: C = 0, $H<em>2$ = 0, $O</em>2$ = 0
• Rule 2: An ion’s oxidation number is its charge.
  • Ex: $Na^+$ = +1, $S^{2-}$= −2, $CO_3^{2-}$= −2
• Rule 3: Hydrogen’s oxidation number is +1, except in metal hydrides where H is the anion (e.g., $CaH_2$ or LiH) and it is –1.
• Rule 4: The oxidation number of oxygen is –2, except in peroxides (i.e., $H<em>2O</em>2$, $Na<em>2O</em>2$) where it is −1 and in combination with fluorine (O = +2).
• Rule 5: In a compound, the sum of all of the oxidation numbers of each part must equal the total charge of that compound or complex ion.

Applying Oxidation Number Rules

• Samples: NaCl, $CaCl<em>2$, $SO</em>4^{2-}$
• Note: Some atoms have more than one oxidation state.
  • First assign oxidation numbers for atoms that have rules.
  • Then, follow rules 2 and 5 for the other atoms.

Oxidation Number Examples

• NaCl: +1 -1
• $CaCl_2$: +2 -1
• $SO_4^{2-}$: +6 -2
• Multiplied:
  • -2
  • -8
• Overall:
  • 0
  • 0
  • -2
• Note: Some atoms have more than one oxidation state.
  • First assign oxidation numbers for atoms that have rules.
  • Then, follow rules 2 and 5 for the other atoms.

Example 1: Determining Oxidation Number for N in $HNO_3$

• Step 1 – Set up a table for the elements: H N $O_3$
• Step 2 – Rule 3 tells us that the oxidation number of $H^+$ = +1 and rule 4 tells us that O = -2. These numbers can be written above the atoms in the table.
  • +1 -2
  • H N $O_3$
• Step 3 – The total oxidation number is calculated on the bottom by multiplying the oxidation number by the number of each atom present (i.e., for H, +1 x 1 atom = +1; for O, -2 x 3 atoms = -6).
  • +1 -2
  • H N $O_3$
  • +1 -6
• Step 4 – Use rule 5 (the sum of the oxidation numbers in the bottom row must equal the total charge of the compound). The sum must equal 0 because this is a neutral molecule. You can use a variable “x” to represent the total oxidation number of N:
  • +1 -2
  • H N $O_3$
  • +1 x - 6
• Solve for x:
  • (+1) + x + (-6) = 0
  • x = +5
  • Since there is only one N atom, the oxidation number of N must be +5.
  • +1 +5 -2
  • H N $O_3$
  • +1 x = +5 - 6

Example 2: Determine the oxidation number for P in $Na<em>3PO</em>4$

• +1 +5 -2
• $Na<em>3$ P $O</em>4$
• +3 +5 -8
• Rule 2 tells us that the oxidation number of $Na^+$ = +1 and rule 4 tells us that O = –2. These numbers can be written above the atoms in the table.
• The total oxidation number is calculated on the bottom by multiplying the oxidation number by the number of each atom present (i.e., for Na, +1 x 3 atoms = +3; for O, –2 x 4 atoms = –8).
• Next, rule 5 tells us the sum of the oxidation numbers in the bottom row must be zero because this is a neutral molecule. This means: (+3) + x + (-8) = 0. Therefore, x = +5. Since there is only one P atom, the oxidation number of P must be +5.

Example 3: Determine the oxidation number for Cr in $Cr<em>2O</em>7^{2-}$

• -2 +6
• $Cr<em>2O</em>7$
• -14 +12
• This is a complex ion with an overall charge of 2–. This time, the bottom oxidation numbers must total –2.
• Rule 4 tells us O = –2. Using to rule 5, (–14) + x = –2, therefore x = +12.
• There are 2 Cr atoms, therefore the oxidation number of each Cr must be +6.