Oxidation-Reduction Reactions

Oxidation Reduction Reactions In chemical reactions, electrons are transferred from one atom to another. Some atoms in a molecule gain electrons while other atoms lose electrons Referred to as redox reactions

Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

Oxidation A species is oxidized when it loses electrons. Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.

Reduction A species is reduced when it gains electrons. Here, each of the H+ gains an electron, and they combine to form H2.

Oxidation and Reduction Oxidation and Reduction always occur together The total number of electrons lost by substances is always the same as the total number gained others Oxidation – loss of electrons Reduction – gain of electrons LEO goes GER

Oxidation and Reduction The oxidizing agent is the substance that is reduced. H+ oxidizes Zn by taking electrons from it. The reducing agent is the substance that is oxidized. Zn reduces H+ by giving it electrons.

Hierarchy of Rules for Assigning Oxidation Numbers Oxidation numbers must add up to charge on molecule, formula unit or ion Atoms of free elements have oxidation numbers of zero Metals in Groups 1A, 2A, and Al have +1, +2, and +3 oxidation numbers, respectively H and F in compounds have +1 and –1 oxidation numbers, respectively Oxygen has –2 oxidation number Group 7A elements have –1 oxidation number

Hierarchy of Rules for Assigning Oxidation Numbers Group 6A elements have –2 oxidation number Group 5A elements have –3 oxidation number When there is a conflict between two of these rules or ambiguity in assigning an oxidation number, apply rule which comes first and ignore conflicting rule Oxidation State Used interchangeably with oxidation number Indicates charge on monatomic ions Iron(III) means +3 oxidation state of Fe or Fe3+

Oxidation–Reduction Reactions Involves 2 processes: Oxidation = Loss of electrons Na  Na+ + e– Oxidation Half-Reaction Reduction = Gain of electrons Cl2 + 2e–  2Cl– Reduction Half-Reaction Net reaction: 2Na + Cl2  2Na+ + 2Cl– Oxidation and reduction always occur together Can't have one without the other

An Example Redox Reactions Net: 2Mg + O2  2MgO Mg  Mg2+ + 2e – Oxidation: Mg  Mg2+ + 2e – Loses electrons = oxidized Reducing agent Reduction: O2 + 4e–  2O2– Gains electrons = reduced Oxidizing agent

Problem Which species functions as the oxidizing agent in the following oxidation-reduction reaction? Zn(s) + Pt2+(aq)  Pt(s) + Zn2+(aq) Pt(s) Zn2+(aq) C. Pt2+(aq) D. Zn(s) E. None of these, as this is not a redox reaction.

Your Turn! Which species gets oxidized in the following reaction? 2Ag+(aq) + Zn(s)  Zn2+(s) + 2Ag(s) Ag(s) Ag+(aq) C. Zn2+(aq) D. Zn(s) E. None of these, as this is not a redox reaction

Ex. 1 Assigning Oxidation Number Li2O Li (2 atoms) × (+1) = +2 (Rule 3) O (1 atom) × (–2) = –2 (Rule 5) sum = 0 (Rule 1) +2 –2 = 0 so the charges are balanced to zero CO2 C (1 atom) × (x) = x O (2 atoms) × (–2) = –4 (Rule 5) sum = 0 (Rule 1) x – 4 = 0 or x = +4 C is in +4 oxidation state

Learning Check Assign oxidation numbers to all atoms: Example 1: ClO4– O (4 atoms) × (–2) = –8 Cl (1 atom) × (–1) = –1 (molecular ion) sum ≠ –1 (violates Rule 1) Rule 5 for oxygen comes before Rule 6 for halogens Cl (1 atom) × (x) = x sum = –1 (Rule 1) –8 + x = –1 or x = 8 – 1 So x = +7; Cl is oxidation state +7

Assign oxidation states to all atoms: MgCr2O7 Learning Check Assign oxidation states to all atoms: MgCr2O7 Mg =+2; O = –2; and Cr = x (unknown) [+2] + [2x] + [7 × (–2)] = 0 2x – 12 = 0 x = +3 Cr is oxidation number of +3 KMnO4 K =+1; O = – 2; so Mn = x [+1] + [x] + [4 × (–2)] = 0 x – 7 = 0 x = +7 Mn is oxidation number of +7

Your Turn! What is the oxidation number of each atom in H3PO4? A. H = –1; P = +5; O = –2 B. H = 0; P = +3; O = –2 C. H = +1; P = +7; O = –2 D. H = +1; P = +1; O = –1 E. H = +1; P = +5; O = –2 +1 -2 H3PO4 +5

Your Turn! What is the oxidation number of each atom in sodium chlorate? A. Na = +1; Cl = +5; O = –2 B. Na = +1; Cl = +7; O = –4 C. Na = +1; Cl = +7; O = –2 D. Na = +1; Cl = +3; O = –1 E. Na = -1; Cl = +5; O = –2 +1 -2 NaClO3 +5

Your Turn! What are the oxidation numbers of potassium and iodine in potassium triiodide, KI3? A. K = +1; I = 0 B. K = +1; I = -1 C. K = -1; I = -3 D. K = +1; I = –1/3 E. K = +1; I = –2/3 So I3 is -1 total +1 Therefore, -1/3 for each I atom KI3 Oxidation numbers can be fractions because they represent an average number of ‘excess’ electrons (or lack thereof) on the atoms – in this case, one extra electron on the three iodide atoms

Your Turn! Assign oxidation numbers to all atoms in the following reaction and use them to determine which species gets reduced. +2 -2 -1/3 +2.5 -2 -1 2S2O32- + I3–  S4O62- + 3I– A. I3– B. I– C. S4O62- D. S2O32- I is reduced from -1/3 to -1