REDOX electrochemistry

Redox reactions involve the transfer of the electron. Spontaneous redox reactions can transfer energy Electrons (electricity) Heat Non-spontaneous redox reactions can be made to happen with electricity.

Trends in Oxidation and Reduction Active metals: Lose electrons easily Are easily oxidized Are strong reducing agents Active nonmetals: Gain electrons easily Are easily reduced Are strong oxidizing agents

Oxidation - is defined as the loss of an electron (LEO - loss of electron; oxidation) when oxidation happens, charge becomes more positive Fe → Fe2+ + 2e- 4 Fe (s) + 3 O2 (aq) → 2 Fe2O3 (s)

Reduction is defined as the gain of an electron (GER- gain of electrons – reduction when reduction happens, charge becomes more negative Fe2+ + 2e- → Fe term is derived from the observation that metal oxides lose mass when refined to produce the pure metal: 2 Fe2 O3 (s) → 4 Fe (s) + 3 O2 (aq)

N2O → NO Has the nitrogen gained or lost electrons? N2O → NO +1 +2 Nitrogen has lost one electron and is therefore being oxidized.

Each sodium atom loses one electron: Lose Electrons = Oxidation Each chlorine atom gains one electron: Gain Electrons = Reduction

Just like the Bronsted-Lowry theory of acids, oxidation and reduction also happens in pairs; a species cannot donate an electron unless another species gains the electron. Since oxidation and reduction always happen together, these are most often called redox reactions.

These pairs of reactions are called half-reactions These pairs of reactions are called half-reactions. For example, in the reaction Cu2+ (aq) + Fe (s) → Cu (s) + Fe2+ (aq) two half-reactions occur; iron is oxidized while the copper (II) ion is reduced. Seen another way, the copper (II) ion causes the oxidation of the iron; it is the oxidizing agent. Likewise, the iron is the reducing agent for the copper (II) ion.

Reducing and oxidizing agents The substance reduced is the oxidizing agent The substance oxidized is the reducing agent Sodium is oxidized – it is the reducing agent Chlorine is reduced – it is the oxidizing agent

There are five basic rules for the determination of oxidation number: Rule 1: The oxidation number for any atom in its elementary state is 0. Rule 2: The oxidation number for any simple ion is the charge on the ion. a. The oxidation number of alkali metals in compounds is 1+ (Li1+ , Na1+ , K1+ , Rb1+ , Cs1+ , Fr1+ ). b. The oxidation number of alkaline-earth metals in compounds is 2+ (Mg2+ , Ca2+ , and Ba2+ ).

Find the oxidation number for magnesium and chlorine

Rule 3:. The oxidation number for oxygen usually is 2- Rule 3: The oxidation number for oxygen usually is 2-. In peroxides, it is 1-. water peroxide

Rule 4: The oxidation number for hydrogen is 1+ in all its compounds except in metallic hydrides like NaH or BaH2 , where it is 1-. Rule 5: All other oxidation numbers are assigned so that the sum of oxidation numbers equals the net charge on the molecule or polyatomic ion.

Find the oxidation numbers H2S Ca(OH)2 2(+1) + (-2) = 0 H O (+2) + 2(-2) + 2(+1) = 0 Ca O H

Find the oxidation numbers S O X + 3(-2) = -1 N O  X = +5  X = +6

Find the oxidation number of Cr: Ex: Cr2O72- 2X + 7(-2) = -2 Cr O  X = +6

Write the oxidation numbers for each atom: SO2 S2O32- SO42- MgSO4 H2SO4 S = +4 and O = -2 S = +2 and O = -2 S = +6 and O = -2 Mg = +2, S = +6 and O = -2 H = +1, S = +6 and O = -2

Find the oxidation numbers P4 P2O5 PCl5 H3PO4 PO43- Na3PO4 P = +5 and O = -2 P = +5 and Cl = -1 H = +1, P = +5 and O = -2 P = +5 and O = -2 Na = +1, P = +5 and O = -2

A summary of terminology for oxidation-reduction (redox) reactions transfer or shift of electrons Y X loses electron(s) Y gains electron(s) X is oxidized Y is reduced X is the reducing agent Y is the oxidizing agent X increases its oxidation number Y decreases its oxidation number

Not all reactions are redox Reactions in which there has been no change in oxidation number are not redox rxns.

FIND THE OXIDATION NUMBERS: 2H2(g) + O2(g) 2H2O(g) +1 -2 The O.N. of H increases; it is oxidized; it is the reducing agent. The O.N. of O decreases; it is reduced; it is the oxidizing agent.

Find the oxidation numbers: 2Al(s)+ 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g) +1 +6 -2 +3 +6 -2 2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) +3H2(g) The O.N. of Al increases; it is oxidized; it is the reducing agent. The O.N. of H decreases; it is reduced; it is the oxidizing agent.

Find the oxidation numbers PbO(s) + CO(g) → Pb(s) + CO2(g) PbO(s) + CO(g) → Pb(s) + CO2(g) +2 -2 +2 -2 +4 -2 The O.N. of C increases; it is oxidized; it is the reducing agent. The O.N. of Pb decreases; it is reduced; it is the oxidizing agent.

a) 6Na (s)+ N2 (g)  2Na3N (s) b) Mg (s)+ Cl2 (g)  MgCl2 (s) Using oxidation states work out what is being oxidized and what is being reduced a) 6Na (s)+ N2 (g)  2Na3N (s) b) Mg (s)+ Cl2 (g)  MgCl2 (s) c) 4Fe (s)+ 3O2 (g)  2Fe2O3 (s) d) Ca (s)+ C (s)  CaC2 (s) e) MnO4-(aq) + S2O32-(aq)  Mn 2+(aq) +S2O62(aq) f) VO2+(aq) + Zn (s)  VO 2+(aq) + Zn2+(aq)