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Have you ever heard about oxidation ? Where? When? When the iron goes rusty? In a combustion?

REDOX. Have you ever heard about oxidation ? Where? When? When the iron goes rusty? In a combustion? When we breathe?. REDOX. Fe + 1/2 O 2  FeO Fe + HCl  FeCl 2 Do you see similarities in these two reactions? In which valence acts the iron, before and after?

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Have you ever heard about oxidation ? Where? When? When the iron goes rusty? In a combustion?

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  1. REDOX • Have you ever heard about oxidation? • Where? When? • When the iron goes rusty? • In a combustion? • When we breathe?

  2. REDOX • Fe + 1/2 O2 FeO • Fe + HCl  FeCl2 • Do you see similarities in these two reactions? • In which valence acts the iron, before and after? • The first one is an oxidation, and the second one?

  3. REDOX • In the reaction with oxygen the Fe converts in Fe2+. • In the second reaction the Fe converts also in Fe2+. • In both cases Fe has been oxidised. • In both cases Fe has lost electrons.

  4. REDOX • CONCEPTS • OXIDATION REDUCTION • Gain of OxigenLoss of Oxigen • Loss of Hydrogen Gain of Hydrogen • Loss of electrons Gain of electrons • Increase in Decrease in oxidation state oxidation state

  5. Oxidation state (number)Rules • Monoatomic ions: its charge • Uncombined elements: 0 • Oxigen: -2 excep in peroxides –1 • Hydrogen: +1 excep in hidrides –1 • Group I ions: +1 (Na+) • Group II ions: +2 (Mg2+) • Group III ions: +3 (Al3+) • Group VI ions: -2 (O2-, S2-,) • Group VII ions: -1 (F-, Cl-, Br-, I-, )

  6. REDOX RULES • 1. The oxidation level of an atom is 0 when the atom is in its elemental form: • Li, C, Na, Mg, etc. • H2, N2, O2, F2, Cl2, etc. • 2.  In ionic compounds, the oxidation state of an atom is the same as its charge: • NaCl = Na+1 + Cl-1, MgO = Mg+2 + O-2 • 3.  The sum of the oxidation levels of all the atoms in a compound must equal 0. • AlCl3 = Al+3 + 3Cl-1

  7. REDOX RULES (cont) • 4. In covalent compounds, the oxidation level of H is +1, O is -2, F is -1. • 5. Rules 3 and 4 allow the calculation of the oxidation levels of other atoms in a molecule: • HNO3  1-H atom = +1, 3-O atoms = 3(-2) = -6 • (+1) + (-6) = -5; N = +5 • What is the oxidation level of the P atom in H3PO4, in PH3? • What is the oxidation level of S in SO2, in SF6?

  8. OTHER CONCEPTS • oxidizing agent • it becomes reduced • its oxidation state decreases • reducing agent • it becomes oxidized • its oxidation state increases • spectator ion • don’t change its oxidation state

  9. SUMMARY • Something is oxidized if • It gains oxygen • It loses electrons • Its oxidation state increases • Something is reduced if • It loses oxygen • It gains electrons • Its oxidation state decreases

  10. EXERCICES • Insert electrons (e-) on the appropiate side of the following half-equations in order to balance and complete them, so that the electrical charges on both sides are equal K  K+ H2  2 H+ O  O2- Cu+  Cu2+ Cr3+  Cr2+

  11. EXERCICES • In each case state which element is oxidised or reduced, and give the oxidation states before and after the reaction Cl2 + 2 Br-  2 Cl- + Br2 2Fe + 3Cl2  2FeCl3 H2 + Cl2  2HCl 2FeCl2 + Cl2  2FeCl3 2 H2O + 2 F2  4HF + O2

  12. EXERCICES • Indicate whether each element has been oxidised, reduced, both, or has remained unchanged. Cu2O + 2H+  Cu2+ + Cu + H2O 3Br2 + 6OH-  BrO3- + 5Br- + 3H2O 4IO3-  3IO4- + I-

  13. BALANCE THE EQUATION • BrO3- + Fe2+  Br- + Fe3+ • State the oxidations states • Write the half-reactions • Balance the number of atoms ox/red • Balance O with H2O • Balance H with H+ • Balance the charge with electrons (e-)

  14. BALANCING REDOX • Make the number of electrons in both half-equations equal • Sum the half-equations • Transform to molecular compounds

  15. Reaction between potassium bromate (KBrO3) and iron (II) sulphate (FeSO4) in presence of sulphuric acid, producing potassium bromide and iron(III) sulphate K+ + BrO3- + Fe2+ + SO42- +H+  K+ + Br- + Fe3+ +SO42- KBrO3 + FeSO4 + H2SO4  KBr + Fe2(SO4)3 +5 -2 +6 -2 +6 -2 S.Ox. Fe2+  Fe3+ + 1 e- )x6 S.Re. BrO3- +6H+ +6 e-  Br- + 3 H2O BrO3- +6H+ +6Fe2+  Br- +6Fe3+ + 3 H2O KBrO3 +3H2SO4 +6 FeSO4  KBr +3Fe2(SO4)3 +3H2O

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