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Oxidation Numbers & Balancing equation Oxidation Number • Oxidation number is defined as The charge an atom has Or appears to have When electrons are distributed according to certain rules Oxidation Number Rules • The oxidation number of • an Element is 0 • group One elements is +1 • group Two elements is +2 in compounds The oxidation number of an ion is equal to the charge on the ion • halogens is -1 (except ……????) (in binary compounds) • The oxidation number of H in a compound is +1 – except in metal hydrides when it is -1 • The oxidation number of O in a compound is -2 – except (x2) in peroxides when it is -1 (H2O2) in OF2 when it is +2 (why?) • Oxidation numbers • add up to zero in a compound • add up to the charge of a complex ion • What is the oxidation number of each element in :- (write down before you go on) H20 MnO4¯ I2 KBrO3 Na2S2O3 H2O2 NaOCl The oxidation number of each element is :H20 +1 -2 MnO4¯ +7 -2 I2 0 KBrO3 +1 +5 -2 Na2S2O3 +1 +2 -2 H2O2 +1 -1 NaOCl +1 -2 +1 Learning Check Can I give the oxidation number RULE for a) Oxygen b) Hydrogen c) free element d) Neutral atom (sum) e) Ion (simple and complex) f) Group 1 element g) Group 2 element h) HALOGEN STILL NOT The End - click to go on Balancing Equations with oxidation numbers STEPS 1. Assign oxidation numbers 2. Identify what is oxidised and reduced 3. Write half equation SIDE by SIDE for each (showing number of electrons on the move for one atom of each) 4. 5. 6. 7. 8. Rewrite for the number of atoms given e.g. Cr2 Balance the electrons REWRITE the original equation using these “prefixes” Balance remainder by inspection CHECK – do the charges on each side cancel out?? Example Assign & Identify Cr2O72- + Fe2+ + H+ Cr3+ + Fe3+ + H20 +6 -2 +2 +1 +3 Oxidised x1 Reduced x3 +3 +1 -2 2 Identify oxidised or reduced & number of electrons lost or gained per atom & as given Reduced Cr + 3e- Cr Cr2 + 6e- 2Cr Oxidised Fe – e- Fe Fe – e- Fe ATOM GIVEN • Balance Electrons Cr2 + 6e- 2Cr 6 Fe – 6 e- 6 Fe • Rewrite and sub back Cr2O72- + 6Fe2+ + H+ 2Cr3+ + 6Fe3+ + H20 • Balance remainder by inspection 14 2Cr3+ + 6Fe3++ H 0 Cr2O72- + 6Fe2++ H+ 2 7 Check • Charges on each side should balance • Cr2O72- + 6Fe2++ 14 H+ 2Cr3+ + 6Fe3++ 7 H20 • LEFT 212+ 14+ 24+ RIGHT 6+ 18+ ___ 24+