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Topic: Redox – Half reactions Assign Oxidation number to H, Cl and O for the following compounds 1. HClO 2. HClO2 3. HClO3 4. HClO4 Now we know how to assign oxidation number…we can look at redox rxns • Haber Process –N2(g) + 3H2(g) 2NH3(g) • Start by assigning oxidation numbers 0 0 -3 +1 • N2(g) + 3H2(g) 2NH3(g) • What was oxidized? Reduced? 0 0 -3 +1 N2(g) + 3H2(g) 2NH3(g) 4• 3• 2 1• 0 -1• -2 -3• -4• O • Oxidation I • Is L • Losing electrons N Began Ended 0 -3 Gained 3 electrons = Reduced R • Reduction I • Is H 0 +1 G • Gaining electrons Lost 1 electron = Oxidized Why use the word “reduced” when electrons are gained? Electrons are Negative! Look how oxidation number changes Ex: Cl gains an electron → Cl-1 • oxidation # ↓ from 0 to -1; the # was reduced Half Reactions • Even though oxidation & reduction reactions occur together we write separate equations for each process and include # of e- gained/lost – known as Half-Reactions Half Reactions • Half-reactions must demonstrate: – conservation of mass & conservation of charge – # atoms on left must = # atoms on right • (always balance mass first!!!!) • Keep 7 diatomics together, everything else write as single ion) – total charge on left must = total charge on right • Add electrons to more positive side 0 0 3H2 N2 + -3 +1 2NH3 • Oxidation = electrons lost OIL (becomes more positive) –So electrons are on the product side But something •H2 H+1 + 1e- is wrong! 0 = +1 1 and -1 equals 0 Remember… • H2 H+1 + 1e- • Total Charge on left = Total Charge on right • # atoms on left = # atoms on right • H2 • Something is still wrong! • Charge is off now • H2 2H+1 + 2e+1 2H + 1e 2 x +1 = +2 and -2 equals 0 0 N2 + 0 3H2 -3 +1 2NH3 • Reduction = electrons gained RIG (becomes more negative) –So electrons are on the reactant side • Each N gained 3 electrons, 6 - 2 N-3 –N2 + 3e Balance mass But not you have -6 0 You Try: • • • • 0 +2 -2 Mg + S MgS What is oxidized? What is reduced? Assign Oxidation Numbers Figure out change in oxidation numbers Mg: 0 to +2 = Oxidation S: 0 to -2 = Reduction Now write the Half-Reactions 0 0 +2 -2 Mg + S MgS Mg is oxidized: S is reduced: Mg Mg+2 + 2e- S + 2e- S-2 0 +1 -1 0 +2 -1 Zn + 2HCl H2 + ZnCl2 What is oxidized? Reduced? • Zn goes from 0 to +2 = oxidation • H goes from +1 to 0 = reduction • Cl goes from -1 to -1; No change 0 +1 -1 0 +2 -1 Zn + 2HCl H2 + ZnCl2 • Write Half Reactions Zn Zn+2 + 2e2H+1 + 2e- H2 In Redox # of electrons gained has to equal number of electrons lost Zn Zn+2 + 2e2H+1 + 2e- H2 H2 2H+1 + 2eN2 + 6e- 2N-3 In Redox # of electrons gained has to equal number of electrons lost 3 H2 +1 2H + 3H2 2e +1 6H + N2 + 6e- 2N-3 3H2 + N2 6H+1 + 2N-3 6e N2(g) + 3H2(g) 2NH3(g) 3H2 + N2 6H+1 + 2N-3 The steps… start with balanced rxn 1. Assign oxidation numbers to all atoms in equation 2. Determine elements changed oxidation number 3. Identify element oxidized & element reduced 4. Write half-reactions (diatomics must stay as is, everyone you can write with their oxidation number only – NH3 = N-3 but Cl2 stays Cl2 not Cl) – Make sure # of atoms on each side is balanced – Make sure charge on each side is balanced 5. Number electrons lost & gained must be equal; multiply half-reactions if necessary 6. Add half-reactions 0 +1 +5 -2 +2 +5 -2 0 Cu + 2AgNO3 Cu(NO3)2 + 2Ag 1. Assign oxidation numbers to all atoms in equation 2. Determine elements changed oxidation number – Cu 0+2 Ag +1 0 – N +5 +5 unchanged O -2 -2 unchanged 3. Identify element oxidized & element reduced – Oxidation (OIL) =Cu – Reduction (RIG) = Ag 0 +1 +5 -2 +2 +5 -2 0 Cu + 2AgNO3 Cu(NO3)2 + 2Ag 4. Write half-reactions – Cu Cu+2 + 2e– Ag+1 + 1e- Ag 5. Number electrons lost & gained must be equal; multiply halfreactions if necessary – 2(Ag+1 + 1e- Ag) – 2Ag+1 + 2e- 2Ag 0 +1 +5 -2 +2 +5 -2 0 Cu + 2AgNO3 Cu(NO3)2 + 2Ag 6. Add half-reactions; Transfer coefficients to skeleton equation Cu Cu+2 + 2e+______________________ 2Ag+1 + 2e- 2Ag Cu + 2Ag+1 + 2e- 2Ag + Cu+2 + 2eCu + 2Ag+1 2Ag + Cu+2 Cu + 2AgNO3 Cu(NO3)2 + 2 Ag