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REDOX REACTIONS

REDOX REACTIONS. OXIDATION NUMBERS. THE APPARENT CHARGE ON AN ATOM RULES The oxidation number on any free element is zero The oxidation number of a monatomic ion is equal to the charge on the ion The oxidation number of Hydrogen in most compounds is +1

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REDOX REACTIONS

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  1. REDOX REACTIONS

  2. OXIDATION NUMBERS • THE APPARENT CHARGE ON AN ATOM • RULES • The oxidation number on any free element is zero • The oxidation number of a monatomic ion is equal to the charge on the ion • The oxidation number of Hydrogen in most compounds is +1 • The oxidation number of Oxygen in most compounds is -2 • The sum of the oxidation numbers of all the atoms in a particle must be equal to the apparent charge of that particle.

  3. Determine the Oxidation Numbers for each of the atoms in the following reactions • H2SO4 + NaOH  Na2SO4 + H2O • Cu + AgNO3  Cu(NO3)2 + Ag • Redox Reaction - a reaction in which electrons are transferred (ox. # change)

  4. REDUCTION = The gain of electrons by atoms or ions • OXIDATION = The loss of electrons from an atom or ion • LEO goes GER • Oxidizing agent – a substance that tends to gains electrons • Reducing agent – a substance that tends to lose electrons

  5. Identifying Oxidizing and Reducing Agents • Cu + HNO3 Cu(NO3)2 + NO + H2O • H2S + Br2  S + 2HBr • 2H2 + O2  2H2O

  6. Rules for writing ionic equations • Binary acids: • HCl, HBr and HI are strong acids and are written as ions All other binary acids are written in molecular form • Ternary acids: • If the number of oxygen atoms exceeds the number of hydrogen atoms by two or more, the acid is strong and is written as ions • Polyprotic acids: • In the second and subsequent ionizations the acid is weak and written in molecular form • Bases: • Hydroxides of group I and II are strong and are written as ions • Salts: • Solubility rules! • Gases: • Gases are always written in molecular form

  7. BALANCING REDOX REACTIONS • Write the skeleton equations for the oxidation and reduction half reactions • Balance the half reactions with respect to electrons • Balance all the atoms except hydrogen and oxygen • Balance oxygen by adding H2O • Balance the hydrogen atoms by adding H+ • Check to see that the charges are balanced • Balance the total number of electrons in the two half reactions by multiplying by the factor finding the least common multiple of electrons lost and gained. • Add the two half reactions and simplify • Perform a final check on the equation to make sure that the number of atoms and the charge are balanced

  8. HNO3 + H3PO3 NO + H3PO4 + H2O

  9. Ag + HNO3 AgNO3 + NO + H2O

  10. VOLTAIC CELLS • A voltaic cell is a device used to produce electric energy from an oxidation-reduction reaction • Anode compartment – oxidation half reaction • Cathode compartment – reduction half reaction • Shorthand method of representing cell reactions • The oxidation half reaction is written first • The reduction half reaction is written second • Two vertical lines separate the half reaction and represents the salt bridge.

  11. STANDARD CELL POTENTIALS • Standard cell potential (Eocell) is the measured cell potential when the ion concentrations in the half cells are 1M and at 25oC and 101 kPa. • Eocell = Eoreduction – Eooxidation • Half cell potentials are read of a table

  12. CALCULATING STANDARD CELL POTENTIALS • A voltaic cell is constructed using the following half reactions • Fe3+(aq) + e-  Fe2+(aq) • Ni2+(aq) + 2e-  Ni(s) • Determine the cell reaction and calculate the standard cell potential • Look up reduction potentials on the table • EoFe3+ = + 0.771 V • EoNi3+ = -0.257 V

  13. The half cell with the more positive reduction potential is the one in which reduction occurs. (this is the cathode) • The oxidation occurs at the anode • Add the half reactions together • Make sure the e- lost equals e- • Calculate the standard cell potential • Eocell = Eoreduction – Eooxidation

  14. Oxidation (at anode) • Ni(s) Ni2+(aq) + 2e- • Reduction (at cathode) • Fe3+(aq) + e-  Fe2+(aq) • Ni(s) + 2Fe3+(aq) Ni2+(aq) + 2Fe2+(aq) • Eocell = Eoreduction – Eooxidation +0.771 V - (-0.257 V) +1.028V *Eo is not multiplied by any number even if the half reaction is!

  15. Try these • Copper (II) ion to copper • Aluminium ions to aluminum atoms • Silver ions to silver • Copper (II) ions to copper

  16. A strip of copper is immersed in a solution containing zinc ions. Does the reaction occur? • Cu + Zn2+ Cu2+ + Zn • Remember: • Eocell = Eoreduction – Eooxidation • A spontaneous reaction must have a positive voltage • Zn2+ + 2e-  Zn -0.7626 V • Cu2+ + 2e-  Cu +0.340 V • Eocell = -0.7626 – (+0.340) • This reaction will not occur because the voltage is negative, the reverse reaction will occur.

  17. A strip of cobolt metal is immersed in a solution containing silver ions. Does the reaction occur? • Co + Ag+ Co2+ + Ag • Ag+ + 1e-  Ag +0.799 V • Co2+ + 2e-  Co -0.277 V • Eocell = Eoreduction – Eooxidation • Eocell = 0.799 – (-0.277) • + 1.076 V

  18. Predict the voltage produced by the following cells • Zn/Zn2+//Fe2+/Fe • Mn/Mn2+//Br2/Br- • Cu/Cu2+//Ag+/Ag

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