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19-105-W01-L16-1 19-105 W-01 WEEK 6 Lectures 16-18 Begin ELECTROCHEMISTRY J&A Chapter 18, 18.1, 18.2 Monday - Lecture 16; Wednesday - Lecture 17 18.1 Half reactions Oxidation States 18.2 Balancing Oxidation-Reduction Equations Friday: Lecture 18 Review class for those who need it on balancing REDOX equations. 19-105-W01-L16-2 OXIDATION Loss of electrons by an atom, molecule, ion, etc. REDUCTION Gain of electrons by an atom, molecule, ion, etc. e.g. 2Na(s) + CR2(g) 6 2NaCR(s) Here the Na metal is oxidized to Na+ (1 electron lost per sodium atom) and CR2 is reduced to 2CR (1 electron gained per chlorine atom) 19-105-W01-L16-3 Oxidation-Reduction or REDOX reactions Reactions which involve BOTH GAIN (reduction) and LOSS (oxidation) of electrons Convenient to think of the overall REDOX reaction as two individual HALF-reactions, namely an Oxidation Half-reaction and a Reduction Half-reaction 2Na(s) 6 2Na+ + 2e- (oxidation 1/2-reaction) - - CR2(g) + 2e 6 2CR (reduction 1/2-reaction) 2Na(s) + CR2(g) 6 2NaCR(s) 19-105-W01-L16-4 Oxidizing AGENT 6 Causes another species to be oxidized by accepting/taking electrons from it. Reducing AGENT 6 Causes another species to be reduced by donating electrons to it. LEORA 6 Loss of Electrons is Oxidation and that GEROA 6 species is a Reducing Agent. Gain of Electrons is Reduction and that species is an Oxidizing Agent. Oxidation results in an increase in the Oxidation State (or Number) by loss of electrons. Reduction results in an decrease in the Oxidation State (or Number) by gain of electrons. 19-105-W01-L16-5 Rules for assigning Oxidation Numbers (apply in this sequence) 1. Pure element, oxidation no. = 0 2. Monoatomic ion, oxidation no. = charge on ion 3. Fluorine in its compounds is always -1 4. Oxygen in its covalent compounds is -2, except in peroxides where it is -1 5. Hydrogen in covalent compounds with non-metals is +1, in metal hydrides it is -1 6. The sum of the oxidation numbers = 0 for a neutral (no charge) compound; for an ion, the sum of the oxidation numbers = ion charge 7. Look at the column number in the periodic table 19-105-W01-L16-6 Examples: Assign Oxidation Numbers (or States) to all the atoms in the following:CO2 PbS NH3 CH4 SF6 Cr2O72MnO4NO3C2H6O H3PO4 H2PO4F2O ARO2CROCRO2- 19-105-W01-L16-7 Absolutely FOOL-PROOF method for balancing ANY redox equation 1. Assign Oxidation Numbers and write separate equations for each half reaction, then work on each half-reaction separately. 2. Balance atoms being oxidized or reduced. 3. From change in oxidation numbers, get the number of electrons involved and add them to the high oxidation number side of the equation. 4. Now balance the charge, using either H+ (if in ACID solution) or OH (if in base solution). 5. Now balance the OXYGEN by adding H2O to the appropriate side of the equation. 6. Check that everything balances - it will! 7. Scale the two half reactions, to equalize the number of electrons in each. 8. Add the two half-reactions. 19-105-W01-L16-8 Example: (all aqueous) MnO4 + Fe2+ 6 Fe3+ + Mn2+ in acid solution. Oxidation numbers Mn in MnO4 is +7, in Mn2+ it is +2 Fe2+ is +2, Fe3+ is +3 Oxidation Step: Fe2+ 6 Fe3+ + 1e (steps 1,2,3) Reduction step: MnO4 6 Mn2+ (steps 1 and 2) 5e + MnO4 6 Mn2+ (step 3) + 8H + 5e + MnO4 6 Mn2+ (step 4) + 8H + 5e + MnO4 6 Mn2+ + 4H2O (step 5) Visual check - all OK (step 6) 5Fe2+ 6 5Fe3+ + 5e (step 7) + 5e + MnO4 6 Mn2+ + 4H2O (step 7) 8H+ Add 8H+ + MnO4 + 5Fe2+ 6 Mn2+ + 4H2O + 5Fe3+ (step 8) 19-105-W01-L16-9 - - In base, NO2 (aq) + AR(s) 6 NH3(g) + ARO2 (aq) - Oxid Nos. N in NO2 is +3, in NH3 is -3 - AR in AR(s) is 0, in ARO2 (aq) is +3 Oxidation: AR 6 ARO2 + 3e (steps 1,2,3) 4(OH ) + AR 6 ARO2 + 3e (steps 4) 4(OH ) + AR 6 ARO2 + 3e + 2(H2O) (steps 5,6) Reduction: NO2 6 NH3 (steps 1 and 2) 6e + NO2 6 NH3 (step 3) 6e + NO2 6 NH3 + 7(OH ) (step 4) 5(H2O) + 6e + NO2 6 NH3 + 7(OH ) (steps 5,6) Scale, add and tidy up (cancel superfluous water) 8(OH ) + 2AR(s) 6 2ARO2 + 6e + 4(H2O) 5(H2O) + 6e + NO2 6 NH3 + 7(OH ) 5(H2O) + NO2 + 8(OH ) + 2AR 6 NH3 + 7(OH ) + 2ARO2 + 4(H2O) H2O + NO2 + 8(OH ) + 2AR 6 NH3 + 7(OH ) + 2ARO2 19-105-W01-L16-10 Give Balanced Half Reactions for the following in aqueous solution:1. H+ + Cr2O72- + C2H6O 6 Cr3+ + CO2 19-105-W01-L16-11 2. As2O3(s) + NO3-(aq) 6 H3AsO4(aq) + NO(g) 19-105-W01-L16-12 3. MnO4- + S2- 6 MnS(s) + S(s) (in base) 19-105-W01-L16-13 4. CR2(g) 6 CR- + CRO- (in base) 19-105-W01-L16-14 5. HO2- + Cr(OH)3- 6 CrO42- (in base) 19-105-W01-L16-15 6. Ag2S(s) + CN-(aq) + O2(g) 6 S(s) + [Ag(CN)2]-(aq) 19-105-W01-L16-16 7. More devious fiendishly-cunning examples on redox balancing will be done in class....