Chapter 8 (CIC) and Chapter 4, 20 (CTCS) Read in CTCS Chapter 4.4 (pgs 120-123), and 20.1-2 Problems in CTCS: 4.37, 39 and 20.3, 5, 7, 9.

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Presentation transcript:

Chapter 8 (CIC) and Chapter 4, 20 (CTCS) Read in CTCS Chapter 4.4 (pgs ), and Problems in CTCS: 4.37, 39 and 20.3, 5, 7, 9

New Energy Sources – Why? Run out of fossil fuels –Hydrocarbons not available for plastics or medicines Fossil fuel wastes –CO 2 generation gives greenhouse –NOx and SOx give acid rain problems –Smog gives tropospheric ozone problems CO and other particulates Nuclear power wastes

H 2 as a Fuel H / 2 O 2  H 2 O ΔH = -286 kJ That’s 143 kJ per gram of H compared to –30 kJ/g for coal –46 kJ/g of octane (11 kcal/g) –54 kJ/g of methane How do we get H 2 ? M + 2 H +  M 2+ + H 2 Very expensive for industrial scale

Most Abundant Element in the Universe H 2 O  H / 2 O 2 ΔH = +286 kJ Electrolysis – but where do we get electricity? Fossil fuel combustion –Pollutants –60% efficiency (at best) – 2 nd law of thermodynamics CH 4 (g) + 2H 2 O (g)  4H 2 (g) + CO 2 (g) ΔH = +165 kJ –Find better catalysts?

Storage and Combustion 1 g of H 2 occupies 12L of volume at atmospheric pressure Condense it at -253ºC Li (s) + 1 / 2 H 2 (g)  LiH (s) –Occupies 1 teaspoon of volume LiH (s) + H 2 O (l)  H 2 (g) + LiOH (s) Hindenburg and space shuttle Challenger

Electrochemistry (Redox) Electricity and chemical reactions Zn (s) + 2 H + (aq)  H 2 (g) + Zn 2+ (aq) –Zinc loses 2e - while each H gains an e - –Zinc is oxidized (oxidation is a loss of e - ) –Hydrogen is reduced (reduction is a gain of e - ) –OIL RIG or LEO goes GER How did oxidation/reduction terms come about? Zn (s) + 1 / 2 O 2 (g)  ZnO (s) ZnO (s) + H 2 (g)  H 2 O (l) + Zn (s)

Based on nonsharing of electrons Oxidation Number (O.N.) Rules 1. O.N. of any element is zero (Li, H 2, P 4, S 8,...) 2. O.N. is equal to the charge on an individual atomic ion (Li + = +1, Cl - = -1, Ca 2+ = +2, P 3- = -3,...) 3. O.N. of oxygen equals -2 except OF 2, H 2 O 2, and other peroxides (O 2 2- ; -1) and superoxides (O 2 - ; - 1 / 2 ) Oxidation Numbers (States)

4. O.N. of fluorine in a compound is O.N. of hydrogen is +1 unless it is combined with an element which has a lower electronegativity than itself (NaH, CaH 2, B 2 H 6,...) 6. Sum of O.N. must equal charge on species. H 3 PO 4 H = +1P = +5O = -2 3(+1) +1(+5) +4(-2) = 0 7. Charges must balance on both sides of the equation

Zn (s) + 2 H + (aq)  H 2 (g) + Zn 2+ (aq) Zn = 0, H + = +1, Zn 2+ = +2, H 2 = 0 Q:Determine the O.N. of each element in the following compounds: NaCl BaI 2 FeBr 3 MnO 2 H 2 O 2 HNO 3 AlPO 4 SO 4 2-

Redox/Travel Agents If one species is oxidized, another must be reduced (electrons have to go somewhere!) The species that is oxidized is a reducing agent (oftentimes contains O) The species that is reduced is an oxidizing agent Zn (s) + 2 H + (aq)  H 2 (g) + Zn 2+ (aq) Q:Which is the reducing agent and which is the oxidizing agent?

Balancing Redox Equations Not only must atoms balance, but e - ’s must too Two methods: half-reaction and O.N. method Write skeletal eqn and separate into 2 half-rxns CH 3 CH 2 OH + Cr 2 O 7 2-  CH 3 CO 2 H + Cr 3+ Oxdn: Redn: Balance atoms by adding H +, OH -, or H 2 O 3. Add electrons and equilize e - for the two reactions 4. Add half-reactions and cancel equalities 5. Verify elemental and charge balance

Balance this reaction in an acid CH 3 CH 2 OH + Cr 2 O 7 2-  CH 3 CO 2 H + Cr 3+

In a basic solution, Br 2, disproportionates to bromate (BrO 3 - ) and bromide ions. Balance this equation.

In a basic solution, iron(III) hydroxide reacts with hypochlorite (OCl - ) ion to produce FeO 4 2- and chloride. Balance this reaction.