Transcript CHM 113 Chemistry and the Environment

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
New Energy Sources – Why?
• Run out of fossil fuels
– Hydrocarbons not available for plastics or
medicines
• Fossil fuel wastes
– CO2 generation gives greenhouse
– NOx and SOx give acid rain problems
– Smog gives tropospheric ozone problems
• CO and other particulates
• Nuclear power wastes
H2 as a Fuel
• H2 + 1/2O2  H2O 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 H2?
• M + 2 H+  M2+ + H2
• Very expensive for industrial scale
Most Abundant Element in the Universe
• H2O  H2 + 1/2O2 ΔH = +286 kJ
• Electrolysis – but where do we get electricity?
• Fossil fuel combustion
– Pollutants
– 60% efficiency (at best) – 2nd law of thermodynamics
5000 C
H 2 O  H 2 +
1
2
O2
• CH4(g) + 2H2O(g)  4H2(g) + CO2(g) ΔH = +165 kJ
– Find better catalysts?
Storage and Combustion
• 1 g of H2 occupies 12L of volume at
atmospheric pressure
• Condense it at -253ºC
• Li(s) + 1/2 H2(g)  LiH(s)
– Occupies 1 teaspoon of volume
• LiH(s) + H2O(l)  H2(g) + LiOH(s)
• Hindenburg and space shuttle Challenger
Electrochemistry (Redox)
• Electricity and chemical reactions
• Zn(s) + 2 H+(aq)  H2(g) + Zn2+(aq)
–
–
–
–
Zinc loses 2e- while each H gains an eZinc 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 O2(g)  ZnO(s)
• ZnO(s) + H2(g)  H2O(l) + Zn(s)
Oxidation Numbers (States)
• Based on nonsharing of electrons
Oxidation Number (O.N.) Rules
1. O.N. of any element is zero (Li, H2, P4, S8, ...)
2. O.N. is equal to the charge on an individual
atomic ion
(Li+ = +1, Cl- = -1, Ca2+ = +2, P3- = -3, ...)
3. O.N. of oxygen equals -2 except OF2, H2O2,
and other peroxides (O22-; -1) and superoxides
(O2- ; -1/2)
4. O.N. of fluorine in a compound is -1
5. O.N. of hydrogen is +1 unless it is combined
with an element which has a lower
electronegativity than itself (NaH, CaH2, B2H6,
...)
6. Sum of O.N. must equal charge on species.
H3PO4 H = +1
P = +5
O = -2
3(+1) + 1(+5) + 4(-2) = 0
7. Charges must balance on both sides of the
equation
Zn(s) + 2 H+(aq)  H2(g) + Zn2+(aq)
Zn = 0, H+ = +1, Zn2+ = +2, H2 = 0
Q:Determine the O.N. of each element in the
following compounds:
NaCl
BaI2
FeBr3
MnO2
H 2O 2
HNO3
AlPO4
SO42-
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)  H2(g) + Zn2+(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
CH3CH2OH + Cr2O72-  CH3CO2H + Cr3+
Oxdn:
Redn:
Balance atoms by adding H+, OH-, or H2O
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
CH3CH2OH + Cr2O72-  CH3CO2H + Cr3+
• In a basic solution, Br2, disproportionates to
bromate (BrO3-) and bromide ions. Balance this
equation.
• In a basic solution, iron(III) hydroxide reacts
with hypochlorite (OCl-) ion to produce FeO42and chloride. Balance this reaction.