MIT1_018JF09_lec03_Redox

MIT1_018JF09_lec03_Redox - Redox Chemistry Review I....

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Redox Chemistry Review I. Oxidation State or Number The oxidation state or number of a compound gives a relative measure of how oxidized (electron-poor) or reduced (electron-rich) a compound is. The number is relative because it is only meaningful when compared to the number for another compound, to determine which one is more oxidized or more reduced. Rules for calculating the oxidation state of an element in a molecule (from Brock, Biology of Microorganisms , 11 th Ed. Appendix A-1). 1. The oxidation state of an element in an elementary substance ( e.g. , H 2 , O 2 ) is zero. 2. The oxidation state of the ion of an element is equal to its charge ( e.g. , Na + = +1, O 2- = -2). 3. The sum of the oxidation numbers of all atoms in a neutral molecule is zero. Thus, H 2 O is neutral because it has two H at +1 each and one O at –2. 4. In an ion, the sum of the oxidation numbers of all atoms is equal to the charge on that ion. Thus, in the OH - ion, O(-2) + H(+1) = -1 5. In compounds, the oxidation state of O is virtually always –2, and that of H is +1 (this gets more complicated in some organic compounds). 6. In simple carbon compounds, the oxidation state of C can be calculated by adding up the H and O atoms present and using the oxidation states of these elements as given in #5, because in a neutral compound, the sum of the oxidation numbers must be 0. Thus, the oxidation state of carbon in methane, CH 4 , is –4 (4 H at +1 = +4). 7. In organic compounds with more than one C atom, it may not be possible to assign a specific oxidation number to each C atom, but it is still useful to calculate the oxidation state of the compound as a whole. Thus, the oxidation state of carbon in glucose, C 6 H 12 O 6 , is zero and the oxidation state of carbon in ethanol, C 2 H 6 O, is –2. i. Calculate the oxidation state of each element in the following compounds (answers at end). A. sulfate – SO 4 2- S: __________ B. hydrogen sulfide – H 2 S S: __________ C. ammonia – NH 3 N: __________ D. nitrite – NO 2 - N: __________ E. nitrate – NO 3 - N: __________ F. CO 2 C: __________ G. iron hydroxide – Fe(OH) 3 Fe: __________ II. Reduction and oxidation reactions The oxidation states you just calculated above provide information about how reduced (electron-rich) or oxidized (electron-poor) each element in a compound is. The smaller (more negative) the oxidation state, the more reduced a compound is; conversely, larger (more positive) oxidation states are associated with more oxidized compounds. Hence, Fe 3+ (oxidation state +3) is more oxidized than elemental Fe (oxidation state 0). 1
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Reduction and oxidation reactions (together known as redox ) involve the transfer of electrons (e - ) from one molecule to another. Since free electrons cannot exist in solution, these reactions are always coupled. Generally, electrons are transferred from more reduced compounds to more oxidized compounds, since the reduced compounds are more electron rich than oxidized compounds. The process of losing electrons is called
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MIT1_018JF09_lec03_Redox - Redox Chemistry Review I....

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