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, 11th Ed. Appendix A-1). 1. The oxidation state of an element in an elementary substance (e.g., H2, O2) is zero. 2. The oxidation state of the ion of an element is equal to its charge (e.g., Na+ = +1, O2-= -2). 3. The sum of the oxidation numbers of all atoms in a neutral molecule is zero. Thus, H2O 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, CH4, 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, C6H12O6, is zero and the oxidation state of carbon in ethanol, C2H6O, is –2. i. Calculate the oxidation state of each element in the following compounds (answers at end). A. sulfate – SO42-S: __________ B. hydrogen sulfide – H2S S: __________ C. ammonia – NH3 N: __________ D. nitrite – NO2-N: __________ E. nitrate – NO3-N: __________ F. CO2 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, Fe3+ (oxidation state +3) is more oxidized than elemental Fe (oxidation state 0). 1Reduction 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 oxidation and the process of gaining electrons is called reduction. A helpful mnemonic: “LEO the lion says GER.” (Loss of Electrons is Oxidation; Gain of Electrons is Reduction). In a redox reaction, the oxidizing agent is the reactant that gains electrons and is therefore reduced through the reaction (causing the other compound to become more oxidized). Conversely, the reducing agent is the compound that loses electrons through the reaction and therefore is oxidized through the reaction (causing the other compound to become more reduced). There are different ways of writing redox reactions. Let’s use the example of acetaldehyde (CH3CHO) getting reduced to ethanol (CH3CH2OH), a process that happens in respiration, using NADH as the electron source. NADH is common biological molecule that is frequently the source of reducing power. In its oxidized form, NAD+, it is a common electron acceptor. 1. Sum of two half reactions: acetaldehyde + 2 H+ + 2 e-  ethanol NADH  NAD+ + H+ + 2 e-   acetaldehyde + H+ + NADH  ethanol + NAD+ 2. Electron-transfer diagram: acetaldehyde NAD+ ethanol NADH 3. Overall reaction (doesn’t show electrons): acetaldehyde + H+ + NADH  ethanol + NAD+ ii. Fill in the blanks. A. Through this redox reaction, ________________ is oxidized to ________________. B. ________________ is reduced to ________________. C. ________________ is the oxidizing agent and ________________ is the reducing agent. 2 e-2D. ________________ and ________________ are the oxidized forms. E. ________________ and ________________ are the reduced forms. Given an overall reaction, how do we know which reactant is the oxidizing agent andwhich one is the reducing agent? For simple compounds, we can use the oxidation states with the general rules in Part I. e.g. H2S + 4 H2O + 8 Fe3+  8 Fe2+ + 10 H+ + SO42-Question: Is the S or the Fe getting reduced or oxidized? Oxidation number of S: H2S = -2, SO42-= +6 Oxidation number of Fe: Fe3+ = +3, Fe2+ = +2 Since the oxidation number for S increases, the sulfur is getting oxidized. Since the oxidation number for Fe decreases, the iron is getting reduced. In an electron-transfer diagram, we would draw this reaction this way: Fe3+ SO4 2-e -Fe2+ H2S However, it is not always this easy to determine the oxidizing and reducing agents. Fortunately, there are tables showing the half-reactions for many common reduction reactions, such as the one at the end of this handout. Let’s use the following reaction from the citric acid cycle (FADH2 is a compound similar to NADH): succinate + FAD  fumarate + FADH2 From Table 1, we can obtain the following 2 half-reactions (don’t worry about E0 ′ (V) yet): (1) fumarate + 2 H+ + 2 e- succinate (2) FAD + 2 H++ 2 e- FADH2 We have to reverse equation (2) to get the reactant on the correct side. If the number of electrons is not the same in both reactions, multiply one or both reactions by the appropriate number so that the e-will cancel when the two equations are added together. Remember, overall equations don’t show e-. 3___________ ___________ ___________ ___________ succinate  fumarate + 2 H+ + 2 e-FAD + 2 H+ + 2 e- FADH2