Redox Chemistry Review

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). 1 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 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+ 2e 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 D. ________________ 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 and which 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: Oxidation number of Fe: H2S = -2, Fe3+ = +3, SO42- = +6 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+ - e Fe2+ SO42- 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) (2) fumarate + 2 H+ + 2 e-  succinate 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 eFAD + 2 H+ + 2 e-  FADH2  succinate + FAD  fumarate + FADH2 iii. Fill in the blanks. A. ___________ 2 e- ___________ ___________ ___________ B. ________________ is the oxidizing agent. C. ________________ is the reducing agent. III. Redox reactions and thermodynamics Will the reaction of succinate and FAD occur spontaneously (exothermic reaction) or will it require energy input (endothermic reaction)? We can use thermodynamics to answer this question. Different molecules have different tendencies to donate or accept e-. The compounds in Table 1 are ranked according to their tendency to be in the oxidized form (on the left) or in the reduced form (on the right). Generally speaking, the more reduced a molecule is, the more potential energy it contains that can be released for biological work. Table 1 contains values for E0′, which is reduction potential. Reduction potential is a measure of how likely the reduction (forward) reaction is to occur. When comparing two reactions, the one with a higher reduction potential (more positive, i.e., higher in the table) is more likely to occur in the forward direction. Essentially, electrons will move up the table, flowing from the reduced compound in the reaction with the lower reduction potential to the oxidized compound in the reaction with the higher reduction potential. Take any two equations. The reactant on the left side of the higher equation will be the oxidizing agent, and accept electrons from the product of the lower equation, which will be the reducing agent (by donating electrons). Generally, oxidized compounds higher in the chart are stronger oxidizing agents than the oxidized compounds lower in the chart. Conversely, the reduced compounds lower in the chart are stronger reducing agents than those higher in the chart. Let’s take an example: E0′(V) (1) ½ O2 + 2 H+ + 2 e-  H2O +0.816 (2) NO2- + 8 H+ + 6 e-  NH4+ + 2 H2O +0.34 4 In this system, since the E0′ value is greater for reaction (1), reaction (1) will proceed forwards and reaction (2) will proceed backwards. In other words, O2 is a stronger oxidizing agent than NO2-, so NH4+ will be oxidized to NO2-, and O2 will be reduced to H2O. O2 NO2 e- releases energy H2O NH4+ This reaction will occur spontaneously and release energy because it is thermodynamically favorable. The measure of this is ΔG0’, or the Gibbs free energy. When ΔG0’ < 0, energy is released and reactions proceed without added energy. (Think of sledding down a hill). When ΔG0’ > 0, energy is required to make a reaction proceed. (Think of pushing a sled uphill). It is possible for NO2- to oxidize H2O to O2, but this would require additional energy. Reduction potential values can be used to calculate ΔG0’ according to the Nernst Equation: ΔG0’ = -nF (ΔE0’) n F ΔE0’ = number of e- transferred = Faraday constant (23 kcal V-1 mol-1) = E0’(oxidizing agent) - E0’(reducing agent) ΔE0’ and ΔG0’ have opposite signs, so when ΔE0’ > 0, then ΔG0’ will be negative. Using NH4+/O2 example, let’s calculate ΔE0’ and ΔG0’. 3/2 O2 + 6 H+ + 6 e-  3 H2O NH4+ + 2 H2O  NO2- + 8 H+ + 6 e- E0’ = +0.816 V * E0’ = -0.34 V **  NH4+ + 3/2 O2  NO2- + 2 H+ + H2O ΔE0’ = +0.476 V ΔG0’ = -(6)*(23 kcal V-1 mol-1)*(0.476 V) = -65.688 kcal/mol Note: * ** Multiplying by a constant to get the right number of e- doesn’t change ΔE0’. When we reversed reaction (2), we had to change the sign on ΔE0’. 5 iv. Going back to the succinate and FAD example, calculate ΔE0’ and ΔG0’. Will this reaction occur spontaneously? v. For an overall reaction with CO2 and H2S as reactants and H2O, S and 1 mole of glucose as products, calculate ΔE0’ and ΔG0’. Will this reaction occur spontaneously? Final notes: This is highly over-simplified. For a more complete description, please consult a thermo dynamics or chemistry book. All of these reactions are affected by pH and concentrations of products and reactants. Furthermore, some products or reactants can be further stabilized by other chemical reactions (e.g., Fe3+ can precipitate as Fe(OH)3), which also alter the equilibrium. Thermodynamics describes a world at steady state, but our world is not at thermodynamic equilibrium. Kinetics, or the rates of reactions, also determines whether a reaction will occur or not. Thermodynamics would predict the oxidation of all carbon on earth to CO2, since O2 is such a strong oxidant (notice its place in Table 1). However, conversion of solar energy into reduced carbon compounds by autotrophs leads to higher energy configurations and means that thermodynamically unstable configurations (such as our bodies) can occur. 6 Table 1. Standard reduction potential (E0') values (at 25ºC and pH 7) Since e- are being added to the reactants on the left sides of the equations, these reactions are showing reduction reactions. + + H 2O 2+ ⇒ Fe ⇒ ½ N 2 + 3 H 2O +0.771 ⇒ NO2 + H2O +0.75 - +0.421 + +0.36 + ⇒ NH4 + 3 H2O ⇒ NH4 + 2 H2O +0.34 CH3OH + 2 H + 2 e- ⇒ CH4 + H2O +0.17 fumarate + 2 H + 2 e+ 2 H + 2 e- ⇒ succinate +0.031 ⇒ H2 (pH 0) +0.00 oxaloacetate + 2 H + 2 e+ CH2O + 2 H + 2 e- ⇒ malate -0.166 ⇒ CH3OH -0.18 pyruvate + 2 H + 2 e- ⇒ lactate -0.185 acetaldehyde + 2 H + 2 e2+ SO4 + 8 H + 6 e- ⇒ ethanol -0.197 ⇒ S + 4 H 2O -0.20 SO4 + 10 H + 8 e+ FAD + 2 H + 2 e- ⇒ H 2S + 4 H 2O -0.21 FADH2 -0.219 CO2 + 8 H + 8 e+ S + 2 H + 2 e- ⇒ ⇒ CH4 + 2 H2O -0.24 ⇒ H 2S -0.243 + + + + + 2- + + N2 + 8 H + 6 e + + ⇒ 2 NH4 NAD + H + 2 e - ⇒ NADH -0.320 NADP + H + 2 e + 2 H + 2 e- ⇒ NADPH -0.324 ⇒ H2 (pH 7) -0.414 CO2 + 4 H + 4 e2+ Fe + 2 e- ⇒ 1/6 glucose + H2O -0.43 ⇒ Fe -0.85 + + + + + -0.28 Electrons moving this way  ΔG0' > 0 NO3 + 10 H + 8 e+ NO2- + 8 H + 6 e - Eo (V) +0.816 Electrons moving this way • ΔG0' < 0 NO3 + 6 H + 6 e+ NO3 + 2 H + 2 e- ' Half-Reaction ⇒ ½ O2 + 2 H + 2 e 3+ Fe + e+ Answers to sample problems. i. A. S: +6 E. N: +5 ii. A B. C. D. E. iii. A. B. S: -2 F. C: +4 C. N: -3 G. Fe: +3 D. N: +3 + NADH is oxidized to NAD Acetaldehyde is reduced to ethanol Acetaldehyde is the oxidizing agent and NADH is the reducing agent. + Acetaldehyde and NAD are the oxidized forms. Ethanol and NADH are the reduced forms. FAD 2e - fumarate B. FAD is the oxidizing agent. C. Succinate is the reducing agent. FADH2 iv. v. ’ succinate 0’ ΔE0 = -0.250 V; ΔG = +11.5 kcal/mol; No. ’ 0’ ΔE0 = -0.187 V; ΔG = +103.2 kcal/mol; No. This reaction describes the production of glucose, a process that we know requires a considerable amount of energy. Since this reaction is a photosynthetic reaction, this necessary energy is provided by sunlight. 7