Redox Chemistry Handout



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Transcription:

Redox Chemistry Handout This handout is intended as a brief introduction to redox chemistry. For further reading, consult an introductory chemistry or microbiology textbook. Redox reactions involve the transfer of electrons (usually abbreviated e ) from one molecule to the other. Reduction is when a molecule gains electrons. Oxidation is when a molecule loses electrons. (One way to remember this is the mnemonic LEO says GER, which translates to Loss of Electrons is Oxidation; Gain of Electrons is Reduction. ) Since electrons cannot exist free in solution, an oxidation must always be paired with a reduction; hence the term redox (reduction and oxidation) reaction. The terminology can be very confusing: oxidation loss of e reduction gain of e oxidizing agent gains e during reaction and is therefore reduced during reaction reducing agent loses e during reaction and is therefore oxidized during reaction oxidized form form of molecule lacking (it s all relative) an e reduced form form of the molecule having an additional (again, relative) e There are three ways to represent a redox reaction; these are shown below with a representative biological redox reaction: (1) Overall reaction: acetaldehyde + NADH + H + ethanol + NAD + (2) Electron-transfer diagram: acetaldehyde NAD + ethanol NADH + H + (3) Half-reactions Acetaldehyde + 2 H + + 2e Ethanol + NADH NAD + + H + + 2e Acetaldehyde + NADH + H + Ethanol + NAD + e In the reaction shown above: NADH is oxidized to NAD + acetaldehyde is reduced to ethanol acetaldehyde is the oxidizing agent NADH is the reducing agent NADH and ethanol are the reduced forms NAD + and acetaldehyde are the oxidized forms 1/6

There are two main ways that redox chemistry is discussed this class: 1) Given a redox reaction and the direction it proceeds, what is the e flow (that is, which molecule is oxidized and which is reduced)? In order to solve this kind of problem, you will need to know how to tell which of the two forms of a given molecule is the reduced form and which is the oxidized form. The easiest way is to: LOOK IT UP. Use a table of standard oxidation or reduction potentials, like the one on page 6 of this handout. These show the two forms of many common molecules and the redox relationship between them. For example, take the following reaction from the citric acid cycle: succinate + FAD fumarate + FADH 2 Looking at the chart on page 6 of this handout, you ll find two half reactions relating the compounds in this reaction: (1) fumarate + 2H + + 2e succinate and (2) FAD + 2H + + 2e FADH 2 (Don't worry about the E values just yet.) These two half reactions add up to the overall reaction if you reverse reaction (1): succinate fumarate + 2H + + 2e + FAD + 2H + +2e FADH 2 succinate + FAD fumarate + FADH 2 Therefore: succinate is oxidized to fumarate FAD is reduced to FADH 2 FAD is the oxidizing agent succinate is the reducing agent succinate and FADH 2 are the reduced forms fumarate and FAD are the oxidized forms Or, to use an electron-transfer diagram: FAD fumarate e FADH 2 succinate 2/6

2) Given a redox reaction, in which direction will it proceed spontaneously? In General: Different molecules have different tendencies to lose e, which reflects the degree to which they are reduced. Their tendency to lose e is reflected by their position in the chart on page 5; more reduced molecules are at the lower right of the chart. The more reduced a molecule is, the more potential it contains to be released for biological work. For example: NH 4 + N e releases NH + 4 is reduced; it has a tendency to release e (it is lower in the chart). is very oxidized; it has a strong tendency to accept e. Therefore, if you transfer e from NH + 4 to ( up the chart ), will be released ( G<0). This process can continue as e are transferred to from NO 2 ; this will also release : e releases On the other hand, the reverse reaction (transfer e from to N ; down the chart ) requires input to run in the direction shown ( G>0): NH 4 + N NO 3 e releases e requires input N NO 3 This can be put in mathematical terms using the E values on the chart. This is analogous to figuring out G for a reaction to determine which direction will proceed spontaneously. In the case of redox reactions, the of the reaction is measured in volts; each reaction has a standard potential (voltage) E. G and E are related by the Nernst equation: G 0 = -nf( E 0) Where: n = number of electrons transferred per reaction F = the Faraday constant kcal 23 Volt mol E 0 = E 0(reducing agent) E 0(oxidizing agent) 3/6

Note: to determine the direction of a redox reaction, you do not need to use the Nernst equation. You only need to know the sign of E 0. E 0< 0 reaction spontaneous E 0> 0 reaction spontaneous (note that this is the reverse of G) This is shown for the previous reaction: In order to arrive at the desired overall reaction: you must reverse the sign of E 0 for a reaction if you reverse the direction of the reaction. Here reaction (2) must be reversed. Even though we multiplied the reactions by a constant (6 reaction 1; 12 reaction 2) to balance the reaction, we did not multiply E 0 even if you multiply the reaction by constant (this is not the same as G). This is shown below: 6C + 24H + + 24e glucose + 6 E 0 = -0.43V + 12H 2 S 12S + 24H+ + 24e E 0 = +0.243V 6C + 12H 2 S glucose (C 6 H 12 O 6 ) + 6 + 12S E 0 = -0.187V Since E 0< 0, the reaction is spontaneous to the left ( ). This means that it requires input to go to the right therefore, input is required get net glucose synthesis. Since this is a photosynthetic reaction, and a substantial amount of light is available, this is no problem. Note that, since we did not need to balance the reaction to calculate E, we could have simplified the calculation as follows: 6C + e glucose E 0 = -0.43V + H 2 S S + e E 0 = +0.243V C + H 2 S glucose + S E 0 = -0.187V In all cases, you do not have to balance the reaction to find E. G 0 can be calculated in the following way: (Note: you will not need to do, it is only included FYI) This is where balancing the reaction comes in you have to balance the reaction to get n, the number of e transferred. In the case of the above reaction n = 24. G 0 = -nf( E 0) G 0 = -24 kcal 23 Volt mol (-0.187V) G 0 = +103 kcals/mol This is a highly endothermic reaction therefore a substantial input of is required to get net production of glucose. Since this is a photosynthetic reaction, and a substantial amount of light is available, this is no problem. 4/6

Final Note: This overview of oxidation-reduction reactions has been grossly oversimplified to provide the level of understanding necessary to cover the basics of photosynthesis and respiration. In reality, you will find that all redox reactions are affected by ph and the concentrations of products and reactants. Furthermore, some products or reactants can be further stabilized by other chemical reactions (e.g. Fe 3+ can precipitate as Fe(OH) 3 ) which will also alter the equilibrium. Because of these various circumstances, you may come across situations that do not seem to make sense on the basis of standard redox potentials alone. The chart on the following page shows the tendency of various compounds to gain or lose e. The chart on the next page shows the tendency of various compounds to gain or lose e. 5/6

Standard Eº values (at 25 C and ph 7) Half-Reaction E 0 (V) 1/2 + 2 H + + 2 e 0.816 Fe 3+ + e Fe 2+ 0.771 NO 3 + 6 H + + 6 e 1/2 N 2 + 3 0.75 NO 3 + 2 H + + 2 e NO 2 + 0.421 NO 3 + 10 H + + 8 e NH + 4 + 3 0.36 NO 2 + 8 H + + 6 e NH + 4 + 2 0.34 CH 3 OH + 2 H + + 2 e CH 4 + 0.17 fumarate + 2 H + + 2 e succinate 0.031 2 H + + 2 e H 2 (ph 0) 0.00 oxaloacetate + 2 H + + 2 e malate -0.166 C + 2 H + + 2 e CH 3 OH -0.18 pyruvate + 2 H + + 2 e lactate -0.185 acetaldehyde + 2 H + + 2 e ethanol -0.197 S 4 + 8 H + + 6 e S + 4-0.20 S 4 + 10 H + + 8 e H 2 S + 4-0.21 FAD + 2 H + + 2 e FADH 2-0.219 C + 8 H + + 8 e CH 4 + 2-0.24 S + 2 H + + 2 e H 2 S -0.243 N 2 + 8 H + + 6 e + 2NH 4-0.28 NAD + + H + + 2 e NADH -0.320 NADP + + H + + 2 e NADPH -0.324 2 H + + 2 e H 2 (ph 7) -0.414 C + 4 H + + 4 e 1/6 glucose + -0.43 Fe 2+ + 2 e Fe -0.85 In general (these rules follow from the preceding discussion): if electrons are transferred from compounds at the bottom of the list to compounds at the top of the list, (e ) is released ( G < 0). For example, in respiration: glucose + C +, e are being transferred from glucose (making CO2 ) to (making ), therefore this releases a large amount of. if electrons are transferred from compounds at the top of the list to compounds at the bottom of the list, (e ) is required ( G > 0). For example, in photosynthesis: C + glucose +, e are being transferred from (making ) to C (making glucose), therefore this requires a large amount of. 6/6