C & EE 255B Prof. M. K. Stenstrom Winter 2010 BALANCING REDOX EQUATIONS Balancing redox (oxidation-reduction) equations is a simple and very useful technique of performing balances from empirical equations describing microbial stoichiometry. Each basic equation (synthesis or growth, respiration, and decay) can all be balanced and added together to describe a process. Moreover, the equations can be balanced for each type of metabolism (aerobic or oxic, anoxic, and anaerobic). This handout describes the techniques for balancing the equations and then shows some common examples. Be careful when using empirical redox techniques; the reactions can be balanced, but other information must be used to determine if the reaction actually occurs. RULES Each redox equation contains two parts -- the oxidation and reduction parts. Each is balanced separately. 1. The first rule is to balance the major atoms with known end products. The end products of the redox equations must be stated or determined from other sources. The redox equations give you no information about the actual end products. Common end products for carbon are CO2 or cells. Other end products can occur as well. Major atoms are defined as all atoms except oxygen and hydrogen. 2. The next step is to balance the oxygen atoms by adding water (H2O) molecules. 3. Next balance the hydrogen with hydrogen ions (H+). 4. Finally balance the charge with electrons (e-). After these four steps one obtains a balanced half-reaction -- either the oxidation reaction or the reduction reaction. Oxidation reactions will produce electrons (electrons appear on the right-hand side of the equation). Reduction reactions will consume electrons (electrons appear on the left-hand side of the equation). SOME EXAMPLES Consider the oxidation of glucose, C6H12O6: Step 1. Balance the major atoms. In this case we will use CO2 as the end product for carbon. C6H12O6 --------> 6 CO2 Step 2. Balance the oxygen with water: C6H12O6 + 6 H2O --------> 6 CO2 Step 3. Balance the hydrogen: C6H12O6 + 6 H2O --------> 6 CO2 + 24 H+ Step 4. Balance the charge with electrons:2 C6H12O6 + 6 H2O --------> 6 CO2 + 24 H+ + 24 e- (1) Note that the charge and all atoms are balanced. This is the balanced half-reaction. It is an oxidation reaction. Note that electrons are produced or are removed from the substrate (glucose). The oxidation state of carbon is zero on the left hand side and + 2 on the right hand side of the equation. For this reaction to occur it must be balanced by a second half-reaction. It contains a reactant that is called the hydrogen acceptor or electron acceptor. Metabolisms are distinguished by which electron acceptors are used. Aerobic metabolism uses oxygen, anoxic metabolism uses sulfate or nitrate or nitrite, and anaerobic metabolism uses CO2 or an organic molecule. Environmental engineers need to differential between anoxic and anaerobic, since the difference can make the difference between a successful and unsuccessful process. Many microbiologists do not make this distinction. Anaerobic reactions are sometimes called fermentations. For aerobic reactions we can write the balanced half-reaction as follows: Step 1. Balance the major atoms. No major atoms! O2 ----------> Step 2. Balance the oxygen with waters: O2 ----------> 2 H2O Step 3. Balance the hydrogen: O2 + 4 H+ ----------> 2 H2O Step 4. Balance the charge with electrons: O2 + 4 H+ + 4 e-----------> 2 H2O (2) Equations (1) and (2) can now be added to balanced the electrons to zero by multiplying equation (2) by 6 and adding them together as follows: C6H12O6 + 6 H2O --------> 6 CO2 + 24 H+ + 24 e- 6 O2 + 24 H+ + 24 e-----------> 12 H2O Summation C6H12O6 + 6 O2 --------> 6 CO2 + 6 H2O (3) This equation represents the respiration or Theoretical Oxygen Demand (ThOD) equation. From this equation one can calculate the ThOD as 6 moles of O2 per mole of glucose, or 1.07 g O2/g glucose. Be careful to calculate the ratios correctly, as follows: ThOD = 6*32 /(6*12 + 12 + 6*16) = 1.07 (4) Equation (1) can be written with different metabolisms (electron acceptors) using the following half equations: Anoxic:3 NO3- + 6 H+ + 5 e- -------> 0.5 N2 +3 H2O (5) NO2- + 4 H+ + 3 e- --------> 0.5 N2 +2 H2O (6) SO4= + 10 H+ + 8 e- -------> H2S + 4 H2O (7) Anaerobic CO2 + 8 H+ + 8 e- -------> CH4 + 2 H2O (8) At this point in the notes and lecture, you should be able to write balanced redox equations for any reactants, given their end products for the major atoms. EMPIRICAL MICROBIAL EQUATIONS So far in this handout we have described redox equations and how to balance them without regard to how they maybe used. The empirical equations are quite useful to describe the stoichiometry of biological processes. Recall that there are three basic paths in the carbon cycle, as follows: Figure 1. Stoichiometric relationships in microbial growth. Equation A above is the synthesis or growth equation. Equation B is the respiration equation, while equation C is the decay equation. Equation B and C always use the highest oxidation state of the end products as their normal condition. An exception relates to ammonia. In a fully nitrifying system, the end-product form of nitrogen is nitrate. In non-nitrifying systems, ammonia is the end product. SYNTHESIS We can write the synthesis equation if we know an empirical formula for cells. In the 50s and later, activated sludge process cells were analyzed with a carbon-hydrogen-nitrogen analyzer and cell empirical formulas were found to be close to C5H7NO2 for a wide range of activated sludge processes. Later work confirmed the earlier results, although for some systems with very long sludge ages the empirical formula maybe different due to the accumulation of biologically inert, organic products. We begin the balance of our equations in the same was as before by balancing the major atoms first. Let us consider sucrose as a substrate. C12H22O11 -------------> C5H7NO2 Balance the major atoms. Use ammonia as the nitrogen source Substrate Nutrients CellsWaste Products (CO2, H2O)Decay (endogenous respiration)Electron AcceptorElectron AcceptorWaste Products (CO2, H2O, NH4, refractory organics)Electron AcceptorABC4 5 C12H22O11 + 12 NH4+ -------------> 12 C5H7NO2 Balance the oxygen with