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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 Step 4 Balance the charge with electrons 6 CO2 24 H 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 2 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 CO2 8 H 8 e 8 Anaerobic CH4 2 H2O 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 Electron Acceptor Nutrients Electron Acceptor Cells A Substrate Electron Acceptor C Decay endogenous respiration B Waste Products CO2 H2O Waste Products CO2 H2O NH4 refractory organics 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 endproduct 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 3 5 C12H22O11 12 NH4 12 C5H7NO2 Balance the oxygen with water 5 C12H22O11 12 NH4 12 C5H7NO2 31 H2O Balance the hydrogen with hydrogen ions 5 C12H22O11 12 NH4 12 C5H7NO2 31 H2O 12 H 9 Note that there are no electrons needed to balance the equation This results because three is no net oxidation or reduction Equation 9 is the synthesis equation and describes the conversion of the substrate to cells There will usually be some reduction or oxidation in the synthesis equation Equation 9 shows 100 conversion of substrate to cells which is not possible since some of it must be oxidized to produce energy We now must combine the synthesis equation with the respiration equation to obtain a net equation RESPIRATION Equation

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