Oxidation-Reduction (Redox) Reactions Conventions for Assigning Charge to Atoms • Rule (can never be violated): Charge must be conserved, i.e., the charge on a molecule must equal the • Def’n: Reactions in which one or more electrons is sum of the charges on the constituent atoms. shifted from one element to another (In acid/base, gas transfer, and precipitation reactions discussed • Conventions: previously, atoms changed partners, but each atom kept its electrons) 1. H has an oxidation number (charge) of +1. 2. O has an oxidation number of −2. • Oxidation = loss of e−, hence higher charge; 3. N has an oxidation number of −3 when bonded Reduction = gain of e−, hence lower charge; only to H or C, as it is in many organic compounds. Redox = transfer of e− (simultaneous oxidation and 4. S has an oxidation number of −2 when bonded only to H or C, as it is in many organic compounds. reduction); requires an ‘electron donor’ and an ‘electron acceptor’ • Application: The above conventions are applied in the order given, but can be violated if necessary to avoid violating the charge conservation rule. Oxidation Number Calculations Redox Changes Can Have Very Large Effects • Oxidation number of C in H2CO3: 2(+1) + C +3(−2) = 0; C = +4 – Methane (CH4 [C=−4]) vs. cellulose (complex, but C≈0) − vs. carbonate species (C=+4); In general, metabolism Oxidation number of C in HCO3 : 1(+1) + C +3(−2) = −1; C = +4 involves oxidation of organic carbon to carbon dioxide, 2− Same result for CO2, CO3 and photosynthesis reverses this process • Oxidation number of C in glucose (C6H12O6): – Corrosion involves oxidation of pure metals (oxid. state 6*C + 12(+1) +6(−2) = 0; C = 0 = 0) to metal ions (>0)* 2− • Oxidation number of S in SO4 , H2S: – Chlorine (Cl , [Cl=0]) is a disinfectant; chloride (Cl− 2− 2 In SO4 : S + 4(−2) = 0; S = +6 [Cl=−1]) is innocuous In H2S: 2(+1) + S = 0; S = −2 – Nitrification oxidizes ammonia (NH3 [N=0]) to nitrate • Oxidation number of Fe in Fe(s), Fe(OH)3(s): − In Fe(s): Fe = 0 ions (NO3 [N=+5]) In Fe(OH)3(s): Fe + 3(−2) + 3(+1) = 0; Fe = +3 1 Balancing Redox Reactions in Water Example. Balance the following redox reaction: • Determine oxidation numbers of all atoms in the reaction and identify atoms that undergo oxidization or reduction. abFe22+−+↔ Cr O c Fe(OH) (s) + d Cr OH + 27 3 ( )2 • Choose one of the oxidized or reduced species to have a stoichiometric coefficient of 1. • Assign other stoichiometric coefficients that are determined unambiguously by the choice made in Step 2. • Fe(II) is oxidized to Fe(III), and Cr(VI) is reduced to Cr(III) • Electrons must be conserved. Determine stoichiometric • Choose b=1. Then d must be 2 to balance the Cr atoms coefficients for other oxidized and reduced species by balancing the number of electrons lost by atoms that are oxidized with the acsFe22+−+↔ 1Cr O Fe(OH) ( ) + 2Cr OH + number gained by those that are reduced. 27 3 ()2 • Add H2O to either side of the reaction to balance the oxygen • Each Cr atom that reacts gains 3 e−, so for the given coefficients, 6 e− atoms. are transferred. Each Fe(II) releases only 1 e− when it is converted to • Add H+ to either side of the reaction to balance charge. Fe(III), so 6 Fe(II) ions must be oxidized. Thus: + • Hydrogen should automatically balance at this point. Check the 6Fe22+−+↔ 1Cr O 6Fe(OH) (s ) + 2Cr( OH) H balance to confirm. 27 3 2 Organic Matter and Oxygen Demand abFe22+−+↔ Cr O c Fe(OH) ( sd ) + Cr OH + 27 3 ( )2 ¾ Organic Matter: Compounds containing carbon and hydrogen, excluding carbonate • The reaction has 7 oxygen atoms on the left and 22 on the right. To species balance the oxygen, we add 15 water molecules to the left: z Millions of organic compounds are generated 6Fe22+−++↔ 1Cr O 15H O 6Fe(OH) (s ) + 2Cr OH + 27 2 3 ( )2 naturally, and tens of thousands by human activity z Many categories of biochemicals known and • The total charge on the left is +10, while that on the right is +2. quantified in various environments, but many more Adding 8 H+ to the right balances the charge and, as it must, also balances the H atoms (30 on each side). The final, balanced reaction unknown or unquantified is as follows: z “Natural organic matter” (NOM) ubiquitous in natural waters 6Fe22+−++↔ 1Cr O 15H O 6Fe(OH) (s ) + 2Cr OH+ + 8H + 27 2 3 ( )2 z Categorized as dissolved, particulate, or total organic carbon (DOC, POC, and TOC, respectively) 2 Measuring Organic Carbon ¾ Organic Carbon Content of Complex Mixtures z Remove (most) inorganic C from sample z Oxidize organic C to CO2 ⎛⎞mg C6126 H O⎛⎞ 1 mol C 6126 H O ⎛⎞6 mol C⎛⎞ 12,000 mg ⎜⎟5.0 ⎜⎟⎜⎟⎜⎟ z Transfer CO2 to gas phase (bubble with gas at low pH) ⎝⎠L⎝⎠ 180,000 mg⎝⎠ mol C6126 H O⎝⎠ mol C z Detect gaseous CO2 mg C = 2.0 L 1 Bio-Redox Reactions of Organic Matter 0.9 α α α [HCO] 0.8 0 1 2 TOTCO = 23 3 0.7 α0 0.6 ¾ Natural Breakdown of Organic Matter α 0.5 pH=pKa2 0.4 pH=pKa1 z Aerobic HCO = H P 0.3 microorganisms [ 2 3] CO22 CO Organics+ O2 ⎯⎯⎯⎯⎯→ 0.2 0.1 CO++ H O New Cells + NO , PO , SO , ... 0 22() 3 4 4 01234567891011121314 pH 3+ z Anaerobic (e.g., with Fe as electron acceptor) Organics+ Fe3+ ⎯⎯⎯⎯⎯microorganisms→ Total dissolved CO in equilibrium with air is ~1.2x10−5 mol/L 3 2+ at pH<4, but can be much greater at higher pH; Fe+++ CO22 H O New Cells +() NH 32 , H S, PO 4 , ... To transfer most of the TOTCO3 from solution to gas: lower pH and bubble with a CO –free gas 2 Some org-C converted to CO2 (releases energy) and some converted to new cells (requires energy) 3 Energy Storage in Organic Molecules Organic Matter and Oxygen Demand ¾ To a first approximation, electrons lost from organic ¾ Aerobic Breakdown of Organic Matter C when it is oxidized all have similar energy, so energy microorganisms available by redox reactions is determined by the Biodegradable Organics+n O2 ⎯⎯⎯⎯⎯→ electron acceptor CO22++ H O New Cells +() NO 3 , PO 4 , SO 4 , ... In order of decreasing energy preference: Oxygen> Nitrate > Mn2+ >Fe3+ > Organics Oxygen “demanded” by the organics, The more energy available from a given amount of the biochemical oxygen demand (BOD). oxidation, the more new biomass that can be generated from a given initial amount of organic z Rate of O2 consumption is a direct indicator of the carbon substrate rate of oxidation of biodegradable organics (and therefore rate of energy release and cell growth): rnr= O2 Organics oxid'n ¾ Assume organic oxidation is a first order reaction: 14 O2 needed for biodegradation, L 12 rkOrganics oxid'n=− 1 [Biodeg. Organics] 10 Assume n corresponds to 0.7 mg O2/mg Biodeg.Org O2 consumed by rnr==− nkBiodeg. Organics 8 biodegradation, BOD OOrganics2 oxid'n1[ ] 6 4 Biodegradable O2 remaining if Organics system is closed Example: 2 Concentration (mg/L) Concentration 0 Biodeg.org.init = 10 mg/L; n = 0.7 mg O2/mg org; [O2]init =12 mg/L 012345678910 Time (d) 14 O2 needed for biodegradation, L 12 ⎡⎤⎡⎤Biodeg. Biodeg. 10 Assume n corresponds to =−exp()kt1 0.7 mg O2/mg Biodeg.Org O2 consumed by ⎢⎥⎢⎥ 8 biodegradation, BOD ⎣⎦⎣⎦Organicst Organics 0 6 4 Biodegradable O2 remaining if L = BOD remaining (not yet exerted) Lt =−Lkt01exp( ) Organics system is closed 2 Concentration (mg/L) Concentration yLLL=−=1exp − − kt 0 y = BOD exerted tt00{ ( 1)} 012345678910 Time (d) OO=−=−−−y OLkt 1exp O2 remaining [ 22]t [ ]00t [ 20] { ( 1)} 4 12 O2 needed for [Biod.org.]init=10mg/L biodegradation, Lt 10 [O2]init =5.5mg/L < L(0) O2 remaining if n = 0.7 mg O2/mg Biodeg.Org 8 system is closed O2 consumed by 6 biodegradation, BODt 4 Biodegradable Organics 2 C o n c e n tra tio n (m g /L ) 0 012345678910 Time (d) If insufficient O2 is present (or subsequently provided) to meet the full demand, the system will go anoxic and BOD exertion will cease or at least dramatically slow down. 5.