Oxidation-Reduction Reactions Carbonate reactions are acid-base reactions Carbonate reactions are acid-base reactions Transfer of protons – H + Transfer of protons – H + Other systems are similar Other systems are similar H 2 SO 4 H 2 SO 4 H 2 PO 3 H 2 PO 3 HNO 3 HNO 3

Redox reactions are analogous, but are transfer of electron Redox reactions are analogous, but are transfer of electron Very important class of reactions Very important class of reactions Elements have variety of valence states Elements have variety of valence states Number of electrons control valence and thus species Number of electrons control valence and thus species

Primary element valence states of some elements Primary element valence states of some elements C = +4 or -4 C = +4 or -4 S +6 or -2 S +6 or -2 N +5 or +3, also +4, +2 N +5 or +3, also +4, +2 Fe +3 or +2 Fe +3 or +2 Mn +3 or +2, also +7, +6, +4 Mn +3 or +2, also +7, +6, +4

Minor elements also have various valence states Minor elements also have various valence states V, Cr, As, Mo, V, Se, Sb, W, Cu… V, Cr, As, Mo, V, Se, Sb, W, Cu… All nasty elements All nasty elements Important environmental controls – e.g., mining Important environmental controls – e.g., mining

Valence state very important for toxicity and mobility Valence state very important for toxicity and mobility Fe 3+ (oxidized) is highly insoluble – precipitate as Fe-oxide minerals (goethite, lepidocrocite, limonite) Fe 3+ (oxidized) is highly insoluble – precipitate as Fe-oxide minerals (goethite, lepidocrocite, limonite) Fe 2+ (reduced) much more soluble – most Fe in solution is +2 valence Fe 2+ (reduced) much more soluble – most Fe in solution is +2 valence

Assignment of oxidation state Oxidation state of oxygen is always -2 except for peroxides, where it is -1. Oxidation state of oxygen is always -2 except for peroxides, where it is -1. E.g., H 2 O 2 and Na 2 O 2 E.g., H 2 O 2 and Na 2 O 2 Oxidation state of hydrogen is +1 in all compounds except metals where it is -1. Oxidation state of hydrogen is +1 in all compounds except metals where it is -1. NaH NaH NaBH 4 NaBH 4 LiAlH 4 LiAlH 4

All other oxidation states are selected to make the compound neutral All other oxidation states are selected to make the compound neutral Certain elements almost always have the same oxidation state Certain elements almost always have the same oxidation state Alkali metals = +1 Alkali metals = +1 Alkaline earths = +2 Alkaline earths = +2 Halogens = -1 Halogens = -1

Examples What are the oxidation states of N in NO 3 - and NO 2 - ? What are the oxidation states of N in NO 3 - and NO 2 - ? 3O 2- + N x = NO x = -1 3O 2- + N x = NO x = -1 2O 2- + N x = NO x = -1 2O 2- + N x = NO x = -1 N = +5 N = +3

What are the oxidation states of H 2 S and SO 4 2- ? What are the oxidation states of H 2 S and SO 4 2- ? 2H + + S x = H 2 S 2+ + x = 0 2H + + S x = H 2 S 2+ + x = 0 4O 2- + S x = SO x = -2 4O 2- + S x = SO x = -2 S = -2 S = +6

Oxidation Reactions Oxidation can be thought of as involving molecular oxygen Oxidation can be thought of as involving molecular oxygen 3Fe 2 O 3 2Fe 3 O 4 + 1/2O 2 3Fe 2 O 3 2Fe 3 O 4 + 1/2O 2 (hematite)(magnetite) 6Fe 3+ 2Fe Fe 3+ High O contentLower O content In this case, the generation of molecular oxygen controls the charge imbalance

Also possible to write these reactions in terms of electrons: Also possible to write these reactions in terms of electrons: 3Fe 2 O 3 + 2H + + 2e - 2Fe 3 O 4 + H 2 O 3Fe 2 O 3 + 2H + + 2e - 2Fe 3 O 4 + H 2 O (LEO)(GER)

Generally easiest to consider reactions as transfer of electrons Generally easiest to consider reactions as transfer of electrons Reactions may not involve molecular oxygen Reactions may not involve molecular oxygen

Problem is that free electrons are not really defined Problem is that free electrons are not really defined Reactions that consume “free electrons” represent only half of the reaction Reactions that consume “free electrons” represent only half of the reaction A complimentary reaction required to produce a “free electron” A complimentary reaction required to produce a “free electron” Concept is two “half reactions” Concept is two “half reactions”

Half Reaction Example of redox reaction without oxygen: Example of redox reaction without oxygen: Here Zn solid releases electron, which is consumed by dissolved Cu 2+. Here Zn solid releases electron, which is consumed by dissolved Cu 2+. Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq)

Physical model of process Ammeter e- anions cations Dissolves Precipitates Increases Decreases

Ammeter shows flow of electrons from Zn to Cu: Ammeter shows flow of electrons from Zn to Cu: Zn rod dissolves – Zn 2+ increases Zn rod dissolves – Zn 2+ increases Cu rod precipitates – Cu 2+ decreases Cu rod precipitates – Cu 2+ decreases

At the rod, the reactions are: At the rod, the reactions are: Zn = Zn 2+ (aq) + 2e- 2e- + Cu 2+ (aq) = Cu Zn + Cu 2+ (aq) = Zn 2+ (aq) + Cu Half reactions

Benefits: Benefits: Half reactions help balance redox reactions Half reactions help balance redox reactions Used to create framework to compare strengths of oxidizing and reducing agents Used to create framework to compare strengths of oxidizing and reducing agents

Rules for writing and balancing half reactions 1. Identify species being oxidized and reduced 2. Write separate half reactions for oxidation and reduction 3. Balance reactions with respect to atoms and electrical charge by adding e- or H+ 4. Combine half reactions to form net oxidation-reduction reactions

Consider reaction Consider reaction First, ID oxidized and reduced species: First, ID oxidized and reduced species: Iodine is being oxidized from -1 to 0 charge Iodine is being oxidized from -1 to 0 charge Oxygen in peroxide is being reduced to water Oxygen in peroxide is being reduced to water H 2 O 2 + I - I 2 + H 2 O I-I2I-I2 H2O2H2OH2O2H2O

Next – balance elements (oxidation half reaction: Next – balance elements (oxidation half reaction: And charge: And charge: 2I-I22I-I2 2I - I 2 + 2e-

Balance reduction half reaction Balance reduction half reaction First balance oxygen, then add H+ to balance hydrogen, then add electrons for electrical neutrality: First balance oxygen, then add H+ to balance hydrogen, then add electrons for electrical neutrality: H2O2H2OH2O2H2O H2O22H2OH2O22H2O 2H + + H 2 O 2 2H 2 O 2e- + 2H + + H 2 O 2 2H 2 O

Combine two half reactions to get net reactions: Combine two half reactions to get net reactions: 2I - I 2 + 2e- 2e- + 2H + + H 2 O 2 2H 2 O 2H + + 2I - + H 2 O 2 2H 2 O + I 2 Flow of electrons – Oxygen is electron acceptor, reduced; I- is electron donor, oxidized

Common reaction in natural waters is reduction of Fe 3+ by organic carbon Common reaction in natural waters is reduction of Fe 3+ by organic carbon With half reactions: With half reactions: 4Fe 3+ + C + 2H 2 O4Fe 2+ + CO 2 + 4H + C + 2H 2 OCO 2 + 4H + + e- 4Fe 3+ + e-4Fe 2+

From thermodynamic conventions, its impossible to consider a single half reaction From thermodynamic conventions, its impossible to consider a single half reaction There is no thermodynamic data for e- There is no thermodynamic data for e- Practically, half reactions are defined relative to a standard Practically, half reactions are defined relative to a standard The standard is the “Standard Hydrogen Electrode (SHE)” The standard is the “Standard Hydrogen Electrode (SHE)”

SHE By definition, a H+ = 1 Allows electrons to flow but chemically inert

SHE Platinum electrode in solution containing H 2 gas at P = 1 Atm. Platinum electrode in solution containing H 2 gas at P = 1 Atm. Assign arbitrary values to quantities that can’t be measured Assign arbitrary values to quantities that can’t be measured Difference in electrical potential between metal electrode and solution is zero Difference in electrical potential between metal electrode and solution is zero  G f º of H + = 0  G f º of H + = 0  G f º of e - = 0  G f º of e - = 0

Example of how SHE used Fe 3+ + e- = Fe 2+ If wire removes electrons, reaction goes to left If wire adds electrons, reaction goes to right SHE: H + + e- = 1/2H 2(g) E = Potential Positive or negative

In cell A, platinum wire is inert – transfers electrons to or from solution only. In cell A, platinum wire is inert – transfers electrons to or from solution only. Pt wire develop an electrical potential – “tendency” for electrons to enter or leave solution Pt wire develop an electrical potential – “tendency” for electrons to enter or leave solution Define the potential as “activity of electrons” = a e- Define the potential as “activity of electrons” = a e- Not a true activity, really a “tendency” Not a true activity, really a “tendency” Define pe = -loga e-, similar to pH Define pe = -loga e-, similar to pH

In Cell A solution, Fe is both oxidized and reduced In Cell A solution, Fe is both oxidized and reduced Fe 2+ and Fe 3+ Fe 2+ and Fe 3+ Reaction is: Reaction is: If reaction goes to left, Fe 2+ gives up e- If reaction goes to left, Fe 2+ gives up e- If reaction goes to right, Fe 3+ acquires e- If reaction goes to right, Fe 3+ acquires e- If no source or sink of e-, (switch closed), volt meter measures the potential (tendency) If no source or sink of e-, (switch closed), volt meter measures the potential (tendency) Fe 3+ + e- = Fe 2+

Since we have a reaction, can write an equilibrium constant Since we have a reaction, can write an equilibrium constant K eq = a Fe2+ a Fe3+ a e-

Rearranged: Rearranged: a e- is proportional to the ratio of activity of the reduced species to activity of oxidized species a e- is proportional to the ratio of activity of the reduced species to activity of oxidized species a e- is electrical potential caused by ratio of reduced to oxidized species a e- is electrical potential caused by ratio of reduced to oxidized species a e- =K eq -1 a Fe2+ a Fe3+

Consider half cell B: Consider half cell B: Direction of reaction depends on tendency for wire to gain or lose electrons Direction of reaction depends on tendency for wire to gain or lose electrons Equilibrium constant Equilibrium constant H + + e- = 1/2H 2(g) Keq = P H2 1/2 a H+ a e-

Switch closed – electrons flow from one half cell to the other Switch closed – electrons flow from one half cell to the other Electron flow from the side with the highest activity of electrons to side with lowest activities Electron flow from the side with the highest activity of electrons to side with lowest activities

Switch open: Switch open: No longer transfer of electrons No longer transfer of electrons Now simply potential (E) generated at Pt wire Now simply potential (E) generated at Pt wire By convention, potential of SHE (E SHE ) = O By convention, potential of SHE (E SHE ) = O Potential called Eh, i.e. E measured relative to SHE Potential called Eh, i.e. E measured relative to SHE Eh > or or or or < that of SHE

Convention Convention Eh > 0 if a e- 0 if a e- < SHE I.e. if electrons flow from the SHE to the fluid I.e. if electrons flow from the SHE to the fluid

Expressions for activities of electrons: Expressions for activities of electrons: Eh or pe Eh or pe Pe = [F/(2.303RT)]*Eh Pe = 25ºC, pe = 16.9 Eh; Eh = 25ºC, pe = 16.9 Eh; Eh = 0.059pe F = Faraday’s constant = 96,485 coul/mol F = Faraday’s constant = 96,485 coul/mol Couomb = charge /electron = quantitiy of electricity transferred by 1 Amp in1 second. Couomb = charge /electron = quantitiy of electricity transferred by 1 Amp in1 second.