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Oxidation-Reduction Reactions
Carbonate reactions are acid-base reactions: Transfer of protons – H+
Other acid-base systems are similar: Sulfuric acid - H2SO4
Phosphoric acid - H2PO3
Nitric Acid HNO3
Redox reactions are analogous, but are transfer of electrons rather than protons
Very important class of reactions Elements may have variable charges –
number of electrons (valence state) Valence state controls speciation of
elements
Oxidation-Reduction Reactions
Examples of primary valence states of some elements C = +4 or -4 S +6 or -2 N +5 or +3, also +4, +2 Fe +3 or +2 Mn +3 or +2, also +7, +6, +4
Minor elements also have various valence states V, Cr, As, Mo, V, Se, Sb, W, Cu… All nasty elements Important environmental controls – e.g.,
mining
Valence state very important for mobility, as well as absorption and thus toxicity Fe3+ (oxidized) is highly insoluble
Precipitate as Fe-oxide minerals (magnetite, hematite, goethite, lepidocrocite, limonite)
Fe2+ (reduced) much more soluble – most Fe in solution is +2 valence
Common precipitates are Fe-sulfides (pyrite, marcasite)
Assignment of oxidation state
Valence state of oxygen is always -2 except for peroxides, where it is -1. E.g., H2O2 and Na2O2
Valence state of hydrogen is +1 in all compounds except when bonded with metals where it is -1. NaH NaBH4
LiAlH4
Valence state of all other elements are selected to make the compound neutral
Certain elements almost always have the same oxidation state Alkali metals = +1 (left most column) Alkaline earths = +2 (second column
from left) Halogens = -1 (2nd column from right)
Examples
What are the oxidation states of N in NO3
- and NO2-?
3O2- + Nx = NO3- 6- + x = -1
2O2- + Nx = NO2- 4- + x = -1
N = +5
N = +3
What are the oxidation states of H2S and SO4
2-? 2H+ + Sx = H2S 2+ + x = 0
4O2- + Sx = SO42- 8- + x = -2
S = -2
S = +6
Oxidation Reactions
Oxidation can be thought of as involving molecular oxygen 3Fe2O3 2Fe3O4 + 1/2O2
(hematite) (magnetite)All as Fe3+ One as Fe2+ + two as Fe3+
High O/Fe ratio Lower O/Fe ratioOxidized Reduced
In this case, the generation of molecular oxygen controls the charge imbalance
Also possible to write these reactions in terms of electrons: 3Fe2O3 + 2H+ + 2e- 2Fe3O4+ H2O LEO – lose electron oxidation – the Fe3+ is
oxidized GER – gain electron reduction – the Fe2+
is reduced OIL – oxidation is loss RIG – reduction is gain
Generally easiest to consider reactions as transfer of electrons Many redox reaction do not involve
molecular oxygen
Problem is that free electrons are not really defined Reactions that consume “free electrons”
represent only half of the reaction A complementary reaction required to
produce a “free electron” Concept is two “half reactions” The half reaction simultaneously create
and consume electrons, so typically not expressed in reaction
Half Reactions
Example of redox reaction without oxygen:
Here Zn solid releases electron, which is consumed by dissolved Cu2+.
Zn(s) + Cu2+(aq) Cu(s) + Zn2+
(aq)
Physical model of processAmmeter
e-
e-
anions
cations
DissolvesPrecipitates
Increases Decreas
es
Salt bridge – keeps charge balance in solution.
Ammeter shows flow of electrons from Zn to Cu: Zn rod dissolves – Zn2+ increases Cu rod precipitates – Cu2+ decreases
At the rod, the reactions are:
Zn = Zn2+(aq) + 2e-
2e- + Cu2+(aq) = Cu
Zn + Cu2+(aq) = Zn2+
(aq) + Cu
Half reactions
Benefits of using half reactions: Half reactions help balance redox
reactions 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 using (1) atoms and (2) electrical charge by adding e- or H+
4. Combine half reactions to form net oxidation-reduction reactions
Consider reaction
First, ID oxidized and reduced species: Iodine is being oxidized from -1 to 0
charge Oxygen in peroxide is being reduced to
water
H2O2 + I- I2 + H2O
I- I2
H2O2 H2O
Next – balance elements (oxidation half reaction:
And charge:
2I- I2
2I- I2 + 2e-
Balance reduction half reaction First balance oxygen, then add H+ to
balance hydrogen, then add electrons for electrical neutrality:
H2O2 H2O
H2O2 2H2O
2H+ + H2O2 2H2O
2e- + 2H+ + H2O2 2H2O
Combine two half reactions to get net reactions:
2I- I2 + 2e-
2e- + 2H+ + H2O2 2H2O
2H+ + 2I- + H2O2 2H2O + I2
Flow of electrons – Oxygen is electron acceptor, reduced; I- is electron donor, oxidized
Common reaction in natural waters is reduction of Fe3+ by organic carbon
With half reactions:
4Fe3+ + C + 2H2O 4Fe2+ + CO2 + 4H+
C + 2H2O CO2 + 4H+ + 4e-
4Fe3+ + 4e- 4Fe2+
From thermodynamic conventions, its impossible to consider a single half reaction There is no thermodynamic data for e-
Practically, half reactions are defined relative to a standard
The standard is the “Standard Hydrogen Electrode (SHE)”
SHE
Platinum electrode in solution containing H2 gas at P = 1 Atm.
Assign arbitrary values to quantities that can’t be measured Difference in electrical potential
between metal electrode and solution is zero
DGfº of H+ = 0 DGfº of e- = 0
SHE
By definition,aH+ = 1
Allows electrons to flow but chemically inert
Half reaction in solution:H+ + e- = 1/2H2(g)
Example of how SHE used
Fe3+ + e- = Fe2+
If reaction goes to left, wire removes electronsIf reaction goes to right, wire adds electrons
SHE:H+ + e- = 1/2H2(g)
E = PotentialPositive or negative
In cell A, platinum wire is inert – transfers electrons to or from solution only.
If wire has no source of electrons Pt wire develop an electrical potential –
“tendency” for electrons to enter or leave solution
Define the potential as “activity of electrons” = ae-
Not a true activity, really a “tendency” Define pe = -logae-, similar to pH
In Cell A solution, Fe is both oxidized and reduced Fe2+ and Fe3+
Reaction is:
If reaction goes to left, Fe2+ gives up e- If reaction goes to right, Fe3+ acquires e- If no source or sink of e-, (switch open),
volt meter measures the potential (tendency)
Fe3+ + e- = Fe2+
Since we have a reaction
can write an equilibrium constant
Keq = aFe2+
aFe3+ ae-
Fe3+ + e- = Fe2+
Rearranged:
ae- is proportional to the ratio of activity of the reduced species to activity of oxidized species
ae- is electrical potential (in volts) caused by ratio of reduced to oxidized species
ae-= Keq-1
aFe2+
aFe3+
Consider half cell B:
Direction of reaction depends on tendency for wire to gain or lose electrons
Equilibrium constant
H+ + e- = 1/2H2(g)
KSHE = PH2
1/2
aH+ ae
-
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
Overall reaction:
Direction of reaction depends on which half cell has highest activity of electrons
Fe3+ + 1/2H2(g) =Fe2+ + H+
Flow of electrons
Switch open: No longer transfer of electrons Now simply potential (E) generated at Pt
wire By convention, potential of SHE (ESHE) =
O Potential called Eh, i.e. E (electromotive
force) measured relative to SHE (thus the “h”)
Eh > or < O depends on whether ae- is > or < that of SHE
Convention Eh > 0 if ae- of the half cell < SHE I.e. if electrons flow from the SHE to the
fluid For thermodynamics:
Is equivalent to:
Fe3+ + 1/2H2(g) =Fe2+ + H+
Fe3+ + e- =Fe2+
Expressions for activities of electrons: Eh or pe pe = [F/(2.303RT)]*Eh @ 25ºC, pe = 16.9 Eh; Eh = 0.059pe
F = Faraday’s constant = 96,485 coul/mol
Coulomb = charge /electron = quantity of electricity transferred by 1 Amp in 1 second.