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.