Oxidation-Reduction (Redox) Reactions

Eh, Ph, and the chemistry of natural waters

Important equations and terms for redox reactions:

ΔGR = nFEΔGR = ΔGR

° + RTlnQEh = E = E° + (2.303RT/nF)logQ = (0.05916/n)logQE° = -(0.05916/n)logKEh = 1.23 + 0.01479log[O2] - 0.05916pH

R = 1.987 x 10-3 kcal/mol°T = 298.15°KF = 23.06 kcal/Vn = number of electrons transferred in the reactionQ = reaction quotientE° = standard potential for a half-reactionEh = electromotive force (emf) generated between an electrode

in any state and the H2 electrode (the referencehydrogen scale Eh) in standard state (theoretical voltage)

Nernst factor at standard state

Zn(s) + Cu 2+ (aq) ←→ Zn 2+ (aq) +Cu(s)

Each metal has a different relative preference for electrons

Electrode Potential

Separate the reaction into half-reactions and add the e-

Zn(s) + Cu 2+ (aq) ←→ Zn 2+ (aq) +Cu(s)

Zn(s) ←→ Zn 2+ (aq) + 2e-

Cu 2+ (aq) + 2e-←→ Cu(s)There will be Energy gained or lost by the exchange of e-

Unfortunately, such measurements are not possible (nor would these reactions occur in the natural environment:electrons are not given up except to another element or species).

This requires the establishment of an arbitrary reference value

Hydrogen Scale Potential Eh

−+→← + eaqHgH )()(2

12

The potential measured for the entire reaction is thenassigned to the half-cell reaction of interest:

Zn(s) ←→ Zn 2+ (aq) + 2e- becomes

++ +→←+ HsZngHaqZn 2)()()( 22

The potential for this reaction = -0.763 V

So the Eh of the reduction of Zn2+ to Zn (s) = -0.763 V

(+) Eh means reaction moves left to right(-) Eh means reaction moves right to left

(+) Eh means the metal ion is reduced by H2(g) to the metal

(-) Eh means the metal will be oxidized to the ion and H+ is reduced

++ +→←+ HsZngHaqZn 2)()()( 22

Eh = E = E° + (2.303RT/nF)logQ = (0.05916/n)logQ

Calculating the Eh of Net Reactions

We can calculate Eh values by algebraic combinations of thereactions and potentials that are listed in Table 14.3 of the text.

There is, however, a “catch”

Calculate the Eh for the reaction:Fe3+(aq) + 3e- ←→ Fe(s)

This reaction is the algebraic sum of two reactions listed in Table 14.3

Fe3+(aq) + e- ←→ Fe2+(aq) Fe2+(aq) + 2e- ←→ Fe(s)

Since the reactions sum, we might assume that we can simply sum the Eh values to obtain the Eh of the net reaction.

Doing so, we obtain an Eh of 0.33 V. However, the true Eh of this reaction is -0.037 V.

What did we do wrong?

Fe3+(aq) + 3e- ←→ Fe(s)

We have neglected to take into consideration the number of electrons exchanged.

In the algebraic combination of Eh values, we need to multiply the Eh for each component reaction by the number of electrons exchanged.

We then divide the sum of these values by number of electrons exchanged in the net reaction to obtain the Eh of the net reaction

∑=i

iinet

Ehzz

netEh1

)(

Fe3+(aq) + e- ←→ Fe2+(aq) Fe2+(aq) + 2e- ←→ Fe(s)

∑=i

iinet

Ehzz

netEh1

)(

where the sum is over the component reactions i.

By Hess’s Law, the ΔG of the net reaction must be the simple sum of the component reaction ΔG’s

…but Eh values are obtained by multiplying ΔG by z.

ΔGR = nFE

∑=i

iinet

Ehzz

netEh1

)(

Eh = 1/3 (1*0.77+2*-0.44) = 1/3 (0.77-0.88) = 1/3(-0.11)=-0.037

Fe3+(aq) + e- ←→ Fe2+(aq) Fe2+(aq) + 2e- ←→ Fe(s)

nF

G-E

reaction-half of potential std. E

C25at volts0.05916 factor Nernst F

2.303RT

electrons ofnumber n

kcal/voltg 23.061 constant faraday F

Kelvinin temp. T

kcal/mol 0.001987 constant gasR

(D)(C)

(B)(A)ln

nF

RT EEh(volts)

ofo

o

o

o

dc

bao

Δ=

=

==

=

==

=

==

+=

eq

reduced oxidized

dD cC e bB aA

lnFG- E

][P

][Hln

F

RT E Eh

-

oR

o

1/2H

o

2

+=++

Δ=

++=

n

Eh = E° + (2.303RT/nF)logQ = (0.05916/n)logQ

From Garrels & Christ (1965)

If an environment is characterized by a certain redoxreaction, it has a characteristic Eh

Oxidizing

Reducing

Eh = 1.23 + 0.01479log[O2] - 0.05916pH

Similar to Fig. 14.4 on p. 238

[O2 ] = 10 -83.1 atm

[O2 ] = 1 atm

[O2] conditions for natural waters

1) write the equation for the reaction

)()()( 2221

2 gHgOlOH +↔

2) “electrode” the equation

−++↔ + egOHlOH 2)(2)( 221

2

2e-

O in H2O gives up 2e-

3)

( )22/12 ][][log

2

05916.0 ++= HOEEh o

4) get Eo from ΔGro

kcalOHGG

eGOGHG

lOHGeGOGHGG

of

or

of

of

of

of

of

of

of

or

687.56

0

)](([)](2)(2/1)(2[

2

2

22

+=−=Δ

=−==

−−++=Δ+

+

Vx

Eo 23.106.232

687.56+=+=

6)

( )pHOEh

HOEh

05916.0]log[01479.023.1

][][log2

05916.023.1

2

22/12

−+=

+= +

This demonstrates the Eh - pH relationship as shown in figure 14.4 p. 238 and is the equation for the lines limitingthe minimum and maximum [O2] for natural waters

Figure 14.11 on p. 248

Oxidizing

Reducing