Water and its Chemistry

Oxidation / Reduction

Rules for Assigning Oxidation States

1. Oxidation state of an atom in an element = 02. Oxidation state of monatomic element = charge3. Oxygen = 2 in covalent compounds (except in peroxides where it = 1)4. H = +1 in covalent compounds5. Fluorine = 1 in compounds6. Sum of oxidation states = 0 in compounds Sum of oxidation states = charge of the ion

Balancing by Half-Reaction Method

1. Write separate reduction, oxidation reactions.2. For each half-reaction:

- Balance elements (except H, O)

- Balance O using H2O

- Balance H using H+

- Balance charge using electrons

Balancing by Half-Reaction Method (continued)

3. If necessary, multiply by integer to equalize electron count.4. Add half-reactions.5. Check that elements and charges are balanced.

Half-Reaction Method - Balancing in Base

1. Balance as in acid.2. Add OH that equals H+ ions (both sides!)3. Form water by combining H+, OH.4. Check elements and charges for balance.

0.54 Li+ + e Li 3.05

17_01TTable 17.1 Standard Reduction Potentials at 25°C (298 K) for Many Common Half-reactions Half-reaction ° (V) Half-reaction

2.87 O2 + 2H2O + 4e 4HO 0.401.99 Cu2+ + 2e Cu 0.341.82 Hg2Cl2 + 2e 2Hg + 2Cl 0.341.78 AgCl + e Ag + Cl 0.221.70 SO4

+ 4H+ 2e H2SO3 + H2SO3 + H2O 0.201.69 Cu2+ + e Cu+ 0.161.68 2H+ + 2e H2 0.001.60 Fe3+ + 3e Fe 0.0361.51 Pb2+ + 2e Pb 0.131.50 Sn2+ + 2e Sn 0.141.46 Ni2+ + 2e Ni 0.231.36 PbSO4 + 2e Pb + SO4

2 0.351.33 Cd2+ + 2e Cd 0.401.23 Fe2+ + 2e Fe 0.441.21 Cr3+ + e Cr2+ 0.501.20 Cr3+ + 3e Cr 0.731.09 Zn2+ + 2e Zn 0.761.00 2H2O + 2e H2 + 2OH 0.830.99 Mn2+ + 2eMn 1.180.96 Al3+ + 3e Al 1.660.954 H2 + 2e 2H 2.230.91 Mg2+ + 2eMg 2.370.80 La3+ + 3e La 2.370.80 Na+ + e Na 2.710.77 Ca2+ + 2e Ca 2.760.68 Ba2+ + 2e Ba 2.90

I2 + 2e 2I

F2 + 2e 2F

Ag2+ + e Ag+

Co3+ + e Co2+

H2O2 + 2H+ + 2e 2H2OCe4+ + e Ce3+

PbO2 + 4H+ + SO42 + 2ePbSO4 + 2H2O

MnO4 + 4H+ + 3eMnO2 + 2H2O

2e + 2H+ + IO4IO3

+ H2OMnO4

+ 8H+ + 5eMn2+ + 4H2OAu3+ + 3e AuPbO2 + 4H+ + 2ePb2+ + 2H2OCl2 + 2e 2Cl

Cr2O72 + 14H+ + 6e2Cr3+ + 7H2O

O2 + 4H+ + 4e 2H2OMnO2 + 4H+ + 2eMn2+ + 2H2OIO3

+ 6H+ + 5e½I2 + 3H2OBr2 + 2e 2Br

VO2 + 2H+ + e VO2+ + H2OAuCl4

+ 3e Au + 4Cl

NO3 + 4H+3e NO + 2H2O

ClO2 + e ClO2

2Hg2+ + 2e Hg22+

Ag+ + e AgHg2

2+ + 2e 2Hg+

Fe3+ + e Fe2+

O2 + 2H+ + 2e H2O2MnO4

+ eMnO42 0.56 K+ + e K 2.92

0.52Cu+ + e Cu

° (V)

The pE ScalepE = -log(ae-)

Low pE means e- are available (reducing)High pE means e- not available (oxidzing)

pE at Standard

Conditionsfor Rxn’sin Natural

Waters

pE and Free EnergyG = -2.303nRT(pE)n = moles of e-R = gas constantT = absolute temperaturepE for the reaction

pE and EquilibriumLog K = n(pEo)Gives insight into which species will be dominate.

From the Nernst Equation

pE = pEo + log(Q)

Q is the reaction quotient

From the Nernst Equation