~redox_reactions_f07

Oxidation-reduction (redox) reactions

• Reactions in which there are changes in oxidation state (oxidation number) between reactants and products

2 MnO4- + 10 Br- + 16 H+ → 2 Mn2+ + 5 Br2 + 8 H2O

• One reactant must be oxidized (lose electrons, the reductant or reducing agent) and another must be reduced (gain electrons, the oxidant or oxidizing agent)

– In some instances a single reactant will be both oxidized and reduced (disproportionation)

– in other cases two species containing the same element in different oxidation states will combine to give a single productwith an intermediate oxidation state (comproportionation).

Oxidation-reduction (redox) reactions, cont’d

• Redox reactions may be separated into oxidation and reduction “half-reactions”.

MnO4- + 8 H+ + 5 e- → Mn2+ + 4 H2O

2 Br- → Br2 + 2 e-

2 MnO4- + 10 Br- + 16 H+ → 2 Mn2+ + 5 Br2 + 8 H2O

– When half-reaction reagents can be isolated into separate compartments connected by a conduit for ion migration, electron flow through an external circuit can be utilized to perform work (basis for batteries and electrochemical cells).

– The potential (Eo) of an electrochemical cell is the sum of the potentials of the reduction and oxidation half-reactions

– The potential for a cell or half-reaction is related to the free energy change for the redox reaction through the relationship ΔGo = -nöEo where n is the number of electrons transferred and ö is the Faraday (9.65 x 104 J V-1 mol-1)

An electrochemical cell

Zn(s) = Zn2+ + 2 e- Eo = -(-0.76) V Cu2+ + 2 e- = Cu(s) Eo = 0.34 V

Zn(s) + Cu2+ = Zn2+ + Cu(s) Eo (cell) = (0.76 + 0.34) =1.10 V

ΔGo = -nöEo

anode(oxidation)

cathode(reduction)

Cu2+ CuZn Zn2+

Galvanic vs electrolytic cells

For a Galvanic Cell, a spontaneous reaction takes place.

ΔE > 0 (positive)

ΔG < 0 (negative)

For the Galvanic cell, the cell performs electrical work on the surroundings (acts as a battery)

A positive voltage means a rxn is thermodynamically favored and thus reactants are “unstable”

For an Electrolytic Cell, a non-spontaneous reaction takes place.

ΔE < 0 (negative)

ΔG > 0 (positive)

For the electrolytic cell, electrical work is performed on the system

A negative voltage means a rxn is not thermodynamically favored and thus reactants are “stable”

Cu2+ CuZn Zn2+

Zn2+ ZnCu Cu2+

Conventions relative to electrode potentials

• All half-reactions are written as reduction reactions

• Potentials of half-reactions are relative to H+(aq) + e- = ½

H2(g) with Eo = 0.00 V

• Potentials are recorded for standard conditions: 1.0 M conc. for soluble reactants/products, 100 kPa (1 atm) for gases, pure solids

• Cell potentials are obtained by adding potentials for half-reactions

– Sign of potential changes when half-reaction is written as oxidation

• Changes in half-reaction or cell potentials resulting from non-standard conditions can be determined using the Nernst equation

E = Eo - RTnF

[products]

[reactants]ln

R = 8.31 V C mol-1 K-1 ö = 9.65 x 104 C mol-1

ö

Half-Reaction E° (volts)Li+(aq) + e- = Li(s) -3.04K+

(aq) + e- = K(s) -2.92Ca2+

(aq) + 2e- = Ca(s) -2.76Na+

(aq) + e- = Na(s) -2.71Mg2+

(aq) + 2e- = Mg(s) -2.38Al3+

(aq) + 3e- = Al(s) -1.662 H2O(liq) + 2e- =

H2(g) + 2 OH-(aq) -0.83

Zn2+(aq) + 2e- = Zn(s) -0.76

Cr3+(aq) + 3e- = Cr(s) -0.74

Fe2+(aq) + 2e- = Fe(s) -0.41

Cd2+(aq) + 2e- = Cd(s) -0.40

Ni2+(aq) + 2e- = Ni(s) -0.23

2 H+(aq) + 2e- = H2(g) 0.00

Half-Reaction E° (volts)Sn4+

(aq) + 2e- = Sn2+(aq) 0.15

Cu2+(aq) + e- = Cu+

(aq) 0.16Cu2+

(aq) + 2e- = Cu(s) 0.34Cu+

(aq) + e- = Cu(s) 0.52I2(s) + 2e- = 2I-(aq) 0.54O2(g) + 2H+(aq) + 2e- =

H2O2(aq) 0.70Fe3+

(aq) + e- = Fe2+(aq) 0.77

Ag+(aq) + e- = Ag(s) 0.80

NO3-(aq) + 4H+

(aq) + 3e- = NO(g) + 2 H2O(liq) 0.96

Br2(liq) + 2e- = 2 Br-(aq) 1.07

O2(g) + 4 H+(aq)+ 4e- = 2 H2O(liq)1.23

Cr2O72-

(aq) + 14 H+(aq) + 6e- =

2 Cr3+(aq) + 7 H2O(liq) 1.33

Cl2(g) + 2e- = 2 Cl-(aq) 1.36Ce4+

(aq) + e- = Ce3+(aq) 1.44

MnO4-(aq) + 8 H+

(aq) + 5e- = Mn2+

(aq) + 4 H2O(liq) 1.51H2O2(aq) + 2 H+

(aq) + 2e- = 2 H2O(liq) 1.78

Co3+(aq) + e- = Co2+

(aq) 1.82F2(g) + 2e- = 2 F-

(aq) 2.87

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Some electrode potentials

A is the reductant (reducing agent); gives up electrons, i.e., A = A+ + e-

B is the oxidant (oxidizing agent); gains electrons, i.e., B + e- = B-

A is oxidized by B

B is reduced by A

A+ is the oxidation product

B- is the reduction product

For this reaction to be favorable as written entries in a table of electrode (half-reaction) potentials would appear as follows:

A+ + e- = A

B + e- = B-

The reduction potential for A+/A will be less positive than that for B/B-

Getting the terminology straight

A + B = A+ + B-

more reducing more oxidizing

Rules for assigning oxidation state

• The sum of the oxidation states for all atoms in a species must equal the charge on the species.

• The oxidation state of an atom in an elemental form is zero.

• The oxidation state of a monatomic ion is equal to its charge.

• Terminal atoms (or groups of atoms) are assigned an oxidation state consistent with the charge in their monatomic anionic form, i.e., filled valence shell.

• The most electronegative element is assigned a negative oxidation state in compounds of hydrogen.

• In compounds such as HO-OH, H2N-NH2, etc. the E-E bond does not contribute to the oxidation state of either atom. However, in compounds such as SSO3

2- (S2O32-, thiosulfate)

assignment of a filled valence shell to the terminal sulfur gives it an oxidation state of -2.

Balancing oxidation-reduction (redox) reactions

• Write separate half-reactions for oxidation and reduction

• Balance each half-reaction for mass utilizing water and protons as necessary, i.e.,– Get rid of oxygen as water– Get rid of hydrogen as H+

– Get hydrogen from H+

– Get oxygen from water • Balance each half-reaction for charge by adding electrons to the right

(oxidation) or left (reduction) as required• Multiply half-reactions by the appropriate factors such that the electrons

lost in oxidation equal the electrons gained in reduction• Add the two half-reactions; if the same reagent appears as both reactant

and product, subtract the smaller amount from both sides of the equation

HNO3(aq) + Cu(s) = NO2(g) + Cu2+(aq)








and product, subtract the smaller amount from both sides of the equation

HClO = Cl2 + ClO3-








and product, subtract the smaller amount from both sides of the equation• If the reaction is conducted in basic medium, add OH- to both sides to

eliminate any H+ (H+ + OH- = H2O)

O

P4 = PH3 + H2PO- (basic solution)

Latimer diagrams – a way to simplify data

• Latimer diagrams summarize several electrode potentials (half-reactions) in abbreviated form for a given element

• Half-reactions are written in sequence, as reductions, with the most highly oxidized species to the left

• Only the species involving the element are shown; other species must be added using rules for balancing half-reactions

• When potential of a half-reaction is more positive than the one to the left the species will disproportionate

• Separate diagrams are required for acidic and basic conditions

Examples of Latimer diagrams

Acidic solution

Basic solution

Acidic solution

MnO4- + 8 H+ + 5 e- = Mn2+ + 4 H2O Eº = 1.49

Acidic solution

1.49

Basic solution

Examples and interpretation of Latimer diagrams

MnO4- + 4 H+ + 3 e- = MnO2 + 2 H2O Eº = 0.59

Examples and interpretation of Latimer diagrams

Acidic solution

HClO2 + 2 e- + 2 H+ = HClO + H2O Eº = 1.674

HClO2 + H2O = ClO3- + 2 H+ + 2 e- Eº = -1.181

2 HClO2 = H+ + ClO3- + HClO Eº = 0.493

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~redox_reactions_f07