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Oxidation-Reduction (Redox) Reactions

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Redox: Introduction

• Electrons (e-) are transferred from one compound to another– e--donors (lose electrons) – e--acceptors (gain electrons)

• Loss of e- by e--donor = oxidation • Gain of e- by e--acceptor = reduction • Oxidation and reduction always

accompany one another – Electrons cannot exist freely in solution

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Balancing Redox Reactions

• Reaction needs to be charge balanced; e- need to be removed– To balance charge, cross multiply reactants

• Redox reactions can also be written as half reactions, with e- included– Multiply so that e- are equal (LCM) and

combine half reactions

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Redox Reactions• Ability of elements to act as e--donors or

acceptors arises from extent to which orbitals are filled with electrons – Property depends on decrease in energy of atoms

resulting from having only incompletely filled orbitals

• Some similarities to acid-base reactions– Transferring electrons (e-) instead of H+

– Come in pairs (oxidation-reduction)– In acid-base reactions, we measure changes in [H+]

(pH); in redox reactions, we measure voltage changes

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Measuring voltage

• Standard potential tables have been created for how much voltage (potential) a reaction is capable of producing or consuming– Standard conditions: P =1 atm, T = 298°K,

concentration of 1.0 M for each product– Defined as Standard Potential (E°)

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Non-Standard Redox Conditions (real life)

• For non-standard conditions, E° needs to be corrected – Temperature– Number of electrons transferred– Concentrations of redox reactants and

products

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Nernst Equation• Nernst Equation

–

• E = electrical potential of a redox reaction• R = gas constant• T = temperature (K)• n = number of electrons transferred• F = Faraday constant = 9.65 x 104 J / V∙mole• Q = ion activity product

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Nernst Equation (example)• Fe(s) + Cd2+ Fe2+ + Cd(s)

– Fe oxidized, Cd reduced• Need to find standard potentials of half

reactions– Fe2+ + 2e- Fe(s)

– Cd2+ + 2e- Cd(s)

• Look up in Table

Standard Potentials

Half-Reaction E0 (V)Li+(aq) + e- → Li(s) -3.05K+

(aq) + e- → K(s) -2.93Ba2+

(aq) + 2 e- → Ba(s) -2.90Sr2+

(aq) + 2 e- → Sr(s) -2.89Ca2+

(aq) + 2 e- → Ca(s) -2.87Na+

(aq) + e- → Na(s) -2.71Mg2+

(aq) + 2 e- → Mg(s) -2.37Be2+

(aq) + 2 e- → Be(s) -1.85Al3+

(aq) + 3 e- → Al(s) -1.66Mn2+

(aq) + 2 e- → Mn(s) -1.182 H2O + 2 e- → H2(g) + 2 OH-

(aq) -0.83Zn2+

(aq) + 2 e- → Zn(s) -0.76Cr3+

(aq) + 3 e- → Cr(s) -0.74Fe2+

(aq) + 2 e- → Fe(s) -0.44Cd2+

(aq) + 2 e- → Cd(s) -0.40PbSO4(s) + 2 e- → Pb(s) + SO4

2-(aq) -0.31

Co2+(aq) + 2 e- → Co(s) -0.28

Ni2+(aq) + 2 e- → Ni(s) -0.25

Sn2+(aq) + 2 e- → Sn(s) -0.14

Pb2+(aq) + 2 e- → Pb(s) -0.13

2 H+(aq) + 2 e- → H2(g) 0

Half-Reaction E0 (V)2 H+

(aq) + 2 e- → H2(g) 0Sn4+

(aq) + 2 e- → Sn2+(aq) 0.13

Cu2+(aq) + e- → Cu+

(aq) 0.13SO4

2-(aq) + 4 H+

(aq) + 2 e- → SO2(g) + 2 H2O 0.20AgCl(s) + e- → Ag(s) + Cl-(aq) 0.22Cu2+

(aq) + 2 e- → Cu(s) 0.34O2(g) + 2 H2 + 4 e- → 4 OH-

(aq) 0.40I2(s) + 2 e- → 2 I-

(aq) 0.53MnO4

-(aq) + 2 H2O + 3 e- → MnO2(s) + 4 OH-

(aq) 0.59O2(g) + 2 H+

(aq) + 2 e- → H2O2(aq) 0.68Fe3+

(aq) + e- → Fe2+(aq) 0.77

Ag+(aq) + e- → Ag(s) 0.80

Hg22+

(aq) + 2 e- → 2 Hg(l) 0.852 Hg2+

(aq) + 2 e- → Hg22+

(aq) 0.92NO3

-(aq) + 4 H+

(aq) + 3 e- → NO(g) + 2 H2O 0.96Br2(l) + 2 e- → 2 Br-

(aq) 1.07O2(g) + 4 H+

(aq) + 4 e- → 2 H2O 1.23MnO2(s) + 4 H+

(aq) + 2 e- → Mn2+(aq) + 2 H2O 1.23

Cr2O72-

(aq) + 14 H+(aq) + 6 e- → 2 Cr3+

(aq) + 7 H2O 1.33Cl2(g) + 2 e- → 2 Cl-(aq) 1.36Au3+

(aq) + 3 e- → Au(s) 1.50MnO4

-(aq) + 8 H+

(aq) + 5 e- → Mn2+(aq) + 4 H2O 1.51

Ce4+(aq) + e- → Ce3+

(aq) 1.61PbO2(s) + 4H+

(aq) + SO42-

(aq) + 2e- → PbSO4(s) + 2H2O 1.70H2O2(aq) + 2 H+

(aq) + 2 e- → 2 H2O 1.77Co3+

(aq) + e- → Co2+(aq) 1.82

O3(g) + 2 H+(aq) + 2 e- → O2(g) + H2O 2.07

F2(g) + 2 e- -----> F-(aq) 2.87

The greater the E°, the more easily the substance reduced

Strong Reducing Agents

Strong Oxidizing Agents

Written as reductions

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Nernst Equation (example)• By convention, half-reactions are shown

as reduction reactions• Fe(s) + Cd2+ Fe2+ + Cd(s)

– Fe2+ + 2e- Fe(s), E° = -0.44 V

– Cd2+ + 2e- Cd(s), E° = -0.40 V

– For oxidation, use negative value (+0.44 V)– Added together, get +0.04 V as standard

potential for complete reaction

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Nernst Equation (example)

• Assume [Fe2+] = 0.0100 M and [Cd2+] = 0.005 M (instead of the standard 1.0 M)–

• For Fe2+

– E = -0.44 – [(8.314)(298)]/[(2)(9.65x10-4)] ln(1/0.001) = -0.50

• Fe Cd2+

– E = -0.40 – (0.128) ln (1/0.005) = -0.47

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Redox and Thermodynamics

• Energy released in a redox reaction can be used to perform electrical work using an electrochemical cell– A device where electron transfer is forced to

take an external pathway instead of going directly between the reactants

– A battery is an example

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Electrochemical Cell

Zn(s) + Cu2+ Zn2+ + Cu(s)

Zn(s) Zn2+ + 2e- Cu2+ + 2e- Cu(s)

PositiveElectrode(e- added)

NegativeElectrode

(e- removed)

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Redox Cell using Platinum

• Platinum is a good inert means of transferring electrons to/from solution

• Consider the half-reaction in the presence of a Pt electrode:– Fe3+ + e- ↔ Fe2+

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Redox Cell

Pt wireelectrode H2 gas (1 atm)

Salt bridge

[H+] = 1Fe2+ andFe3+

Fe3+ + e- ↔ Fe2+

←: Pt wire removes electrons from half cell A

→: Pt wire provides electrons to the solution

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Redox Cell using Platinum

• If Pt wire not connected to source/sink of electrons, there is no net reaction– But wire acquires an electrical potential

reflecting tendency of electrons to leave solution

– Defined as activity of electrons [e-]• pe = -log [e-]

– Can be used in equilibrium expressions

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Redox Cell using Platinum

• Fe3+ + e- ↔ Fe2+

–

–

• [e-] proportional to the ratio of activities of the reduced to the oxidized species

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Redox Cell

Pt wireelectrode H2 gas (1 atm)

Salt bridge

[H+] = 1Fe2+ andFe3+

H+ + e- ↔ ½ H2(g)

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Redox Cell using Platinum

• Other half-cell (B)• H+ + e- ↔ ½ H2(g)

• Can write an equilibrium expression:•

– SHE = standard hydrogen electrode– By convention, [e-] =1 in SHE

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Redox Cell using Platinum

• If switch is closed, electrons will move from solution with higher activity of e- to the solution with lower activity of e-

• Energy is released (heat)

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Redox Cell using Platinum

• Combine half reactions:• Fe3+ + ½ H2(g) ↔ Fe2+ + H+

• Direction of reaction depends on which half-cell has higher activity of electrons

– Now open switch: no transfer of e-

• Voltage meter registers difference in potential (E) between the 2 electrodes– Potential of SHE = 0, so E = potential of electrode in half-cell A– Defined as Eh– Measured in volts

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Eh

• Eh is positive when:– [e-] in solution A less than [e-] in SHE

• Eh is negative when:– [e-] in solution A greater than [e-] in SHE

•

– At 25°C, pe = 16.9– Eh = 0.059 pe

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Eh as Master Variable

• Fe3+ + ½ H2(g) ↔ Fe2+ + H+

• Recall: G° = -RT ln Keq

• Standard state

• At non-standard state: GR = G° + RT ln Keq

–

–

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Eh as Master Variable

• From electrochemistry: GR = -nF Eh– n = number of electrons– By convention, sign of Eh set for half-reaction

written with e- on left side of equation

– Divide through by –nf

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Eh as Master Variable•

• Rewrite to put oxidized species on top

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Eh (example)

• SO42- + 8e- + 10H+ ↔ H2S + 4 H2O

– 8 electrons transferred–

•

• E° = ( -1 / (8 x 96.5)) x (-231.82) = +0.30 V