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Chapter 9  Oxidation-Reduction (Redox) Reactions

Reducing agent = electron donor (left side of the periodic table, metals)

Oxidizing agent = electron acceptor/taker (right side of the periodic table, "like" its namesake oxygen, polyatomic ions with high oxidation state central atoms, e.g. NO3 and ClO4 which is used in solid rocket fuel)

Oxidation = removal of electrons by an oxidizing agent

Reduction = addition of electrons by a reducing agent

Oxidizing agent = agent of oxidation, the compound that is reduced (gained electrons) having oxidized something else, oxidant

Reducing agent = agent of reduction, the compound that is oxidized (lost electrons) having reduced something else

9.1  Balancing Redox Reactions

Must balance based on the number of electrons, e, transferred.

9.1.1  Balancing in pure water with neutral pH using Half-reactions

Co (s) + Br2 (aq) ⇋ Co2+ (aq) + 2 Br (aq)
Oxidation half-rxn:

  1. Balance the central atom: Co (s) ⇋ Co2+ (aq)
  2. Balance the charge by adding electrons:

Reduction half-rxn:

  1. Br2 (aq) ⇋ 2 Br (aq)
  2. Br2 (aq) + 2 e ⇋ 2 Br (aq)

Both reactions involve 2 electrons, so we add them together to get the whole reaction:

Co (s)+Br2 (aq)+ 2 eCo2+ (aq)+ 2 e +2 Br (aq)
reducing oxidizing conjugate conjugate
agent agent oxidizing reducing
    agent agent

9.1.2  Balancing in acidic solution using half-reactions

  1. Balance the central atom
  2. Balance O atoms with H2O
  3. Balance H atoms with H+
  4. Balance charge by adding e
CrO42− (aq)+HAsO2 (aq)Cr3+ (aq)+H3AsO4 (aq)
(6+)(2-)4 (1+)(3+)(2-)2 (3+) (1+)3(5+)(2-)4

Oxidation half-rxn:

  1. HAsO2 ⇋ H3AsO4
  2. HAsO2 + 2 H2O ⇋ H3AsO4
  3. HAsO2 + 2 H2O ⇋ H3AsO4 + 2 H+
  4. HAsO2 + 2 H2O ⇋ H3AsO4 + 2 H+ + 2 e

Reduction half-rxn:

  1. CrO42− ⇋ Cr3+
  2. CrO42− ⇋ Cr3+ + 4 H2O
  3. CrO42− + 8 H+ ⇋ Cr3+ + 4 H2O
  4. CrO42− + 8 H+ + 3 e ⇋ Cr3+ + 4 H2O

The least common multiple of 2 e and 3 e is 6 e:

 3 HAsO2 + 6 H2O ⇋ 3 H3AsO4 + 6 H+ + 6 e
+2 CrO42− + 16 H+ + 6 e ⇋ 2 Cr3+ + 8 H2O
 3 HAsO2 + 0←6 H2O + 2 CrO42− + 10←16 H+ + 6 e ⇋ 3 H3AsO4 + 0←6 H+ + 6 e + 2 Cr3+ + 2←8 H2O
 3 HAsO2 + 2 CrO42− + 10 H+ ⇋ 3 H3AsO4 + 2 Cr3+ + 2 H2O

9.2  Balancing in basic solution using half-reactions

Cr(OH)3+HXeO4CrO42−+Xe
(3+)(1-)3 (1+)(6+)(2-)4 (6+)(2-)4 0
  1. Balance the central atom
  2. Balance O atoms with OH
  3. Balance H atoms with H+
  4. Balance erroneous H+ by adding enough OH to both sides to turn all H+ into H2O
  5. Balance charge by adding e

Reduction half-rxn:

  1. HXeO4 ⇋ Xe
  2. HXeO4 ⇋ Xe + 4 OH
  3. HXeO4 + 3 H+ ⇋ Xe + 4 OH
  4. HXeO4 + 3 H+ + 3 OH ⇋ Xe + 4 OH + 3 OH

    HXeO4 + 3 H2O ⇋ Xe + 7 OH

  5. HXeO4 + 3 H2O + 6 e ⇋ Xe + 7 OH

Oxidation half-rxn:

  1. Cr(OH)3 ⇋ CrO42−
  2. Cr(OH)3 + OH ⇋ CrO42−
  3. Cr(OH)3 + OH ⇋ CrO42− + 4 H+
  4. Cr(OH)3 + OH + 4 OH ⇋ CrO42− + 4 H+ + 4 OH

    Cr(OH)3 + 5 OH ⇋ CrO42− + 4 H2O

  5. Cr(OH)3 + 5 OH ⇋ CrO42− + 4 H2O + 3 e

The least common multiple of 3 e and 6 e is 6 e:

 HXeO4 + 3 H2O + 6 e ⇋ Xe + 7 OH
+2 Cr(OH)3 + 10 OH ⇋ 2 CrO42− + 8 H2O + 6 e
 HXeO4 + 0←3 H2O + 6 e + 2 Cr(OH)3 + 3←10 OH ⇋ Xe + 0←7 OH + 2 CrO42− + 5←8 H2O + 6 e
 HXeO4 + 2 Cr(OH)3 + 3 OH ⇋ Xe + 2 CrO42− + 5 H2O

9.3  Electrochemical Cells

  1. Voltaic (galvanic) cells produce electricity from a spontaneous redox reaction.
  2. Electrolytic cells consume electricity to drive a non-spontaneous reaction.

9.3.1  Voltaic cells

In order to "capture" electricity, we have to separate the half-rxns physically into half-cells.

The oxidation and reduction half-cells will each contain half of the reactants and half of the products.

Electrodes are conductive materials (pieces of metal or graphite), that mediate e transfer from the oxidation half-cell to the reduction half-cell.

The electrode itself might be a reactant or product in the half-rxn.

Cu (s) + AgNO3 (aq) ⇋ Cu(NO3)2 (aq) + Ag (s)

Cu (s) + 2 Ag+ (aq) ⇋ Cu2+ (aq) + 2 Ag (s)

What happens?

  1. Cu (s) ⇋ Cu2+ (aq) + 2 e occurs at the anode (oxidation electrode, Cu (s) loses mass)
  2. e spontaneously flow through the wire to achieve equilibrium
  3. Ag+ (aq) + e ⇋ Ag (s) occurs at the cathode (reduction electrode, Ag (s) gains mass)
  4. current = rate of flow of e through the wire, measured in amperes

    1 A = 1 C/s

  5. Potential difference (electromotive force, emf) = difference of potential energy that would drive 1 ampere against 1 ohm of resistance

    Volt, 1 V = 1 J/C

  6. Ecell = +0.46 V > 0 indicating that we have correctly determined which way the e are flowing through the wire (and also correctly hooked up the voltmeter)
  7. Ag+ is the oxidizing agent taking Cu (s) e; Cu (s) is the reducing agent giving up its electrons
  8. Salt bridge completes the circuit

9.3.2  Cell Notation

Chemical equations separate reactants from products with an arrow

Cell notation separates half-rxns/half-cells using a || = salt bridge

Alphabetical order and vowel/consonant pairing:

Cu (s) | Cu2+ (aq, 1 M) || Ag+ (aq, 1 M) | Ag (s) has Ecell = +0.46 V

If there is no solid form to act as the electrode, an inert electrode like Pt (s) can be used:

Pt (s) | H2 (g, 1 atm) | H+ (aq 1 M) || Fe3+ (aq, 1 M), Fe2+ (aq, 1 M) | Pt (s) has Ecell = +0.77 V

H2 (g) + 2 Fe3+ (aq) ⇋ 2 H+ (aq) + 2 Fe2+ (aq)

9.3.3  Standard Hydrogen Electrode and Standard Reduction Potentials, Ered

By selecting a standard anode, we can compare the reactivities of different half-rxns/cells to each other

Standard Hydrogen Electrode, SHE

Pt (s) | H2 (g, 1 atm) | H+ (aq 1 M) || variable reduction/cathode reaction Ecell = Ered
SHE || F2 (g, 1 atm) | F (aq, 1 M) | Pt (s) Ered = +2.87 V because F2 is a great oxidizing agent, it wants those electrons

SHE || Mg2+ (aq, 1 M) | Mg (s) Ered = −2.37 V because the electrons are going the other way, Mg2+ is a terrible oxidizing agent because Mg (s) is a great reducing agent (just like conjugate acid-base pair strength, one is strong the other is weak)

For any other general electrochemical cell, Ecell = Ered, cathodeEred, anode

You subtract Ered, anode because it is being oxidized, not reduced.

half-celloxidizing agent + e → reducing agentEred
Cl2/ClCl2 + 2 e ⇌ 2 Cl1.36 V
H+/H22 H+ + 2 e ⇌ H20 V
Pb2+/PbPb2+ + 2 e ⇌ Pb-0.13 V
Co2+/CoCo2+ + 2 e ⇌ Co-0.28 V
Mg2+/MgMg2+ + 2 e ⇌ Mg-2.37 V


From this short list we can make statements regarding RELATIVE reactivities

Problem

What is the cell notation that correctly indicates the spontaneous reaction between Pb2+/Pb and Co2+/Co half-cells?

Pb2+ wants to be reduced more than Co2+ so the product is Pb (s), Co (s) must be oxidized:

Pb2+ + Co (s) ⇋ Pb (s) + Co2+ (aq)

Co (s) | Co2+ (aq, 1 M) || Pb2+ (aq, 1 M) | Pb (s) has Ecell = −0.13 V − −0.28 V = +0.15 V

———————————————————————–

Non-electrochemical cell reactions can be predicted based on whether Ecell is positive or negative.

Mg (s) | Mg2+ (aq, 1 M) || Cl2 (g, 1 atm) | Cl (aq, 1 M) has Ecell = 1.36 V − −2.37 V = 3.73 V

Mg (s) + Cl2 (g) ⇋ Mg2+(Cl2−)2 (s) = MgCl2 (s) is product favored, spontaneous
Co (s) | Co2+ (aq, 1 M) || Mg2+ (aq, 1 M) | Mg (s) has Ecell = −2.37 V − −0.28 V = −2.09 V

Co (s) + Mg2+ (aq) ⇋ Co2+ (aq) + Mg (s) is reactant favored, non-spontaneous

9.3.4  Electrochemical Equilibrium

Standard Gibbs Free Energy is the maximum amount of non-expansion work, e.g. electrical work, that can be done.

wmax = Δ Grxn 

How much work can be done is related to the potential difference of the electrochemical cell, Ecell, with units of V = J/C.

wmax = Δ Grxn = −nFEcell = 

mol e



96485 C
mol e






J
C



    (1)

n is the number of electrons transferred in the balanced chemical reaction.

Mg (s) | Mg2+ (aq, 1 M) || Cl2 (g, 1 atm) | Cl (aq, 1 M) has Ecell = 1.36 V − −2.37 V = 3.73 V

wmax = Δ Grxn = 

mol e



96485 C
mol e






3.73 J
C



 = −720000  J = −720  kJ 

Equating two things that are equal to Δ Grxn:

nFEcell = −RT lnK 
Ecell = 
RT
nF
 lnK     (2)

Non-standard electrochemical cells have Δ G = −nFEcell:

nFEcell = −nFEcell + RT lnQ 

Which yields the Nernst equation for non-standard electrochemical cells:

Ecell = Ecell − 
RT
nF
 lnQ     (3)
Problem

What is Ecell for the following electrochemical cell?

Cr (s) | Cr3+ (aq, 0.000875 M) || Cu2+ (aq, 1.28 M) | Cu (s)

half-cellEred
Cu2+/Cu+0.34 V
Cr3+/Cr-0.73 V
Ecell = 0.34 − −0.73 = 1.07  V 

Cr (s) + Cu2+ (aq, 1.28 M) ⇋ Cr3+ (aq, 0.000875 M) + Cu (s)

2 Cr (s) + 3 Cu2+ (aq, 1.28 M) ⇋ 2 Cr3+ (aq, 0.000875 M) + 3 Cu (s)

Q = 
[Cr3+]2
[Cu2+]3
 = 
(0.000875)2
(1.28)3
 = 3.651 × 10−7 
Ecell = 1.07 − 
(8.314 
J
mol K
)(298.15  K)
(6  mol e)(96485 
C
mol e
)
 ln
3.651 × 10−7
= 1.13  V 

Notice that Ecell > Ecell because of the large concentration difference at initial time; there is very little product Cr3+ and a lot of the reactant Cu2+.

———————————————————————–

Concentration cells use only concentration differences to drive the current.

Problem

What is Ecell for the following electrochemical cell?

Zn (s) | Zn2+ (aq, 0.0320 M) || Zn2+ (aq, 1.07 M) | Zn (s)

Ecell = −0.76 − −0.76 = 0.0  V 
Q = 
[Zn2+]anode
[Zn2+]cathode
 = 
(0.0320)
(1.07)
 = 0.02991 
Ecell = 0.0 − 
(8.314 
J
mol K
)(298.15  K)
(2  mol e)(96485 
C
mol e
)
 ln
0.02991 
= 0.0451  V 

———————————————————————–

9.3.5  Mixing Pressure and Concentrations

What about a concentration cell of Hydrogen electrodes?

Pt (s) | H2 (g, 0.8 atm) | H+ (aq, 0.2 M) || H+ (aq, 0.6 M) | H2 (g, 0.4 atm) | Pt (s)

Ecell = 0.0 − 0.0 = 0.0  V 

H2 (g) → 2H+ (aq) + 2 e

ne = 2  mol e 

H2anode (g, 0.8 atm) + 2 Hcathode+ (aq, 0.6 M) ⇋ 2 Hanode+ (aq, 0.2 M) + H2cathode (g, 0.4 atm)

Q = 
[H+]anode2PH2cathode
[H+]cathode2PH2cathode
 = 
(0.2  M)2(0.4  atm)
(0.6  M)2(0.8  atm)
 = 0.0556 
Ecell = 0.0 − 
(8.314 
J
mol K
)(298.15  K)
(2  mol e)(96485 
C
mol e
)
 ln
0.0556 
= 0.0371  V 

9.4  Electrolytic Cells

Electrolytic Cells use a voltaic cell (i.e. battery) or other source of electricity to drive a non-spontaneous redox reaction.

9.4.1  Electroplating

Takes advantage of the reduction at the cathode to add mass (solid metal) to the cathode.

Problem

What current in amperes is needed to electroplate 0.277 g Co (s) in 842 seconds from a Co2+ solution?

Co2+ + 2 e ⇋ Co (s)

0.277  g Co 


1 mol Co
58.9332  g Co
 





2  mol e
1  mol Co
 


= 9.40 × 10−3  mol e 
I = 
neF
t
 = 
(9.40 × 10−3  mol e)(96485  C/mol e)
842  s
 = 1.08  C/s = 1.08  A     (4)

———————————————————————–

9.4.2  Molten Salts

Molten salts can be electrolyzed to reform the elements.

2 NaCl (l) ⇋ 2 Na (l) + Cl2 (g)

Problem

A mixture of molten LiF and NaI is electrolyzed; what is oxidized and reduced first?
The most positive reduction potential of the reducible species will be reduced first.

The most negative reduction potential of the oxidizable species will be oxidized first.

half-cellEred
F2 |F+2.87 V
I2 |I+0.54 V
Na+ |Na-2.71 V
Li+ |Li-3.04 V


Na+ and Li+ can be reduced (F and I are already reduced); Na+ is easier to reduce because it has the more positive Ered so it will reduce first.

F and I can be oxidized (Na+ and Li+ are already oxidized); I is easier to oxidize because it has the more negative Ered so it will oxidize first.

———————————————————————–

9.4.3  General Aqueous Electrolysis

Voltaic cell: Mg (s) | Mg2+ (aq) || Cl2 (g) | Cl (aq) | Pt (s) has Ecell = +3.73 V

Mg (s) is being oxidized and Cl2 (g) is being reduced.

It would take 3.73 V to drive this reaction backwards to reduce Mg2+ and oxidize Cl.

In this case, the electrolytic cell is powered by a voltaic cell:

Problem

What is reduced and oxidized first?

What can be reduced?

Mg2+ but also the H in H2O. The oxidation state of H is 1+ in H2O.

half-rxnEred
2 H2O (l) + 2 e ⇋ H2 (g) + 2 OH (aq)-0.83 V (basic solution, -0.41 V pure water)
2 H+ (aq) + 2 e ⇋ H2 (g)0.00 V (acidic solution)
Mg2+ (aq) + 2 e ⇋ Mg (s)-2.37 V


Instead of reducing Mg2+, you reduce H2O because it has a more positive reduction potential!
What can be oxidized?

Cl but also the O in H2O. The oxidation state of O is 2- in H2O.

half-rxnEred
Cl2 (g) + 2 e ⇋ 2 Cl (aq)+1.36 V
O2 (g) + 4 H+ (aq) + 4 e ⇋ 2 H2O (l)+1.23 V (acidic solution, +0.82 V pure water)
O2 (g) + 2 H2O (l) + 4 e ⇋ 4 OH (aq)+0.40 V (basic solution)


Instead of oxidizing Cl, you oxidize H2O because it has a more negative reduction potential!

The products of this electrolytic cell are H2 (g) and O2 (g), not Mg (s) and Cl2 (g).
Except that this theoretical prediction is wrong due to the overpotential of the oxidation of H2O, putting the actual reduction potential at +1.48 V for O2 making Cl2’s more negative.

Therefore the products would actually be H2 (g) and Cl2 (g).

———————————————————————–

Problem

What are the products of the electrolysis of a NiI2 solution?

Ni2+ could be reduced:

Ni2+/Ni Ecell = −0.23 V is more positive than H2O/H2 Ecell = −0.83 (−0.41) V so Ni2+ will be preferentially reduced.

I could be oxidized:

I2/I Ecell = +0.54 V is more negative than O2/H2O Ecell = +1.23 (+0.82) V so I will be preferentially oxidized.

Half-rxns:

Products = Ni (s) and I2 (g)

———————————————————————–

Problem

What are the products of the electrolysis of a NaF solution?

Na+ could be reduced:

Na+/Na Ecell = −2.71 V is NOT more positive than H2O/H2 Ecell = −0.83 (−0.41) V so H2O will be preferentially reduced to H2.

F could be oxidized:

F2/F Ecell = +2.87 V is NOT more negative than O2/H2O Ecell = +1.23 (+0.82) V so H2O will be preferentially oxidized to O2.

Half-rxns:

Products = H2 (g) and O2 (g)

———————————————————————–

Problem

What are the products of the electrolysis of a Al(NO3)3 solution?

Al3+ could be reduced, but so can NO3:

Al3+/Al Ecell = −1.66 V is NOT more positive than H2O/H2 Ecell = −0.83 (−0.41) V but NO3/NO Ecell = +0.96 V is more positive so NO3 will be preferentially reduced to NO.

There is nothing in solution other than H2O that has the potential to be oxidized (Al3+ and NO3 are both already oxidized) so H2O will be oxidized to O2.

Half-rxns:

Products = NO (g) and O2 (g)

———————————————————————–


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