RO13.  Solutions to Selected Problems

Problem RO1.1.

a) Ag⁺    b) Ni²⁺   c) Mn⁷⁺    d) Cr⁶⁺   e) Cu³⁺   f) Fe³⁺   g)Os⁸⁺    h) Re⁵⁺

Problem RO1.3.

a) Pb²⁺ & S^2-   b) Sn⁴⁺  & 2 O^2-    c) Hg²⁺ & S^2-   d) Fe⁴⁺ & 2 S^2-    e) 2 Fe³⁺ & 3 O^2-     f) 2 Fe³⁺ & 1 Fe²⁺ & 4 O^2-

Problem RO1.4.

Probably Fe²⁺, to replace Zn²⁺ ions.

Problem RO1.5.

a) C ²⁺   b) C⁴⁺    c) ^4-   d) C⁰   e) C³⁺

Problem RO1.7.

a) Mg²⁺   b) Cu²⁺   c) Mn³⁺    d) Ca²⁺   e) Mn²⁺   f) Mn²⁺

Problem RO1.8.

a) Cu(II), Fe(II)    b) Zn(II), Fe(III)    c) Be(II), Al(III)    d) Cu(I), Fe(III)      e) Cu(II), Al(III)

Problem RO2.1.

a)  Cu    →   Cu(⁺   +   e^-

b)  Fe³⁺   +   3 e^-→   Fe

c)  Mn   →   Mn³⁺   +   3 e^-

d)  Zn²⁺   +   2 e^-   →   Zn

e)  2 F^-   →   F₂   +   2 e^-

f)  H₂   →   2 H⁺   +   2 e^-

Problem RO2.2.

a) Cu(I)    +    Fe(III)      →  Cu(II)    +    Fe(II)

b)  Cu(I)    +    Ag(0)      →  Cu(0)    +    Ag(I)

c)  3 F₂   +    2  Fe      →  6 F^-    +    2 Fe(III)

d)  2 Mo³⁺   +    3 Mn      →  2 Mo    +    3 Mn²⁺

Problem RO2.3.

a) MnO₂   +   2 H⁺   + 2 e^-    →   Mn(OH)₂

b)  2 NO    +    2 e^-   +    2 H⁺   →   N₂O    +   H₂O

c)  HPO₃^2-   +    2 e^-   +    3 H⁺     →   H₂PO₂^-   +    H₂O

d)  Sn(OH)₆^2-   +   2 e^-   +   3 H⁺     →     HSnO₂^-   +  4  H₂O

Problem RO2.4

a)  yes    b)  no       c) yes    d)  no    e) yes    f) yes    g) yes    h) no

Problem RO2.5.

a)  E_rxn = - 0.153 V - 0.33 V = - 0.483 V

b)  No.

c) 

d)  The O₂ is activated as an electrophile.  Addition of an electron may become easier.

e)  The reduction potential is 0.2 V more positive when the resulting superoxide ion is stabilised by binding a proton.  A similar shift could occur when coordinated to copper.

f)

g)  The two aspartate ions would make the copper complex less cationic.  That may make it easier to remove an electron from the copper complex.

Problem RO2.6.

a)  Fe²⁺  +  ClO₂   →    Fe³⁺   +    ClO₂^-

b)  E_rxn = 0.95 V - 0.77 V = 0.18 V

c)  yes.

d) 

e)  ClO₂ has an unpaired electron.  ClO₂^- has electrons paired.

f)  It could be inner sphere: the oxygen in the ClO₂ could coordinate to the iron.

g) The mixture quickly forms (H₂O)₅Fe(ClO₂)²⁺ via inner sphere electron transfer.

Problem RO3.1.

MnO₄^- :     4 x O^2- (= 8^-) + Mn⁷⁺ = 1^- overall

MnO₂ :     2 x O^2- (= 4^-) + Mn⁴⁺ = neutral overall

difference = 3 e^- 

Problem RO3.2.

a)     SO₄^2- :     4 x O^2- (= 8^-) + S⁶⁺ = 2^- overall

S₂O₈^2- :     8 x O^2- (= 16^-) + 2 x S⁷⁺ (= 14⁺) = 2^- overall

difference = 1 e^- per S, or 2 e^- overall

S₂O₈^2- + 2 e^- → 2 SO₄^2-

b)    HPO₃^2- :     3 x O^2- (= 6^-) + H⁺ + P³⁺ = 2^- overall

P :     P(0)

difference = 3 e^- 

HPO₃^2- + 3 e^- + 5 H⁺ → P + 3 H₂O

c)     Ti₂O₃ :     3 x O^2- (= 6^-) + 2 x Ti³⁺ (= 6⁺)  = neutral overall

TiO :     O^2-  + Ti²⁺ = neutral overall

difference = 1 e^- per Ti, or 2 e^- overall

Ti₂O₃  + 2 e^-  + 2 H⁺   → 2 TiO + H₂O    

d)     N₂ : N(0)

NH₂OH :     O^2- + 3 x H⁺ (= 3⁺) +  N^-   = neutral overall

difference = 1 e^- per N, or 2 e^- overall

N₂  + 2 e^-  + 2 H⁺  + 2 H₂O   → 2 NH₂OH    

Problem RO3.3.

Lithium is an alkali metal, in the first column of the periodic table.  It has a relatively low ionization energy because it has a noble gas configuration as a cation.  That noble gas configuration is stable because of the relatively large number of nuclear protons and a relatively short distance between the nucleus and the outermost shell of electrons.  In lithium metal, the outermost electron is  relatively far from the nucleus and so it is at a relatively high energy, and easily lost.

Problem RO3.4.

Fluorine is a halogen, with a relatively high electron affinity.  It easily gains an electron to get to a noble gas configuration as a fluoride anion.    That noble gas configuration is stable because of the relatively large number of nuclear protons and a relatively short distance between the nucleus and the outermost shell of electrons. 

Problem RO3.5.

From most easily oxidized to least easily oxidized: Li > Al > Fe > Cu > Au

Problem RO3.6.

a)  E⁰ = + 0.796 (Ag⁺/Ag) - 1.83 (Au/Au⁺) = -1.034 V (no forward reaction)

b)  E⁰ = - 0.44 (Fe²⁺/Fe) + 0.762 (Zn/Zn²⁺) = + 0.0322 V (forward reaction)

c)  E⁰ = + 0.52 (Cu⁺/Cu) + 3.04 (Li/Li⁺) = + 3.56 V (forward reaction)

d)  E⁰ = + 0.77 (Fe³⁺/Fe²⁺) - 0.796 (Ag/Ag⁺)  = - 0.026 V (no forward reaction)

Problem RO3.7.

When the table of standard reduction potentials is displayed with the most negative value at the top and the most positive value at the bottom, any given half-reaction will go forward if it is coupled with the reverse of a half-reaction that lies above it in the table.  The opposite is not the case; no half reaction will go forward if it is coupled with the reverse of a half-reaction below it in the table.

Problem RO4.1.

a)

a

b)

c)

Problem RO4.2.

a)  2 Li + F₂ → 2 Li⁺ + 2 F^-

b)  E^o = +5.91 V

c)  Things look pretty grim.

d) 

e)  This scheme would result in the release of a small amount of energy at each stage.  Each step could be harnessed to perform a task more efficiently, with less heat loss.

Problem RO4.3.

Problem RO4.4.

Problem RO4.5.

There are really two significant departures from expectation here.  Lithium is much more active than expected based on electronegativity.  The larger alkali metals, cesium, rubidium and francium, are all less active than expected on that basis.

We will see that another factor the influences activity in redox is the stability of ions in aqueous solution.  Lithium cation is a small ion; water molecules bind very strongly to the ion because the electrons get relatively close to lithium's nucleus.  That strong binding stabilizes this ion especially, tipping the malance of the reaction more strongly towards oxidation of lithium.  The larger alkali metal ions are not nearly as stabilized by water ligands in aqueous solution, so the balance of their reactions does not tilt as strongly towards aqueous ions.

Problem RO5.1.

Li⁺/Li: E⁰ = - 3.04 V; ΔH_vap = 147 kJ/mol; IE = 520 kJ/mol; ΔH_h =  -520 kJ/mol       

Na⁺/Na:  E⁰ = - 2.71 V; ΔH_vap = 97 kJ/mol; IE = 495 kJ/mol; ΔH_h =  -406 kJ/mol  

K⁺/K:  E⁰ = - 2.931 V; ΔH_vap = 77 kJ/mol; IE = 419 kJ/mol; ΔH_h =  -320 kJ/mol  

Potassium should be the easiest of the three to oxidize.  It is easier to oxidize than sodium.  However, lithium's high heat of hydration reverses the trend and tips the balance of reaction in favour of ion formation.

Problem RO5.2.

Cu²⁺/Cu: E⁰ = + 0.340 V; ΔH_vap = 300 kJ/mol; IE = 745 kJ/mol & 1958 kJ/mol; ΔH_h =  - 2099 kJ/mol 

Ni²⁺/Ni: E⁰ = - 0.25 V; ΔH_vap = 377 kJ/mol; IE = 737 kJ/mol & 1753 kJ/mol; ΔH_h =  - 2096 kJ/mol

Zn²⁺/Zn: E⁰ = - 0.7618 V; ΔH_vap = 123 kJ/mol; IE = 906 kJ/mol & 1733 kJ/mol; ΔH_h =  - 2047 kJ/mol

In this case, zinc may be considered the outlier.  Copper should be easier to reduce than nickel based solely on electronegativity.  However, zinc's very low heat of vaporization suggests that formation of the solid metal is less favoured in that case, helping to tilt the balance toward zinc ion instead.

Problem RO5.3.

a)

b)

c)

d)

Problem RO5.4.

a)

b)

c)  K_eq = k_f / k_r = (2.0 x 10^-9 s^-1) / (1.9 x 10^-11 s^-1) = 105

d)  Δg = - RTlnK = - (1.987 x 10^-3 kcal K^-1mol^-1(298 K)[ln(105)] = -17.5 kcal/mol

e)

f)

g)  Anywhere between 0.70 eV and 0.82 eV, or 0.76 eV as an average.

h)  ΔG = - nFE^o = - (1)(23 kcal V^-1 mol^-1)(0.76 V) = -17.5 kcal // mol 

(would be + 17.5 kcal/mol in the other direction)

j)

Problem RO6.1.

Both reduction potentials are very positive.

Ag⁺ + e^- →  Ag (s)  E⁰ = + 0.796 V

Au⁺ + e^- →  Au (s) E⁰ = + 1.83 V

That means both metals are likely to be found in the reduced state.

Problem RO6.2.

SnO (s) + 2H⁺ + 2 e^- → Sn (s) + H₂O     E⁰ = -0.10 V

 CO(g) + H₂O → CO₂(g) + 2H⁺ + 2 e-   E⁰ = - (-0.11 V)

SnO (s) + CO (g)  → Sn (s) + CO₂ (g)  ΔE⁰ = + 0.01 V

Problem RO6.3.

Fe₃O₄(s) + 8 H⁺ + 8 e^- → 3Fe(s) + 4 H₂O    E⁰ = +0.085 V

CO(g) + H₂O → CO₂(g) + 2H⁺ + 2 e-   E⁰ = - (-0.11 V)

Fe₃O₄(s) + 4CO (g)  → 3Fe(s)  + 4CO₂ (g)  ΔE⁰ = + 0.195 V

Problem RO6.4.

Al³⁺_(aq) + 3e^- →   Al (s)      E⁰ = -1.662 V

CO(g) + H₂O → CO₂(g) + 2H⁺ + 2 e-    E⁰ = - (-0.11 V)

2 Al³⁺(aq) + 3CO (g) + 3H₂O → 2Al(s)  + 3CO₂ (g)  ΔE⁰ = - 1.772 V

Problem RO6.5.

Al³⁺  + 3e^-    →   Al (s)      E⁰ = -1.662 V

 2 O^2-    →    O₂(g)  + 4e^-     E⁰ = - (.40 V, estimated)

4 Al³⁺(aq) + 6 O^2- (aq)  →   4 Al (s)  + 3O₂ (g)  ΔE⁰ = - 2.062 V

Problem RO6.6.

Even if the aluminum ions from the cryolite are reduced to Al⁰, they will be replenished by new aluminum ions from the bauxite ore.

Problem RO7.1.

a) The potential of the zinc reaction is too negative to be spontaneous, so the zinc must be oxidized.

 Zn_(s) + 2OH^- → ZnO_(s) + H2O_(l) + 2e^-    E^o = + 1.28 V

2 MnO_2(s) + H₂O_(l) + 2e^- → Mn₂O_3(s) + 2OH^-_(aq)    E^o = + 0.15 V

The net reaction is:

 2 MnO_2(s) +   Zn_(s)  → ZnO_(s) + Mn₂O_3(s)     E^o = + 1.43 V

 b)  This reaction involves a pair of electrons.

ΔG = - n F E⁰_cell 

ΔG =  - 2 x 96,485 J V^-1 mol^-1  x 1.43 V

ΔG =  - 275,947 J mol^-1 = -276 kJ mol^-1

Problem RO7.2.

a)  We are told that citric acid gives up protons that form hydrogen gas.  The protons are getting reduced.

2 H⁺   +  2 e^-     →      H_2(g)            E^o = 0.0 V

That leaves either the copper or the zinc to be oxidized.

Zn²⁺   +   2e^-    →    Zn_(s)            E^o = - 0.762 V

Cu²⁺   +   2e^-    →    Cu_(s)            E^o = + 0.34 V

The oxidation of zinc would be spontaneous, but the oxidation of copper would not.  The zinc is oxidized.

  Zn_(s)     →    Zn²⁺   +   2e^-         E^o = + 0.762 V

2 H⁺   +  2 e^-     →      H_2(g)            E^o = 0.0 V

2 H⁺   +   Zn_(s)     →    Zn²⁺   +    H_2(g)            E^o = +  0.762 V

b) This reaction involves a pair of electrons.

ΔG = - n F E⁰_cell 

ΔG =  - 2 x 96,485 J V^-1 mol^-1  x 0.762 V

ΔG =  - 147,043 J mol^-1 = -147 kJ mol^-1

Problem RO7.3.

a)  The reactions are:

oxidation:  Pb    →  Pb²⁺  + 2 e^-

reduction:  PbO₂  +  4 H⁺   + 2 e^-   →  Pb²⁺   +  2  H₂O

b)  The reactions are:

oxidation:  Cd     →  Cd²⁺  + 2 e^-

reduction:  NiO(OH) +  3 H⁺   + 3 e^-   →  Ni    +  1.5  H₂O

c)  The reactions are:

oxidation:  Zn    →  Zn²⁺  + 2 e^-

reduction:  MnO₂  +   H⁺   +  e^-   →  MnO₂H  

To learn about balancing reactions, see here.

Problem RO7.4.

a)  The reactions are:

Ag₂O (s) +  H₂O  + 2 e^-   →   2 Ag (s) + 2 ^-OH (aq)  E⁰ = +0.342 V

Zn (s)       →      Zn²⁺ (aq)    +   2e^-         E⁰ = -(-0.762 V)

Ag₂O (s) +  H₂O  + Zn (s)   →   2 Ag (s) + 2 ^-OH (aq) + Zn²⁺ (aq)    ΔE⁰ = +1.104 V

b)  Zn | Zn²⁺, HO^- || HO^- | Ag | Ag₂O

Note the presence of the "extra" vertical line because hydroxide is in solution in the electrolyte but the silver metal would be a solid, in contrast to the zinc ion that is in solution with the hydroxide.

Problem RO7.5.

a)  Charge on complex is calculated as follows:

charge on metal: 3+

charge on ligand donors (4 neutral N, 2 anionic N donors): 2-

other charges on ligand (carboxylate arm):  1-

overall: 0

b) ΔG = RT log K = RT (-pK_a) = - 8.314 J K^-1 mol^-1 x 300 K x (-20.2) = 50,382 J mol^-1 = 50.4 kJ/mol

c)  It is more difficult to add the electron to the more negatively charged anion, so the reduction potential is more negative in that case.

d)  ΔG = -n F E° = - 1 x 96,485 J V^-1 mol^-1 x (-0.545 V) = 52,583 J mol^-1 = 52.6 kJ/mol

e)   ΔG = -n F E� = - 1 x 96,485 J V^-1 mol^-1 x (-0.575 V) = 55,479 J mol^-1 = 55.5 kJ/mol

f)

ΔG_{Fe(II)CO2H/CO2-} = 50.4 + 55.5 -52.6 kJ mol^-1 =  53.3 kJ/mol

ΔG = -RT logKa

K_a = e^(-ΔG/RT) = e^{((-53,300/(8.314 J /K mol x 300 K))} = e^-21.4

pKa = 21.4

g)  The Fe(III) complex is easier to deprotonate because it yields an anionic species, whereas the Fe(II) complex yields a dianionic species.  The greater charge buildup in the latter would cost more energy.

Problem RO8.1.

a)  Cu     -->    Cu₂O                                     MoO₂       -->       Mo               

2 Cu     -->    Cu₂O                                         MoO₂       -->       Mo               

2 Cu  +  H₂O     -->    Cu₂O                             MoO₂       -->       Mo      +   2 H₂O   

2 Cu  +  H₂O     -->    Cu₂O   + 2 H⁺                   4H⁺    +       MoO₂       -->       Mo      +   2 H₂O    

2 Cu  +  H₂O     -->    Cu₂O   + 2 H⁺   + 2e^-           4e^-    +     4H⁺    +       MoO₂       -->       Mo      +   2 H₂O 

2x (2 Cu  +  H₂O     -->    Cu₂O   + 2 H⁺   + 2e^- )          4e^-    +     4H⁺    +       MoO₂       -->       Mo      +   2 H₂O

adding:

4 Cu  + 2  H₂O     -->   2 Cu₂O   + 4 H⁺   + 4e^-         

 4e^-    +     4H⁺    +       MoO₂       -->       Mo      +   2 H₂O  

equals

4 Cu  +   MoO₂          -->   2 Cu₂O    +         Mo  

b)  NH₂OH    -->      N₂                                        Ag₂O -->   Ag

 2 NH₂OH    -->      N₂                                            Ag₂O -->   2 Ag

 2 NH₂OH    -->      N₂     +  2 H₂O                           Ag₂O -->   2 Ag    +   H₂O  

2 NH₂OH    -->      N₂     +  2 H₂O   + 2 H⁺               2 H⁺   +     Ag₂O -->   2 Ag    +   H₂O 

2 NH₂OH    -->      N₂     +  2 H₂O   + 2 H⁺  + 2 e-             2 H⁺   +     Ag₂O + 2 e-  -->   2 Ag    +   H₂O 

adding:

 2 NH₂OH                   -->      N₂     +  2 H₂O   + 2 H⁺  + 2 e-           

 2 H⁺   +     Ag₂O + 2 e-      -->   2 Ag    +   H₂O

equals

2 NH₂OH  +     Ag₂O   -->   N₂     +  2 Ag    +  3 H₂O

c)  Fe₃O₄  -->   Fe                                                             CO    -->      CO₂

Fe₃O₄  -->   3 Fe                                                           CO    -->      CO₂

Fe₃O₄  -->   3 Fe      + 4 H₂O                                        H₂O    +             CO    -->      CO₂

8 H⁺      + Fe₃O₄  -->   3 Fe      + 4 H₂O                         H₂O    +             CO    -->      CO₂       +     2 H⁺    

8 e^-   +  8 H⁺      + Fe₃O₄  -->   3 Fe      + 4 H₂O                         H₂O    +             CO    -->      CO₂       +     2 H⁺     +  2 e^-   

8 e^-   +  8 H⁺      + Fe₃O₄  -->   3 Fe      + 4 H₂O                      4x   (H₂O    +             CO    -->      CO₂       +     2 H⁺     +  2 e^-  ) 

adding:

8 e^-   +  8 H⁺      + Fe₃O₄  -->   3 Fe      + 4 H₂O                        

4 H₂O    +           4  CO    -->     4 CO₂       +     8 H⁺     +  8 e^-   

equals

Fe₃O₄  +  4  CO      -->   3 Fe      + 4   CO₂             

d)  I₂      -->       IO₃^-                                                   MnO₄^-    -->         MnO₂     

 I₂      -->      2  IO₃^-                                                   MnO₄^-    -->         MnO₂    

6 H₂O   +    I₂      -->      2  IO₃^-                                        MnO₄^-    -->         MnO₂     +  2 H₂O

6 H₂O   +    I₂      -->      2  IO₃^-    + 12 H⁺                           4H⁺    +      MnO₄^-    -->         MnO₂     +  2 H₂O

6 H₂O   +    I₂      -->      2  IO₃^-    + 12 H⁺   + 10 e^-            3 e^-    +     4H⁺    +      MnO₄^-    -->         MnO₂     +  2 H₂O

3x (6 H₂O   +    I₂      -->    2  IO₃^-    + 12 H⁺   + 10 e^-)          10x (  3 e^-    +     4H⁺    +      MnO₄^-    -->   MnO₂     +  2 H₂O)

adding

18 H₂O   +  3  I₂      -->      6  IO₃^-    + 36 H⁺   + 30 e^-           

30 e^-    +     40 H⁺    +    10  MnO₄^-    -->      10   MnO₂     +  20 H₂O

equals

 3  I₂   +   4 H⁺    +    10  MnO₄^-     -->      6  IO₃^-   +   10   MnO₂     +  2 H₂O          

e)  H₃Mo₇O₂₄    -->     Mo                                        S₂O₃^2-   -->         SO₃^2-   

H₃Mo₇O₂₄    -->    7  Mo                                                S₂O₃^2-   -->        2 SO₃^2-   

H₃Mo₇O₂₄    -->    7  Mo    + 24 H₂O                                3H₂O   +       S₂O₃^2-   -->        2 SO₃^2-   

45 H⁺   +   H₃Mo₇O₂₄    -->    7  Mo    + 24 H₂O                        3H₂O   +       S₂O₃^2-   -->        2 SO₃^2-    +  6 H⁺    

45 e^- +  45 H⁺   +   H₃Mo₇O₂₄    -->    7  Mo    + 24 H₂O                3H₂O   +       S₂O₃^2-   -->        2 SO₃^2-    +  6 H⁺  + 4 e^-  

4 x (45 e^- +  21 H⁺   +   H₃Mo₇O₂₄    -->    7  Mo    + 24 H₂O)       45 x (3H₂O   +     S₂O₃^2-   -->      2 SO₃^2-    +  6 H⁺  + 4 e^- )

adding:

180 e^- +  180 H⁺   +   4 H₃Mo₇O₂₄    -->    28  Mo    + 96 H₂O              

135 H₂O   +     45  S₂O₃^2-   -->        90 SO₃^2-    +  270 H⁺  + 180 e^-  

equals

  4 H₃Mo₇O₂₄  +   45 S₂O₃^2-    +    39 H₂O    -->    28  Mo   +  90 SO₃^2-    +  90 H⁺                    

checking:

28 Mo --> 28 Mo; 90 S --> 90 S; 90 H --> 90 H; 270 O  --> 270 O

Problem RO8.2.

a)    Fe(OH)₂     -->  Fe₂O₃                                                 N₂H₄   -->      NH₄⁺  

 2 Fe(OH)₂     -->  Fe₂O₃                                                     N₂H₄   -->    2  NH₄⁺  

   2 Fe(OH)₂     -->  Fe₂O₃  +   H₂O                                          N₂H₄   -->    2  NH₄⁺  

2 Fe(OH)₂     -->  Fe₂O₃     +   H₂O   +   2 H⁺                      4 H⁺   +    N₂H₄   -->    2  NH₄⁺  

  2 Fe(OH)₂     -->  Fe₂O₃     +   H₂O +   2 H⁺  +     2 e^-                 2 e^-   +    4 H⁺   +    N₂H₄   -->    2  NH₄⁺       

adding:

 2 Fe(OH)₂     -->  Fe₂O₃     +   H₂O +   2 H⁺  +     2 e^-                

2 e^-   +    4 H⁺   +    N₂H₄   -->    2  NH₄⁺    

equals:

 2 Fe(OH)₂  +     2 H⁺   +    N₂H₄    -->  Fe₂O₃     +   H₂O  +    2  NH₄⁺        

in basic conditions:      

 2 Fe(OH)₂  +     2 H⁺  +   2 ^-OH   +    N₂H₄    -->  Fe₂O₃     +   H₂O  +    2  NH₄⁺     +   2 ^-OH   

2 Fe(OH)₂  +     2  H₂O   +    N₂H₄    -->  Fe₂O₃     +   H₂O  +    2  NH₄⁺     +   2 ^-OH

2 Fe(OH)₂  +     H₂O   +    N₂H₄    -->  Fe₂O₃     +    2  NH₄⁺     +   2 ^-OH     

b)  MnO₄^-    -->  HMnO₄^-                                  V³⁺    -->    VO²⁺         

MnO₄^-    -->  HMnO₄^-                                  V³⁺  +   H₂O   -->    VO²⁺    

H⁺   +    MnO₄^-    -->  HMnO₄^-                                  V³⁺  +   H₂O   -->    VO²⁺     + 2 H⁺    

e^-    +   H⁺   +    MnO₄^-    -->  HMnO₄^-                                  V³⁺  +   H₂O   -->    VO²⁺     + 2 H⁺     + e^-     

adding:

e^-    +   H⁺   +    MnO₄^-    -->  HMnO₄^-                                 

 V³⁺  +   H₂O   -->    VO²⁺     + 2 H⁺     + e^-     

equals:

MnO₄^-  +    V³⁺  +   H₂O -->  HMnO₄^-    +   VO²⁺     +  H⁺                                  

under basic conditions:

MnO₄^-  +    V³⁺  +   H₂O  +  ^-OH      -->  HMnO₄^-    +   VO²⁺     +  H⁺    +  ^-OH  

MnO₄^-  +    V³⁺  +   H₂O  +  ^-OH      -->  HMnO₄^-    +   VO²⁺    +   H₂O

MnO₄^-  +    V³⁺  +     ^-OH      -->  HMnO₄^-    +   VO²⁺  

Problem RO9.1.

The bonds to iron would contract because the increased charge on the iron would attract the ligand donor electrons more strongly.  The bonds to copper would lengthen because of the lower charge on the copper.

Problem RO9.2.

a)  Most likely there are repulsive forces between ligands if the bonds get too short.

b)  Insufficient overlap between metal and ligand orbitals would weaken the bond and raise the energy.

c)  The range of possible bond lengths gets broader as energy is increased.  The bond has more latitude, with both longer and shorter bonds allowed at higher energy.

Problem RO9.3.

Problem RO9.4.

Problem RO9.5.

The water molecules may pivot toward the more highly charged Fe(III), or they may shift closer to it because of the attraction between the ion and the dipole of the water molecule.

Keep in mind that such adjustments would happen in non-polar solvents, too, although they would involve weaker IMFs such as ion - induced dipole interactions.

Problem RO9.6.

The reactants and products are very similar in this case.  However, the Fe(III) complex has shorter bonds than the Fe(II) complex because of greater electrostatic interaction between the metal ion and the ligands.  These changes in bond length needed in order to get ready to change from Fe(III) to Fe(II) (or the reverse) pose a major barrier to the reaction.

Problem RO9.7

a)  The drawing is an oversimplification, but in general the water molecules are shown reorienting after the electron transfer because of ion-dipole interactions.  In this case, the waters are shown orienting to present their negative ends to the more positive iron atom after the electron transfer.  In reality, in a protein there are lots of other charges (including charges on the ligand) that may take part in additional ion-dipole interactions.

b)  Because electron transfer is so fast, atomic and molecular reorganisations are actually thought to happen before the electron transfer.  The water molecules would happen to shift into a position that would provide the greatest possible stabilisation for the ions and then the electron would be transferred.  A less polar solvent than water would be less able to stabilize ions and the electron would be slower to transfer as a result.  In addition, a less polar solvent than water would be a poor medium to transmit an electron, which is charged and therefore stabilized by interactions with polar solvents.

Problem RO9.8.

a) Here is a plot of the data.

b)  It doesn't look linear.  If we plot the y axis on a log scale, things become a little more linear.

It looks closer to a logarithmic relationship than a linear one.

c)  Assuming one electron transfer:

The graph takes the same form but in the opposite direction along the x axis.

Problem RO9.9.

a)

b)  We can see two sides of an inverted curve.  The reaction gets much faster as the free energy becomes more negative, but at some point the rate begins to decrease again.

Problem RO9.10.

The acceptor compound becomes an anion when it accepts an electron.  The first three compounds do not appear to be strongly electrophilic; they can accept electrons simply because of resonance stability of the resulting anion.  The last three have electron withdrawing groups (chlorines and oxygens) that would stabilize the anion even further.

Problem RO9.11.

a) octahedral; bpy is a bidentate ligand.

b)  Co is first row; Co(I) and Co(II) have relatively low charge.  Usually we would expect them to be high spin.  Co(III) is at a cut-off point in the first row; it is just electronegative enough that isit is udually low spin.

c) 

d)  In a transfer from Co(II) to Co(III), there is additional reorganization needed because the metal changes between high and low spin.  Not only does one electron have to move from one metal to another metal, but additional electrons have to shuffle from one orbital to another on the same metal to accommodate the change.  These reorganizations have a barrier, slowing the reaction.

Problem RO10.1.

a) octahedral

b)  In the first row, 2⁺ complexes are almost always high spin.  However, 3⁺ complexes are sometimes low spin.

c)

d)  The Co(II) complex is high spin and labile.  The ligands are easily replaced by water.

e)  The Cr(III) complex is only d³; it is inert.

Problem RO10.2.

Problem RO10.3.

Problem RO10.4.

a)

b)

Problem RO10.5.

a)

b) The electron transfer between Os(II) and Ru(III) will not involve any electron reorganization because both are low spin to begin with.  However, the electron transfer between Os(II) and Co(III) will result in cobalt changing from low spin to high spin.  The need to move electrons between different d orbitals on the cobalt will add to the barrier, slowing down the reaction.

c)  The pathway is probably inner sphere because of the bridging ligand.  Furthermore, the conjugation in the bridging ligand would help in conducting an electron from one end of the ligand to the other, either through an electron mechanism or a hole mechanism.

Problem RO10.6.

a) The rate decreases exponentially as distance increases.

b)  You might keep the concentration low in order to increase the distance between molecules, reducing the likely hood of an outer-sphere electron transfer.

c)  If you ran the experiment at a series of dilutions, intramolecular electron transfer would be unaffected but outer sphere electron transfer would not.  If the rates were the same across a number of different concentrations, the reaction would probably be intramolecular.

Problem RO10.7.

a)  Rings are frequently used to introduce conformational rigidity (or decrease conformational flexibility), limiting the range of potential shapes a molecule could adopt.  If the molecule can't wiggle around as much, then the distance between the ends of the molecule should be more constant.

b)  Although the ligand is bridging, it would be difficult to picture either an electron or hole mechanism of inner sphere electron transfer.  There are few pi bonds or lone pairs to use as places to put electrons or temporarily remove electrons from, shuttling the electrons from place to place along the ligand.  A conjugated system would be much more likely to carry out inner sphere electron transfer.

c)

d)  The data is not linear.

e)  The data appear to show two lines that cross. That's a classic symptom of two competing mechanisms.  The faster mechanism, to the left, is probably an intramolecular electron transfer.  The slower mechanism, to the right, may be an intermolecular electron transfer.

Problem RO1.8.

a) 

b) With an open t2g level, Ti(III) is associatively labile.

With a filled t2g level and empty eg level, Ru(II) is inert to substitution.

With an occupied eg level, Co(II) is dissociatively labile.

c)  Outer sphere.  Ruthenium lacks a bridging ligand, so it would have to be supplied by Co.  Even if Co had such a ligand, it would be unable to bind to the ruthenium until another ligand left, which is unlikely as the ruthenium is inert.

d)  The ones that fall on the line undergo the same mechanism.  Since ruthenium undergoes outer sphere electron transfer, the titanium complexes along this line also undergo OSET.

e)  These points are above the line, so they represent cases that are much faster with titanium compared to what is expected by looking at ruthenium.

f)  These last one smust undergo inner sphere electron transfer.  ISET would be possible with titanium, as it could easily bind a bridging ligand provided by the cobalt.

Problem RO11.1.

The compound undergoes a two-electron oxidation.  The oxidized species then undergoes a two-electron reduction.

Problem RO12.2.

a) CH₂CHCH₃

C-H bonds	6-
C-X bonds	0
Oxidation Level	6-

CH₃CH₂CH₃

C-H bonds	8-
C-X bonds	0
Oxidation Level	8-

Propane results from a two-electron reduction of propene.

b) CH₃CH₂CHO

C-H bonds	6-
C-X bonds	2+
Oxidation Level	4-

CH₃CH₂CH₂OH

C-H bonds	7-
C-X bonds	1_
Oxidation Level	6-

Propanol  results from a two-electron reduction of propanal.

c) CH₃CH(OH)CH₂CH₂OH

C-H bonds	8-
C-X bonds	2+
Oxidation Level	6-

CH₃COCH₂CHO

C-H bonds	6-
C-X bonds	4+
Oxidation Level	2-

3-oxobutanal  results from a four-electron oxidation of 1,3-butanediol.

Problem RO12.4.

Problem RO12.5.

Multiple answers may be possible.  However, (a) and (e) require aqueous oxidizing conditions whereas (b) excludes aqueous oxidizing conditions.

Problem RO12.6.

Problem RO12.7.

This site was written by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (retired) with other authors as noted on individual pages.  It is freely available for educational use.

Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License. 

Send corrections to cschaller@csbsju.edu

This material is based upon work supported by the National Science Foundation under Grant No. 1043566.

Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.

Navigation:

Back to Reactivity Index

Back to Structure & Reactivity