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Solutions 3: Electrochemistry, Thermodynamics and Coordination Complexes

These are solutions to select problems from homework set #3.

Q3.1 

  1. \(\mathrm{\Delta G^\circ= -nFE^\circ_{cell}}\)

\(\mathrm{E^\circ_{cell} = (E^\circ_{NO_3^-/NO^-}) - (E^\circ_{Al^{3+}/Al})}\)
\(\mathrm{E^\circ_{cell} = (0.956) -(-1.676)}\)
\(\mathrm{E^\circ_{cell} = 2.63\,V}\)

\(\mathrm{N = 3}\) since:

\(\ce{N}\) oxidation: reactant = +5    product= +2
\(\ce{Al}\) oxidation: reactant = 0      product= +3

\(\mathrm{F=96,485\,C/mol\,e^-}\)

\(\mathrm{\Delta G^\circ=-[(3)(96,485\,C/mol\,e^-)(2.63\,V)]}\)
\(\mathrm{\Delta G^\circ= -761\, kJ\, Spontaneous}\)

  1. \(\mathrm{E^\circ_{cell} = (E^\circ_{F_2/F^-})- (E^\circ_{Li^+/Li})}\)

\(\mathrm{E^\circ_{cell} = 2.866 + 3.040 = 5.906\,V}\)

\(\mathrm{N = 2}\) since:
\(\mathrm{F_2 + 2e^- \rightarrow 2F^-}\)
\(\mathrm{2Li^+ + 2e^- \rightarrow 2Li(s)}\)

\(\mathrm{\Delta G^\circ= -[(2\,mol\,e^-)(96,485\,C/mol\,e^-)(5.906\,V)]}\)
\(\mathrm{\Delta G^\circ=-1,140\,kJ}\)
\(\mathrm{Spontaneous}\)

Q3.2

No solution given (at least here).

Q3.3

\(\begin{align}
& \textrm{Reduction: } \ce{Ag+ (aq) + e- \rightarrow Ag(s)}
& & \mathrm{E^{\circ}=+0.800\:V} \\
& \textrm{Oxidation: } \ce{HNO2(aq) \rightarrow NO2(g) + H+ (aq) + e-}
& & \mathrm{E^{\circ}=-1.07\:V} \\ 
& \textrm{Overall: }\ce{Ag+ (aq) + HNO2 (aq) \rightarrow Ag(s) + NO2(g) + H+ (aq)}\end{align}\)

\(\begin{align}
\mathrm{E^{\circ}_{cell}} & = \mathrm{E^{\circ}_{reduction}+ E^{\circ}_{oxidation}} \\
& = \mathrm{0.8 - 1.07} \\
& = \mathrm{- 0.27\:V}
\end{align}\)

Apply Nernst Equation:

\[\mathrm{E_{cell}=E{^\circ}_{cell}-\dfrac{0.0592\,V}{n}\log Q}\]

\[\mathrm{3.1=-0.27-\dfrac{0.0592}{1} \log \dfrac{p_{NO_2}[H^+]}{[Ag^+][HNO_2]}}\]

\[\mathrm{56.9=\log(p_{NO_2}[10^{-6}])-\log0.35}\]

\[\mathrm{\log(p_{NO_2}[10^{-6}])=56.44}\]

\[\mathrm{(p_{NO_2}[10^{-6}])=2.8 \times10^{56}}\]

\[\mathrm{p_{NO_2}=2.8 \times10^{62}\:bar}\]

Q3.4

Solve for \(\mathrm{E_{cell}}\) in the Nernst Equation:

  • Oxidation reaction: \(\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \hspace{68 pt}\quad E^\circ=0.763}\)
  • Reduction reaction: \(\mathrm{(Ag^+(aq)+e^- \rightarrow Ag(s))\times2 \hspace{47 pt}\quad E^\circ=0.800\,V}\)
  • Net Ionic Equation: \(\mathrm{Zn(s)+2Ag^+(aq) \rightarrow Zn^{2+}(aq) +2Ag(s) \quad E^\circ_{cell}=1.563\,V}\)

Use Nernst equation to solve for \(\ce{[Ag+]}\):

\[\mathrm{E=E^\circ_{cell}-\dfrac{0.0592}{2}\log\dfrac{[Zn^{2+}]}{[Ag^+]^2}}\]

\[\mathrm{1.500\,V = 1.563-\dfrac{0.0592}{2}\log\dfrac{1.5\,M}{x^2}}\]

\[\mathrm{\log\dfrac{1.5\,M}{x^2} = \dfrac{-2(1.500-1.563)}{0.0592}}\]

\[\mathrm{\log\dfrac{1.5\,M}{x^2} = 2.13}\]

\[\mathrm{x = 0.4222}\]

\[\mathrm{[Ag^+]= 0.4222\,M}\]

Q3.5

First must solve for \(\mathrm{[Ag^+]}\) in saturated solution

\(\mathrm{K_{sp}= [Ag^+]^2[CrO_4^{2-}] = (2s)^2(s) = 4s^3}\)

\(\mathrm{s = \sqrt[\Large 3]{\dfrac{1.2\times10^{-11}}{4}} = 1.44\times10^{-4}\, M}\)

\(\mathrm{[Ag^+]_{anode} = 2s = 2.88\times10^{-4}\,M}\)

\(\mathrm{Ag(s)+Ag^+(0.125\,M)\rightarrow Ag(s)+Ag^+(2.88\times 10^{-4}\,M)}\)

\(\mathrm{E_{cell}=E^\circ_{cell}-\dfrac{0.0592}{2}\log\dfrac{2.88\times10^{-4}\,M}{0.125\,M}}\)

\(\mathrm{E_{cell}=0.000+0.078\,V =0.078\,V}\)

Q3.6

First calculate charge passed through and number of moles of electrons transferred:

\(\mathrm{Mol\:e^- = 60\:min \times\dfrac{60\:s}{1\:min}\times\dfrac{2.15\:C}{1\: s}\times\dfrac{1\:mol\:e^-}{96485\:C} = 0.08\:mol\:e^-}\)

  1. \(\ce{Zn^2+}\)

\(\mathrm{0.08\:mol\:e^- \times \dfrac{1\:mol\:Zn^{2+}}{2\:mol\:e^-}\times\dfrac{1\:mol\:Zn}{1\:mol\:Zn^{2+}}\times\dfrac{65.39\:g\:Zn}{1\:mol\:Zn} = 2.62\:g\:Zn}\)

  1. \(\ce{Al^3+}\)

\(\mathrm{0.08\:mol\: e^- \times \dfrac{1\:mol\: Al^{3+}}{3\: mol\: e^-} \times\dfrac{1\: mol\: Al}{1\:mol\: Al^{3+}}\times\dfrac{26.98\:g\:Al}{1\:mol\:Al} = 0.72\:g\:Al}\)

  1. \(\ce{Ag+}\)

\(\mathrm{0.08\:mol\: e^- \times\dfrac{1\: mol\: Ag^+}{1\: mol\: e^-}\times\dfrac{1\: mol\: Ag}{1\: mol\:Ag^+}\times\dfrac{107.9\:g\: Al}{1\: mol\: Ag}= 8.63\:g\:Ag}\)

  1. \(\ce{Ni^2+}\)

\(\mathrm{0.08\:mol\: e^- \times\dfrac{1\: mol\: Ni^{2+}}{2\: mol\: e^-} \times\dfrac{1\: mol\: Ni}{1\: mol\: Ni^{2+}}\times\dfrac{58.69\:g\: Ni}{1 \:mol\: Ni}= 2.35\:g\: Ni}\)

Q3.7

  1. The time must be converted from minutes to seconds, and the dimensional analysis uses Faraday’s constant. The Dimensional analysis is set up as follows:

\[\mathrm{Time\;(h) \times \dfrac{60\;min}{1\;h} \times \dfrac{60\;sec}{1\;min} \times \dfrac{amount\;of\;current\;C}{1\;sec} \times \dfrac{1\;mol\;e^-}{96,485\;C}}\]

Then we will account for the mass by using stoichiometry of the ratio of the moles of metal to moles of electrons along with the molar mass of each element to find the final number of grams. Part a sets up as follows:

\[\mathrm{60\;min \times \dfrac{60\;seconds}{1\;min} \times \dfrac{2.00\;A}{1\;s}\times\dfrac{1\;mol\;e^-}{96,485}\times\dfrac{1\;mol\;Cu^{2+}}{2\;mol\;e^-}\times\dfrac{63.5\;grams\;Cu}{1\;mol\;Cu}}\]

The answer is 2.37 grams

  1. \[\mathrm{60\;min \times \dfrac{60\;seconds}{1\;min} \times \dfrac{2.00\;A}{1\;s}\times\dfrac{1\;mol\;e^-}{96,485}\times\dfrac{1\;mol\;Li^{+}}{1\;mol\;e^-}\times\dfrac{6.941\;grams\;Li}{1\;mol\;Li}}\]

The answer is 0.518 grams

  1. \[\mathrm{60\;min \times \dfrac{60\;seconds}{1\;min} \times \dfrac{2.00\;A}{1\;s}\times\dfrac{1\;mol\;e^-}{96,485}\times\dfrac{1\;mol\;Pb^{2+}}{2\;mol\;e^-}\times\dfrac{207.2\;grams\;Cu}{1\;mol\;Pb}}\]

The answer is 7.74 grams

  1. \[\mathrm{60\;min \times \dfrac{60\;seconds}{1\;min} \times \dfrac{2.00\;A}{1\;s}\times\dfrac{1\;mol\;e^-}{96,485}\times\dfrac{1\;mol\;Mg^{2+}}{2\;mol\;e^-}\times\dfrac{24.3050\;grams\;Mg}{1\;mol\;Mg}}\]

The answer is 0.907 grams.

For review on this topic, visit the page "Electrochemistry 8: Electrolytic cells and electrolysis".

Q3.8

  1. Oxidation reaction: \(\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \hspace {37 pt}\quad E^\circ_{cell}= -0.763\:V}\)

Reduction reaction: \(\mathrm{Sn^{2+}(aq) + 2e^- \rightarrow Sn(s) \hspace {34 pt}\quad E^\circ_{cell}= -0.137\:V}\)

Net reaction: \(\mathrm{Zn(s) + Sn^{2+}(aq) \rightarrow Zn^{2+}(aq) + Sn(s) \quad E^\circ_{cell}= 0.626\:V}\)

This reaction is spontaneous 

  1. Oxidation reaction: \(\mathrm{2Fe^{2+} \rightarrow 2Fe^{3+} + 2e^- \hspace{74 pt}\quad E^\circ_{cell}= 0.771\:V}\)

Reduction reaction: \(\mathrm{Hg^{2+}(aq) + 2e^- \rightarrow Hg(l) \hspace{58 pt}\quad E^\circ_{cell}= 0.854\:V}\)

Net reaction: \(\mathrm{2Fe^{2+}(aq) + Hg^{2+}(aq) \rightarrow 2Fe^{3+}(aq) + Hg(l) \quad E^\circ_{cell}= 0.083\: V}\)

This reaction is spontaneous

  1. Oxidation reaction: \(\mathrm{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^- \hspace{53 pt}\quad E^\circ_{cell}= 0.337\:V}\)

Reduction reaction: \(\mathrm{Sr^{2+}(aq) + 2e^-(aq) \rightarrow Sr(s) \hspace{37 pt}\quad E^\circ_{cell}= -2.89\:V}\)

Net reaction: \(\mathrm{Cu(s) + Sn^{4+}(aq) \rightarrow Cu^{2+}(aq) + Sn^{2+}(aq)\quad E^\circ_{cell}= -3.227\:V}\)

Reaction requires voltages \(\mathrm{> 3.227\: V}\)

Q3.9

Calculate \(\mathrm{\Delta G^\circ_f= -nFE^\circ_{cell} = -4\:mol\:e^- \times \dfrac{96485\: C}{1\: mol\: e^-}\times1.559\:V = -6.017\times10^5\: J = -601.7\: kJ}\)

Find \(\mathrm{\Delta G^\circ_f}\) for (\(\ce{N2H4}\)):

\(\mathrm{-601.7\:kJ = \Delta G^\circ_f[N_2(g)] + 2 \Delta G^\circ_f[H_2O(l)] - \Delta G^\circ_f[N_2H_4(aq)] - \Delta G^\circ_f[O_2(g)]}\)

\(\mathrm{-601.7\:kJ = 0.00\:kJ + 2(-237.2) + 601.7 = 127.3\: kJ}\)

Q3.10

Potassium and magnesium because they are more active than zinc on the activity series. Lead and cobalt are less active so they cannot be used. Sacrificial anode – an active metal that protects the metal on the cathode from corrosion. "The Activity Series Table"

Q3.11

  1. \(\ce{[Co(OH)2(H2O)4]}\): tetraaquadihydroxocobalt(II)
  2. \(\ce{[Ni(OH)3(CO)3]}\): tricarbonyltrihydroxonickel(III)
  3. \(\ce{[Pt(H2O)(NH3)3][PtBr6]}\): monoamminetricyanoplatinum (II) hexabromoplatinate (IV)
  4. \(\ce{[Cu(CN)4(en)2]-}\): tetracyanobis(ethylenediammine)cuprate (III)
  5. \(\ce{Au2[SnCl4]}\): Gold(I) tetrachlorostannate(II)

For extra practice, refer to "Nomenclature of Coordination Complexes".

Q3.12

No solution given (at least here). All of the above are monodentate ligands. Monodentate ligands are essentially Lewis bases that donate a lone (“mono”) pair of electrons to the central metal. These ligands can be neutral or ionic.

Q3.13

To review isomerism, visit "Structural Isomers".

Q3.14

  1. \(\ce{[Cr(H2O)2(CO)2Cl2]}\): diaquadicarbonyldichlorochromium (II)
  2. \(\ce{[Cr(NH3)3(H2O)3]Cl3}\): triamminetriaquachromium (III) chloride
  3. \(\ce{[PtCl(H2O)3]+}\): triaquachloroplatinum (II) ion
  4. \(\ce{[Al(OH)(H2O)8]Cl2}\): octaaquahydroxoaluminum (III) chloride
  5. \(\ce{[Co(NH3)4]Br3}\): tetraamminecobalt (III) bromide

Q3.15

  1. tetraaquadihydroxocobalt(II)
  2. tricarbonyltrihydroxonickel(III)
  3. monoamminetricyanoplatinum (II) hexabromoplatinate (IV)
  4. tetracyanobis(ethylenediammine)cuprate (III)
  5. Gold(I) tetrachlorostannate(II)

For extra practice, refer to "Nomenclature of Coordination Complexes".

Q3.16

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Q3.17

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Q3.18

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Q3.19

In a tetrahedral structure, one ligand is not on the other side of the central atom. In a linear structure, there are no options on how to bind the ligand to the central atom.

Q3.20

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Q3.21

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Q3.22

example 3.jpg  

Q3.23

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Q3.24

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