HW Solutions #3
 Page ID
 2846
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These homework problems are suggested and will not be turned in for review. However, answers will be available for them the following week by your class TAs. For more homework feel free to go to the Homework page.
Week 3:Electrochemistry: Exercises 19  28, Coordination Chemistry 68 + extra eight below (no solutions for these)
Electrochemistry
19. For an electrochemical cell in which the spontaneous reaction is
3Cu^{2+} + 2Al → 2Al^{3+} + 3Cu
what will be the qualitative effect on the cell potential if we add ethylenediamine, a ligand that coordinates strongly with Cu^{2+} but not with Al^{3+}?
If ethylenediamine is added, it will bind to Cu^{2+} and reduce the concentration of Cu^{2+}. Using the Nernst equation,
E_{cell} = E^{o}_{cell}  (0.0592 V/n)logQ
Q = [product]/[reactant] = [Al^{3+}]^{2}/[Cu^{2+}]^{3}
If we reduce the concentration of Cu^{2+}, Q will become larger and logQ will also become larger. Looking at the Nernst equation above, a larger logQ would result in a larger number being subtracted from E^{o}_{cell} and ultimately result in a smaller E_{cell} for the above spontaneous reaction.
20. A copperzinc battery is set up under standard conditions with all species at unit activity. Initially, the voltage developed by this cell is 1.10 V. As the battery is used, the concentration of the cupric ion gradually decreases, and that of the zinc ion increases. According to Le Chatelier’s principle, should the voltage of the cell increase or decrease? What is the ratio, Q, of the concentrations of zinc and copper ions when the cell voltage is 1.00 V?
E_{cell} = 1.00V E^{0}_{cell} = 1.10V n = 2
E_{cell} = E^{0}_{cell}  (0.0592 V/n)logQ
logQ = [n(E^{o}_{cell}  E_{cell})]/0.0592V = [2(1.10V  1.00V)]/0.0592V = 0.20V/0.0592V = 3.38
Nernst equation
21. A galvanic cell consists of a rod of copper immersed in a 2.0M solution of CuSO_{4} and a rod of iron immersed in a 0.10M solution of FeSO_{4}. Using the Nernst equation and the reduction potentials for
Fe^{2+} + 2e^{} → Fe
Cu^{2+} + 2e^{} → Cu
Calculate the voltage for the cell as described.
22. What voltage will be generated by a cell that consists of a rod of iron immersed in a 1.00M solution of FeSO_{4}, and a rod of manganese immersed in a 0.10M solution of MnSO_{4}?
At first glance see that the question involves electric cell concentrations, we know immediately that we will need to solve this using the Nernst equation.
E = E^{o}  (0.0592/n)logQ
Start by picking a direction and writing half reactions to give a balanced equation. Then look up the Standard E values and write them next to the half reactions, flipping signs to match the direction. Add them together to give the Standard Ecell. (Alternatively if using the Ecell = Ecat  Eanode equation do not flip any signs, use the standard reduction potentials for both. The subtraction flips the sign for you).
Fe(s) → Fe^{2+}(aq) + 2e^{} E = +0.41V
Mn^{2+}(aq) + 2e^{} → Mn(s) E = 1.19V

Fe(s) + Mn^{2+}(aq) → Mn(s) + Fe^{2+}(aq) E_{cell} = 0.78V
Because the E_{cell} is negative we know the reaction will run in the opposite direction. The reaction will run in the direction that produces a positive voltage.
Mn(s) + Fe^{2+}(aq) → Fe(s) + Mn^{2+}(aq) Ecell = +0.78V
We can now figure out Q:
Q = Products/reactants = [Mn^{2+}] / [Fe^{2+}] = 0.1M / 1.0M = 0.1
also n = # of electrons traded = 2
Now plugging E_{cell}, Q, and n into the Nernst equation we get:
E = 0.78  (0.0592/2)log(0.1)
= 0.78  0.0296(1)
= 0.8096V
Does the cell voltage, E, increase, decrease, or remain unchanged when each of the following changes is made?
 Excess 1.0M ammonia is added to the cathode compartment.
Cu^{2+} is diluted, lowering the voltage.  Excess 1.0M ammonia is added to the anode compartment.
Zn^{2+} is diluted, increasing the voltage  Excess 1.0M ammonia is added to both compartments at the same time.
No change will occur  H_{2}S gas is bubbled into the Zn^{2+} solution.
ZnS precipitates, voltage increases.  H_{2}S gas is bubbled into the Zn^{2+} solution at the same time that excess 1.0M ammonia is added to the other solution.
ZnS precipitates, Cu^{2+} is diluted. voltage increases slightly. Precipitation is stronger.
24. Consider the galvanic cell: Zn  Zn^{2+}  Cu^{2+}  Cu
Calculate the ratio of [Zn^{2+}] to [Cu^{2+}] when the voltage of the cell has dropped to 1.05 V, 1.00 V, and 0.90 V. Notice that a small drop in voltage parallels a large change in concentration. Therefore, a battery that registers 1.00 V is quite "run down."
E_{cell }= E^{0}_{cell}  (0.0592V)(logQ)/n ; n=2
E^{0}_{anode} = 0.763 , E^{0}_{cathode} = 0.340 > E^{0}_{cell} = 1.103V
1.05V = 1.103V  (0.0592)(logQ)/2 > Q = 61.736
1.00V = 1.103V  (0.0592)(logQ)/2 > Q = 3018.07
0.90V = 1.103V  (0.0592)(logQ)/2 > Q = 7.21 x 10^{6}
23. Consider the cell: Zn  Zn^{2+}(0.0010M)  Cu^{2+} (0.0010M) l Cu for which E° = +1.10 V.
Competitive reactions
25. Show that hydrogen peroxide (H_{2}O_{2}) is thermodynamically unstable and should disproportionate to water and oxygen.
O_{2} + 4 H^{+} + 4e^{} → 2 H_{2}O E_{red} = 1.23V
H_{2}O_{2} + 2 H^{+} + 2e^{} → 2 H_{2}O E_{red} = 1.78V
Combine halfreactions. (Multiply the peroxide half reaction by two to balance electrons.) Then cancel out
2 H_{2}O_{2} + 4 H^{+} + 4 e^{} + 2 H_{2}O → 4 H_{2}O + O_{2} + 4H^{+} + 4e^{}
2 H_{2}O_{2} → 2 H_{2}O + O_{2}
E_{cell} = 1.78V  1.23V = .55V
ΔG = nFE E_{cell} is positive, therefore ΔG is negative and this reaction is spontaneous.
Voltage and pH
26. A silver electrode is immersed in a 1.00M solution of AgNO_{3}. This halfcell is connected to a hydrogen halfcell in which the hydrogen pressure is 1.00 atm and the H^{+} concentration is unknown. The voltage of the cell is 0.78 V. Calculate the pH of the solution.
1.) Setup a cell diagram in the spontaneous direction and write out what is given and what you need to solve for:
Pt(s)H_{2} (g, 1 atm)H^{+ }(x M)Ag^{+}(1.00M)Ag(s)
Ag^{+} is more readily reduced than H^{+}
2.) Use the Nernst Equation to figure out [H+] which you can solve the pH for using pH=log[H^{+}]
Nernst Equation:
E_{cell} = E^{o}_{cell} – (0.0592/n)log Q
 We are given the voltage of the cell (E_{cell})= 0.78
 Need to find E^{o}_{cell}?
E^{o}_{cell}=E^{o}(cathode)E^{o}(anode)
E^{o}_{cell}=E^{o}(right)E^{o}(left)
E^{o}_{cell}=E^{o}(Ag^{+}/Ag)E^{o}(H_{2}/H^{+})
E^{o}_{cell}=+0.800V – (0) = +0.800V

Find n, the number of electrons
2Ag^{+} + 2e à 2Ag(s)
H_{2} (g, 1 atm) à 2H^{+} + 2e
Overall: 2Ag^{+} + H_{2}(g, 1atm) à 2Ag(s) + 2H^{+}
n=2

Find Q
Q= [H^{+}]^{2}/[Ag^{+}]^{2}
Q= [x]^{2}/[1.00M]^{2}
*In substitutions for Q, a=1 for activities of pure solid and liquids, partial pressures (atm) for activities of gases and molarities for the activities of solution components (Ch20, p.837)
E_{cell} = E^{o}_{cell} – (0.0592/n)log Q
E_{cell} = E^{o}_{cell} – (0.0592/n)log [H^{+}]^{2}/[Ag^{+}]^{2}
0.78V = 0.800V – (0.0592/2)log [x]^{2}/[1.00M]^{2}
0.020 = –(0.0592/2)log [H^{+}]^{2}
0.020 = –(0.0592/2) * 2log [H^{+}]
0.020 = –(0.0592) log [H^{+}]
0.0385 V = log [H^{+}]
Solubility product
27. Using halfreactions, show that Ag^{+} and I^{} spontaneously form AgI(s) when mixed directly at unit initial activity. Show that K_{sp} for Agl is 10^{16}.
E(Ag^{+}/Ag) = 0.7996 V; E(AgI/Ag^{+}) = 0.15224 V
E(AgI/Ag^{+}) = E(Ag^{+}/Ag) + 0.0592logK_{sp}(AgI)
K_{sp}(AgI) = 8.3 X 10^{17}
28. A standard hydrogen halfcell is coupled to a standard silver halfcell. Sodium bromide is added to the silver halfcell, causing precipitation of AgBr, until a concentration of 1.00M Br^{} is reached. The voltage of the cell at this point is 0.072 V. Calculate the K_{sp} for silver bromide.
E = E^{o} +0.0592log[Ag^{+}] = E^{o} + 0.0592log(K_{sp}/[Br^{}])
0.072 V = 0.7996 V + 0.0592log(K_{sp}/1.0M)
K_{sp}(AgBr) = 5.1 X 10^{13}
Coordination Chemistry
6. Give the systematic names of [Co(NH_{3})_{4}Cl_{2}]Br, K_{3}Cr(CN)_{5}, and Na_{2}CoCl_{4}.
[Co(NH_{3})_{4}Cl_{2}]Br dichlorotetraamminecobalt(III) bromide
K_{3}Cr(CN)_{5} potassium pentacyanochromate(II)
Na_{2}CoCl_{4} sodium tetrachlorocobaltate(III)
7. Write the formulas for each of the following compounds by using brackets to distinguish the complex ion from the other ions:
 hexaaquonickel(II) perchlorate [Ni(H_{2}O)_{6}](ClO4)_{2}
 trichlorotriammineplatinum(IV) bromide [PtBr(NH_{3})_{3}] (Cl)_{3}
 dichlorotetraammineplatinum(IV) sulfate [PtCl_{2}(NH3)_{4}]SO_{4}^{2}
 potassium monochloropentacyanoferrate (III) K_{3}[FeCl(CN)_{5}]
8. Write the formula for each of the following by using brackets to distinguish the complex ion:
 hydroxopentaaquoaluminum(III) chloride [Al(OH)(H_{2}O)_{5}]Cl_{2}
 sodium tricarbonatocobaltate(III) Na_{3}[Co(CO_{3})_{3}]
 sodium hexacyanoferrate(II) Na_{4}[Fe(CN)_{6}]
 ammonium hexanitrocobaltate(III) (NH_{4})_{3}[Co(NO_{3})_{6}]