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Week 3:Electrochemistry: Exercises 19 - 28, Coordination Chemistry 6-8 + extra eight below (no solutions for these)
19. For an electrochemical cell in which the spontaneous reaction is
3Cu2+ + 2Al → 2Al3+ + 3Cu
what will be the qualitative effect on the cell potential if we add ethylenediamine, a ligand that coordinates strongly with Cu2+ but not with Al3+?
If ethylenediamine is added, it will bind to Cu2+ and reduce the concentration of Cu2+. Using the Nernst equation,
Ecell = Eocell - (0.0592 V/n)logQ
Q = [product]/[reactant] = [Al3+]2/[Cu2+]3
If we reduce the concentration of Cu2+, 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 Eocell and ultimately result in a smaller Ecell for the above spontaneous reaction.
20. A copper-zinc 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?
Cu2+ + Zn(s) -> Zn2+ + Cu(s) Ecell = 1.10V
As the Cu2+ decreases, the voltage of the cell should decrease as explained in question 19. Initially Q is equal to 1 since both species are at unit activity. If the cell voltage is 1.00 V, Q is:
Ecell = 1.00V E0cell = 1.10V n = 2
Ecell = E0cell - (0.0592 V/n)logQ
logQ = [n(Eocell - Ecell)]/0.0592V = [2(1.10V - 1.00V)]/0.0592V = 0.20V/0.0592V = 3.38
Q = 103.38 = 2399
21. A galvanic cell consists of a rod of copper immersed in a 2.0M solution of CuSO4 and a rod of iron immersed in a 0.10M solution of FeSO4. Using the Nernst equation and the reduction potentials for
Fe2+ + 2e- → Fe
Cu2+ + 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 FeSO4, and a rod of manganese immersed in a 0.10M solution of MnSO4?
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 = Eo - (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) → Fe2+(aq) + 2e- E = +0.41V
Mn2+(aq) + 2e- → Mn(s) E = -1.19V
Fe(s) + Mn2+(aq) → Mn(s) + Fe2+(aq) Ecell = -0.78V
Because the Ecell 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) + Fe2+(aq) → Fe(s) + Mn2+(aq) Ecell = +0.78V
We can now figure out Q:
Q = Products/reactants = [Mn2+] / [Fe2+] = 0.1M / 1.0M = 0.1
also n = # of electrons traded = 2
Now plugging Ecell, Q, and n into the Nernst equation we get:
E = 0.78 - (0.0592/2)log(0.1)
= 0.78 - 0.0296(-1)
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.
Cu2+ is diluted, lowering the voltage.
- Excess 1.0M ammonia is added to the anode compartment.
Zn2+ is diluted, increasing the voltage
- Excess 1.0M ammonia is added to both compartments at the same time.
No change will occur
- H2S gas is bubbled into the Zn2+ solution.
ZnS precipitates, voltage increases.
- H2S gas is bubbled into the Zn2+ solution at the same time that excess 1.0M ammonia is added to the other solution.
ZnS precipitates, Cu2+ is diluted. voltage increases slightly. Precipitation is stronger.
24. Consider the galvanic cell: Zn | Zn2+ || Cu2+ | Cu
Calculate the ratio of [Zn2+] to [Cu2+] 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."
Ecell = E0cell - (0.0592V)(logQ)/n ; n=2
E0anode = -0.763 , E0cathode = 0.340 -> E0cell = 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 106
23. Consider the cell: Zn | Zn2+(0.0010M) || Cu2+ (0.0010M) l Cu for which E° = +1.10 V.
25. Show that hydrogen peroxide (H2O2) is thermodynamically unstable and should disproportionate to water and oxygen.
O2 + 4 H+ + 4e- → 2 H2O Ered = 1.23V
H2O2 + 2 H+ + 2e- → 2 H2O Ered = 1.78V
Combine half-reactions. (Multiply the peroxide half reaction by two to balance electrons.) Then cancel out
2 H2O2 + 4 H+ + 4 e- + 2 H2O → 4 H2O + O2 + 4H+ + 4e-
2 H2O2 → 2 H2O + O2
Ecell = 1.78V - 1.23V = .55V
ΔG = -nFE Ecell 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 AgNO3. This half-cell is connected to a hydrogen half-cell 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)|H2 (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+]
Ecell = Eocell – (0.0592/n)log Q
- We are given the voltage of the cell (Ecell)= 0.78
- Need to find Eocell?
Eocell=+0.800V – (0) = +0.800V
Find n, the number of electrons
2Ag+ + 2e- à 2Ag(s)
H2 (g, 1 atm) à 2H+ + 2e-
Overall: 2Ag+ + H2(g, 1atm) à 2Ag(s) + 2H+
*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)
Ecell = Eocell – (0.0592/n)log Q
Ecell = Eocell – (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+]
27. Using half-reactions, show that Ag+ and I- spontaneously form AgI(s) when mixed directly at unit initial activity. Show that Ksp for Agl is 10-16.
E(Ag+/Ag) = 0.7996 V; E(AgI/Ag+) = -0.15224 V
E(AgI/Ag+) = E(Ag+/Ag) + 0.0592logKsp(AgI)
Ksp(AgI) = 8.3 X 10-17
28. A standard hydrogen half-cell is coupled to a standard silver half-cell. Sodium bromide is added to the silver half-cell, 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 Ksp for silver bromide.
E = Eo +0.0592log[Ag+] = Eo + 0.0592log(Ksp/[Br-])
0.072 V = 0.7996 V + 0.0592log(Ksp/1.0M)
Ksp(AgBr) = 5.1 X 10-13
6. Give the systematic names of [Co(NH3)4Cl2]Br, K3Cr(CN)5, and Na2CoCl4.
[Co(NH3)4Cl2]Br dichlorotetraamminecobalt(III) bromide
K3Cr(CN)5 potassium pentacyanochromate(II)
Na2CoCl4 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(H2O)6](ClO4)2
- trichlorotriammineplatinum(IV) bromide [PtBr(NH3)3] (Cl)3
- dichlorotetraammineplatinum(IV) sulfate [PtCl2(NH3)4]SO42-
- potassium monochloropentacyanoferrate (III) K3[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)(H2O)5]Cl2
- sodium tricarbonatocobaltate(III) Na3[Co(CO3)3]
- sodium hexacyanoferrate(II) Na4[Fe(CN)6]
- ammonium hexanitrocobaltate(III) (NH4)3[Co(NO3)6]