HW Solutions #3

UC Davis - Spring 2021
Chemistry 2C

Unit 1: Electrochemistry  ●  Unit 2: Coordination Chemistry  ●  Unit 3: Descriptive Chemistry  ●  Unit 4: Chemical Kinetics  ●  Unit 5: Nuclear Chemistry
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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 6-8 + extra eight below (no solutions for these)

Electrochemistry

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,

E_cell = E^o_cell - (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 E^o_cell and ultimately result in a smaller E_cell 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) E_cell = 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:

E_cell = 1.00V E⁰_cell = 1.10V n = 2

E_cell = E⁰_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

Q = 10^3.38 = 2399

Nernst equation

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 = 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) → Fe²⁺(aq) + 2e^- E = +0.41V
Mn²⁺(aq) + 2e^- → Mn(s) E = -1.19V
----------------------------------------------------
Fe(s) + Mn²⁺(aq) → Mn(s) + Fe²⁺(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) + 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 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?

1. Excess 1.0M ammonia is added to the cathode compartment.
   Cu²⁺ is diluted, lowering the voltage.
2. Excess 1.0M ammonia is added to the anode compartment.
   Zn²⁺ is diluted, increasing the voltage
3. Excess 1.0M ammonia is added to both compartments at the same time.
   No change will occur
4. H₂S gas is bubbled into the Zn²⁺ solution.
   ZnS precipitates, voltage increases.
5. 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."

E_cell = E⁰_cell - (0.0592V)(logQ)/n ; n=2

E⁰_anode = -0.763 , E⁰_cathode = 0.340 -> E⁰_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 10⁶

23. Consider the cell: Zn | Zn²⁺(0.0010M) || Cu²⁺ (0.0010M) l Cu for which E° = +1.10 V.

Competitive reactions

25. Show that hydrogen peroxide (H₂O₂) is thermodynamically unstable and should disproportionate to water and oxygen.

O₂ + 4 H⁺ + 4e^- → 2 H₂O E_red = 1.23V

H₂O₂ + 2 H⁺ + 2e^- → 2 H₂O E_red = 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

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 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⁺]

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(H₂/H⁺)

E^o_cell=+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⁺

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)

• Plug these numbers into the Nernst Equation:

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⁺]

• Solve for pH = -log[H⁺] = (-0.0385 pH) V

Solubility product

27. Using half-reactions, 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 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 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(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:

1. hexaaquonickel(II) perchlorate [Ni(H₂O)₆](ClO4)₂
2. trichlorotriammineplatinum(IV) bromide [PtBr(NH₃)₃] (Cl)₃
3. dichlorotetraammineplatinum(IV) sulfate [PtCl₂(NH3)₄]SO₄^2-
4. potassium monochloropentacyanoferrate (III) K₃[FeCl(CN)₅]

8. Write the formula for each of the following by using brackets to distinguish the complex ion:

1. hydroxopentaaquoaluminum(III) chloride [Al(OH)(H₂O)₅]Cl₂
2. sodium tricarbonatocobaltate(III) Na₃[Co(CO₃)₃]
3. sodium hexacyanoferrate(II) Na₄[Fe(CN)₆]
4. ammonium hexanitrocobaltate(III) (NH₄)₃[Co(NO₃)₆]