Extra Credit 2

Q17.1.2

For the scenario in the previous question, how many electrons moved through the circuit?

If the total charge passing through the cell is equal to 5.3 × 10³ C (from the previous question), then the number of electrons passing can be obtained by dividing the total charge (5.3x 10³ C in this case) by the charge per electron (1.6 X 10^-19 C). Charge per electron is always 1.6 x 10^-19 C.

This is how you find the charge per electron:

\(\frac{96485\text{ Coulombs}}{\text{mole e}^-}\times\frac{1\text{ mole e}^-}{6.022\times10^{23}\text{ Electrons}} = 1.6\times10^{-19} C/\)e^-

Therefore; number of electrons = (5.3 × 10³ C)/(1.6 × 10^-19 C/e^-)

= 3.31 × 10²² number of electrons moved through the circuit.

Q17.7.5

An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a Zn(NO₃)₂ solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123-mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is 7.140 g/cm³. Assume the efficiency is 100%.

This question is referring to the process of electroplating.

step (1) Calculate the moles of electrons passing through the cell.

number of electrons = total charge/charge on electron

A (Ampere) is the SI unit for current which is the flow of electric charge across a surface at the rate of one coulomb per second. Thus, total charge in Coulombs can be obtained by multiplying by the amount of time in seconds the current was used. We are given minutes so a conversion of minutes to seconds is required. Number of electrons can then be obtained by dividing total charge by charge per electron.

=(2.599 A × 60 min × 60 seconds) / 1.6× 10^-19 C

= 5.85 X 10²² number of electrons

Dividing number of electrons by Avogadro's number (6.02 X 10²³) gives us the moles of electrons passing through the cell.

=(5.85× 10²² number of electrons)/ 6.02× 10²³

= 0.0971 mol of electrons

step (2) Write the balanced equation

Zn²⁺+ 2e^- → Zn which tells us the stoichiometry of the reaction.

In order to get moles of Zn we divide .0971 mol of electrons by 2.

Therefore moles of Zn = 0.0486 mol

Next, we find the mass of Zn by using moles of Zn and Molar mass of Zn.

mass of Zn = 0.0486 mol × 65.39 g/mol

= 3.17 grams of Zinc

We are given that the density of zinc is 7.140 g/cm³.

Therefore volume of zinc = (3.17 g) / (7.140 g/cm³ )

= .444 cm³

Converting given value of 0.01123-mm layer of zinc to cm we get (0.01123mm) X (1m / 1000mm) X (100cm / 1m) = 0.001123 cm layer of zinc.

Therefore the surface area = (.444 cm³ ) / (0.001123 cm) = 395.4 cm²

Q19.2.2

Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate:

1. tetrahydroxozincate(II) ion (tetrahedral)
2. hexacyanopalladate(IV) ion
3. dichloroaurate ion (note that aurum is Latin for "gold")
4. diaminedichloroplatinum(II)
5. potassium diaminetetrachlorochromate(III)
6. hexaaminecobalt(III) hexacyanochromate(III)
7. dibromobis(ethylenediamine) cobalt(III) nitrate

1. 4, [Zn(OH)₄]^2-; Tetrahedral so has a coordination number of 4. A single OH ligand has a -1 charge, and Zinc has +2 (From the oxidation state). So (OH)₄ (-4) and Zn(+2) would have a charge of -2. Hydroxo stands for OH and Zincate stands for Zn.
2. 6, [Pd(CN)₆]^2-; Coordination number of 6 because (CN)₆ is bonded to Pd. Hexa stands for 6 and cyano stands for CN. As Pd has a charge of +4 and (CN)₆ has a charge of 6(-1), the overall charge is 2-.
3. 2, [AuCl₂]^-; Coordination number of 2 because Cl₂ is bonded to Au. No Oxidation State is listed after the metal name so the oxidation state is +1. Negative overall charge as Au is +1 and Cl₂ is 2(-1)
4. 4, [Pt(NH₃)₂Cl₂]; Coordination number of 4 because (NH₃)₂ and Cl₂ are bonded to Pt. Diamine stands for (NH₃)₂ and dichloro stands for Cl₂.
5. 6, K[Cr(NH₃)₂Cl₄]; Coordination number of 6 because (NH₃)₂ and Cl₄ are being bonded. Diamine stands for (NH₃)₂ and tetrachloro stands for Cl₄.
6. 6, [Co(NH₃)₆][Cr(CN)₆]; Coordination number of 6 as (NH₃)₆ is bonded to Co and (CN)₆ is bonded to Cr.
7. 6, [Co(en)₂Br₂]NO₃; Coordination number of 6 as en is a bidentate ligand. Ethylenediamine is a neutral ligand and Br₂ contributes a charge of 2(-1) and cobalt has charge +3.

Q12.3.14

The original solution was incorrect.

From the following data, determine the rate equation, the rate constant, and the order with respect to A for the reaction A⟶2C.

Order:

Order with respect to A is 2.

This is found by

(3.8 X 10^-7)/(1.33 X 10^-2) = 2.86 X 10^-5

(1.52 X 10^-6)/(2.66 X 10^-2) = 5.71 X 10^-5

You can see that the ratio of Concentration to Rate is 2 so the order is 2.

Rate equation:

\(rate = k[A]^2\)

Solving for k(rate constant) we set the equation to

\(k = \frac{rate}{[A]^2}\)

k = 2.15 X 10^-3 M^-1h^-1

Q12.6.6

The original solution was incorrect.

Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same?

(a) Cl₂+CO⟶Cl₂CO

rate=k[Cl₂]^3/2[CO]

The rate of the overall reaction is equal to the rate of the slow step elementary reaction. From the rate law we know that the reaction for the elementary step is:

1.5Cl₂+CO⟶?Product(s)

This formula does not match with the given overall reaction formula so elementary reaction and the overall reaction cannot be the same.

(b) PCl₃+Cl₂⟶PCl₅

rate=k[PCl₃][Cl₂]

Might be the same.

(c) 2NO+H₂⟶N₂+H₂O

rate=k[NO][H₂]

The rate of the overall reaction is equal to the rate of the slow step elementary reaction. From the rate law we know that the reaction for the elementary step is:

NO+H₂⟶?Products

This formula does not match with the given overall reaction formula so elementary reaction and the overall reaction cannot be the same.

(d) 2NO+O₂⟶2NO₂

rate=k[NO]²[O₂]

Might be the same.

(e) NO+O₃⟶NO₂+O₂

rate=k[NO][O₃]

Might be the same.

Q21.4.18

The original solution was incorrect.

The isotope \( \ce{^{90}_{38}Sr} \) is one of the extremely hazardous species in the residues from nuclear power generation. The strontium in a 0.500-g sample diminishes to 0.393 g in 10.0 y. Calculate the half-life.

k=(-1/t)ln(A/A₀)

k=(-1/10)ln(.393/.500)

k=.02408 \(year^{-1}\)

k=.693/t_1/2

t_1/2=.693/k

t_1/2=.693/.02408

t_1/2= 28.78 years.

Q20.3.6

It is often more accurate to measure the potential of a redox reaction by immersing two electrodes in a single beaker rather than in two beakers. Why?

The two different beakers may have different ion concentrations which would affect rate flow. Or the different beakers may have different ions with different mobilities. A large difference in cation/anion diffusion rates would increase resistance in the salt bridge and limit electron flow through the circuit. Cell potential is measured in volts, and volts is defined as 1 Joule of work per coulomb of charge transferred. A resistance would decrease the amount of energy doing work. Therefore the greater the resistance, the greater the difference between measured potential and actual potential

Q20.5.17

The original solution was incorrect.

What is the standard change in free energy for the reaction between Ca²⁺ and Na(s) to give Ca(s) and Na⁺? Do the sign and magnitude of ΔG° agree with what you would expect based on the positions of these elements in the periodic table? Why or why not?

Finding E^o we get E^o_cell= E_Cathode - E_Anode

Use Standard Electrode Potentials

E^o_cell= -2.84 - (-2.713)

E^o_cell= -.127

Standard change in free energy for the reaction would be ΔG° = -nFE^o_cell. Here F is Faraday's constant of 96485 C/mole.

ΔG° = -(2)(96485)(-.127)

ΔG° = 24.5 kJ

The magnitude ΔG° of is relatively small which is expected because the two elements are so close on the periodic table. The sign is also expected because the elements being so close makes it likely that a redox reaction is nonspontaneous.