Extra Credit 2

Q17.1.2

For the following scenario in question 17.1.1:

A 2.5 A current is run through a circuit for 35 minutes where 5.3 × 10³ C of charge moved through the current

How many electrons moved through the circuit?

S17.1.2

Given: Charge (in coulombs)

Asked For: Number of electrons

Strategy:

A. First, we recognize that the charge of a single electron is 1.6 × 10^-19 C.

B. Using stoichiometry, we can derive the number of electrons by dividing the total charge given by this value.

\[5.3×10^3C \times \frac{1 electron}{1.6 × 10^{-19} C} = 3.31 × 10^{22} electrons\]

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%.

S17.7.5

Given: Current (in amperes)

Asked For: Surface area (in cm²)

Strategy:

A. First, we calculate the total charge from the current by converting amperes to coulombs (1A=1C/s)

\[total charge = 2.599 A \times 60 minutes \times 60 seconds = 9356.4 C\]

B. Then we divide this total charge by the charge of an electron given by 1.6 x 10^-19C.

\[9356.4 C \times \frac{1 electron}{1.6 × 10^{-19} C} = 5.85 \times 10^{22}\]

C. Now that we have the total number of electrons, we can write a balance equation and use stoichiometry to solve for the moles of Zn.

Zn²⁺ +2e^- →Zn

\[.971 mol electrons \times \frac{1 mol Zn}{2 mol electrons} = .486 mol Zn\]

D. Convert mols of Zn to grams of Zn using its molar mass

\[.486 mol Zn \times \frac{65.39 g Zn}{1 mol Zn} = 31.78 g Zn\]

E. Find the volume of Zn using the given density value

\[31.78 g \times \frac{ 1 cm^3}{7.140 g} = 4.45 cm^3\]

F. Divide this density by the thickness of layer (given) to obtain area

\[ \frac{4.45 cm^3}{.1123 cm} = 39.6 cm^2\]

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

S19.2.2

Given: compound nomencalture

Asked For: coordination number; formulas

Strategy:

To determine coordination numbers we must count the total number of ligands bonded to the central metal and distinguish monodentate and polydentate ligands. To determine the formulas, we use the nomenclature rules and work backwards

1. "tetrahydroxo" = 4 hydroxide ligands, since hydroxide is a monodentate ligand, we have a total of 4 bonds to the central metal
   Coordination Number: ​​​​​​4
   We review the basics of nomenclature and see that "tetra" = 4 and "hydroxo" = OH^-. Since the charge on Zinc is 2+ which is given in the nomenclature by the roman numerals, we can calculate the total charge on the complex is 2-
   Formula: [Zn(OH)₄]²⁻
2. "hexacyano" = 6 cyanide ligands, since cyanide is a monodentate ligand, we have a total of 6 bonds to the central metal
   Coordination Number: 6
   We review the basics of nomenclature and see that "hexa" = 6 and "cyano" = CN^-. Since the charge on Pd is 4+ which is given in the nomenclature by the roman numerals, we can calculate the total charge on the complex is 2-
   Formula: [Pd(CN)₆]²⁻
3. "dicholor" = 2 chloride ligands, since chloride is a monodentate ligand, we have a total of 2 bonds to the central metal
   Coordination Number: 2
   We review the basics of nomenclature and see that "di" = 2 and "chloro" = Cl^-. Since the charge on Au is always 1+, we can calculate the total charge on the complex is 1-
   Formula: [AuCl₂]⁻
4. "diamine" = 2 ammonia ligands and "dichloro" = 2 chloride ligands, since both ammonia and chloride ligands are monodentant, we have a total of 4 bonds to the central metal
   Coordination Number: 4
   We review the basics of nomenclature and see that "di" = 2, "chloro" = Cl^- and "amine" = NH₃. Since the charge on Pt is 2+ which is given in the nomenclature by the roman numerals, we can calculate that the total charge is 0 so the complex is neutral.
   Formula: [Pt(NH₃)₂Cl₂]
5. "diamine" = 2 ammonia ligands and "tetrachloro" = 4 chloride ligands since both ammonia and chloride ligands are monodentant, we have a total of 6 bonds to the central metal
   Coordination Number: 6
   We review the basics of nomenclature and see that "di" = 2, "amine" = NH₃, "tetra" = 4 and "chloro" = Cl^-. Since the charge on the central metal, Cr, is 3+ which is given in the nomenclature by the roman numerals, we can calculate that the total charge of the complex is 1-. The "potassium" at the front of the nomenclature indicates that it is the corresponding cation to this anionic complex.
   Formula: K[Cr(NH₃)₂Cl₄]
6. Both of the metal complexes have "hexa" monodentate ligands which means both have coordination numbers of 6.
   Coordination Number: 6
   We review the basics of nomenclature and see that "hexa" = 6, "amine" = NH₃, and "cyano" = CN^-. Since the charge on Cr is 3+ and Co is 3+ which is given in the nomenclature by the roman numerals, we find that these complexes charges balance out.
   Formula: [Co(NH₃)₆][Cr(CN)₆]
7. "dibromo" = 2 bromide ligands "bis(ethylenediamine)" = 2 (en) ligands; Bromide is a monodentate ligand while (en) is a bidentate ligand. Therefore; we have a coordination number of 6.
   Coordination Number: 6
   We review the basics of nomenclature and see that "di" = 2, "bromo" = Br^-, "bis" = 2, and "ethyllenediamine" = en. Since the charge on Co is 3+ which is given in the nomenclature by the roman numerals. Nitrate is the corresponding anion to this cationic complex.
   Formula: [Co(en)₂Br₂]NO₃

Q12.3.14

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

S12.3.14

Given: Experimental data (concentrations of A and rate), reaction equation

Asked For: rate equation, rate constant

Strategy:

A. Using the experimental data, we can compare the effects of changing [A] on the rate of reaction by relating ratios of [A] to ratios of rates

\[ \frac{2.66 \times 10^{-2}}{1.33 \times 10^{-2}} = 2\] and \[ \frac{1.52 \times 10^{-6}}{3.8 \times 10^{-7}} = 4\]

B. From this we know that doubling the concentration of A will result in quadrupling the rate of reaction. The order of this reaction is 2.

C. We can now write the rate equation since we know the order:

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

D. By plugging in one set of experimental data into our rate equation we can solve for the rate constant, k:

\[3.8 \times 10^{-7} = k \times (1.33 \times 10^{-2})^{2}\]

\[k = \frac{3.8 \times 10^{-7}}{1.769 \times 10^{-4}}\]

\[k= .00215 M^{-1}s^{-1}\]

Q12.6.6

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?

S12.6.6

Given: Reactions, rate equations

Asked To: Identify which equations could be both the elementary and overall reaction

Strategy:

We know that for a reaction to be considered both the elementary and overall reaction, the coefficients of the balanced equation must correspond with the exponents of the rate equation.

Equations B, D and E meet this criteria.

Q21.4.18

The isotope \[_{38}^{90}Sr\] 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.

S21.4.18

Given: masses, time

Asked For: half life

Strategy:

A. Relate the concentration or Sr to the rate constant k through the following equation:

\[k=-\frac{1}{t}ln\frac{[Sr]}{[Sr_0]}\]

\[k=-\frac{1}{10}ln\frac{.393}{.5}\]

\[k=.0241\]

B. Since this reaction is unimolecular, t_1/2 is related to k by

\[k=\frac{.693}{t_\frac{1}{2}}\]

\[.0241=\frac{.693}{t_\frac{1}{2}}\]

\[t_\frac{1}{2} = 28.78 years\]

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?

S20.3.6

Yes. Immerging both electrodes in the same beaker decreases the resistance of the system which increases the accuracy of the measured potential.

Q20.5.17

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?

S20.5.17

The standard reduction potential values for Ca²⁺ is -2.87 V. The standard reduction potential for Na+ , Na+(aq)+e- → Na(s) is equal to -2.71 V. Balancing the two equations gives 2 electrons being transferred.

/[E^o_cell = E_cathode - E_anode /] and since Calcium ion has the more negative reduction potential it is the cathode.

Eocell = -2.71-(-2.87) =0.16 V

Therefore use the the equation to calculate Gibbs free energy.

Change in G= -nFE^o_cell

change in G= -(2) x (96,485 C)x (0.16V)

= -30.9 KJ/ mol

Q20.5.16

For each reaction, calculate E°_cell and then determine ΔG°. Indicate whether each reaction is spontaneous.

1. 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)
2. K₂S₂O₆(aq) + I₂(s) → 2KI(aq) + K₂SO₄(aq)
3. Sn(s) + CuSO₄(aq) → Cu(s) + SnSO₄(aq)

S20.5.16

Given: overall equations

Asked For: E°_cell , ΔG°, and spontaneity

Strategy:

To solve for E°_cell of Equation 1 we will use the following strategy and then apply the same steps to the Equations 2 & 3:

A. Derive the oxidation and reduction half reactions

Oxidation: Na(s) → Na⁺(aq) +e^-

Reduction: 2H₂O(l) +2e^- → OH^-(aq) + H₂(g)

B. Use E°_cell =E°_reduction + E°oxidation ,and values from a table of standard reduction potentials

1: E°_cell =-.828 + -(-2.174) = 1.346 V

Repeating this 2-step process for equations 2 & 3 we get:

2: E°_cell = .786 V

3: E°_cell = .187 V

To solve for ΔG° of Equation 1 we will use the following strategy and then apply the same steps to the Equations 2 & 3:

A. Write out both half reaction equations to determine n, how many electrons are transferred

Oxidation: 2(Na(s) → Na⁺(aq) +e^- )

Reduction: 2H₂O(l) +2e^- → OH^-(aq) + H₂(g)

n= 2 electrons

B. Use the equation ΔG°=-nFE°_cell and Faraday's constant (F = Faraday's constant = 96,485 C/mol)

ΔG°=-(2)(96,485)(1.346)

1: ΔG°=-259737.62 J = -260 kJ

Repeating this 2-step process for equations 2 & 3 we get:

2: ΔG°= 303348.8 J = 303 kJ

3: ΔG°= 36085.4 J = 36.1 kJ

To determine spontaneity we look at the sign of ΔG°

1: ΔG°=-260 kJ <0; Spontaneous

2: ΔG°= 303 kJ > 0; Not spontaneous

3: ΔG°= 36.1 > 0; Not Spontaneous