Extra Credit 48

17.7.2

What mass of each product is produced in each of the electrolytic cells of Q17.7.1 if a total charge of \(3.33 \times 10^{5} C\) passes through each cell? Assume the voltage is sufficient to perform the reduction.

NOTE:

Faraday's Constant (F) = \(96485 \frac{coulomb}{mol}\)

Use \(n_e = \frac{Q}{F}\) to find the mol of electrons. This will be needed in order to determine the mass of each product using stoichiometry.

\(n_e = mol\,of\,electrons\)

\(Q = charge (C, coulombs)\)

Since \(I \times t\) is given as \(3.33 \times 10^{5} C\), plug it into \(n_e = \frac{It}{F}\).

\(3.33\ast 10^{5}\frac{col}{F}=3.4513\) mol e^- passing through the cell

a) \(CaCl_2\)

\(CaCl_{2}\rightarrow Ca+Cl_{2}\)

Determine the redox reactions.

i) \(Ca^{2+}+2e^{-}\rightarrow Ca\)

ii) \(Cl_{2}+2e^{-}\rightarrow 2Cl^{-}\)

(2 electrons transferred)

Overall: \(Ca + Cl_{2} \rightarrow Ca^{2+} + 2Cl^{-}\)

With the mol of electrons calculated earlier, use stoichiometry to determine mass of each product.

i) \(3.4513 mol\,e^- \times \frac{1 mol Ca}{2 mol\,e^-}\) = 1.72565 mol Ca

\(1.72565 mol\,Ca \times \frac{40.078 g }{ 1 mol Ca}\) = 69.16 g Ca

ii) \(3.4513 mol\,e^- \times \frac{1 mol Cl_2}{2 mol\,e^-}\) = 3.4513 mol \(Cl_2\)

\(3.4513 mol\,Cl \times \frac{70.9 g Cl_2}{1 mol Cl_2}\) = 122.349 g \(Cl_2\)

b) \(LiH\)

Determine the redox reactions.

i) \(Li^{+}+e^{-}\rightarrow Li(s)\)

ii) \(2H^{+}+2e^{-}\rightarrow H_{2}(g)\)

Overall: \(2Li + 2H^{+} \rightarrow 2Li^{+} + H_2\)

With the mol of electrons calculated earlier, use stoichiometry to determine mass of each product.

i) \(3.4513 mol\,e^- \times \frac{2 mol Li}{ 2 mol\,e^-}\) = 3.4513 mol Li

\(3.4513 mol\,Li \times \frac{6.94 g\,Li}{1 mol\,Li}\) = 23.95 g Li

ii) \(3.4513 mol\,e^- \times \frac{1 mol H_2}{2 mol\,e^-}\) = 1.72565 mol \(H_2\)

\(1.72565 mol\,H_2 \times \frac{2.0158 g\,H_2}{1 mol\,H_2}\) = 3.4789 g \(H_2\)

c) \(AlCl_3\)

Determine the redox reactions.

i) \((Al^{3+}+3e^{-}\rightarrow Al)\times 2\)

ii) \((Cl^{-}\rightarrow Cl_{2}+2e^{-})\times 3\)

\(2Al^{3+}+3Cl^{^{-}}\rightarrow 2Al+3Cl_{2}\)

(6 electrons transferred)

With the mol of electrons calculated earlier, use stoichiometry to determine mass of each product.

i) \(3.4513 mol\,e^- \times \frac{2 mol\,Al}{6 mol\,e^-}\) = 1.1504 mol Al

\(1.1504 mol\,Al \times \frac{26.9815 g\,Al}{1 mol \,Al}\) = 31.04 Al

ii) \(3.4513 mol\,e^- \times \frac{3 mol\,Cl_2}{6 mol \,e^-}\) = 1.72565 mol \(Cl_2\)

\(1.72565 mol\,Cl_2 \times \frac{70.9 g Cl_2}{ 1 mol \,Cl_2}\) = 122.348 g \(Cl_2\)

d) \(CrBr_3\)

Determine the redox reactions.

i) \(Br^{^{-}} + e^{-} \rightarrow Br\)

ii) \(Cr \rightarrow Cr^{3+} + 3e^{-}\)

\(3Br^{-}+Cr \rightarrow 3Br+Cr^{3+}\)

(3 electrons transferred)

With the mol of electrons calculated earlier, use stoichiometry to determine mass of each product.

i) \(3.4513 mol\,e^- \times \frac{3 mol}{6 mol \,e^-}\) = 1.72565 mol \(Br_2\)

\(1.72565 mol\,Br_2 \times \frac{159.808 g Br_2}{1 mol Br_2}\) = 275.77 g \(Br_2\)

ii) \(3.4513 mol\,e^- \times \frac{2 mol Cr}{6 mol \,e^-}\) = 3.4513 mol Cr

\(3.4513 mol\,Cr \times \frac{51.9961 g Cr}{1 mol Cr}\) = 59.8 g Cr

12.3.11

Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by 25–30%. Women metabolize alcohol a little more slowly than men:

Using the data, determine the rate equation and the overall order for this reaction to find the rate constant.

\(rate=k[C_{2}H_{5}OH]\)

*The rate, \(2.0 \times 10^{-2}\), stays constant despite the change in concentration, therefore it's zero order because \([C_{2}H_{5}OH]^{0}=1\). There is no correlation between the concentration and reaction rate.

Plug in a set of data into the rate equation in order to determine the rate constant.

\(rate=k\)

\(2.0\times 10^{-2}=k[4.4\times 10^{-2}]^{0}\)

\(k=2.0\times 10^{-2} M/h\)

12.6.3

Phosgene, COCl₂, one of the poison gases used during World War I, is formed from chlorine and carbon monoxide. The mechanism is thought to proceed by:

1. Write the overall reaction equation.
2. Identify any reaction intermediates.
3. Identify any catalysts.

a) Add step 1 and step 2 and balance the chemical equation:

Cl + CO + COCl + Cl₂ → COCl + COCl₂ + Cl

CO + Cl₂ → COCl₂

b) Reaction intermediates: COCl (it appears as a product in the first step and as a reactant in the subsequent step)

c) Any catalysts: Cl (since a catalyst speeds up a reaction, it is added as a reactant but returns as a product unchanged; therefore it will appear as a reactant and as a product later.)

21.4.15

\(^{239}_{94}Pu\) is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the \(^{239}_{94}Pu\) present today will be present in 1000 y?

NOTE: Radioactive decay is first order.

Use the equation: \(A = A_{0} \times 2^{-\frac{t}{t_{1/2}}}\)

\(t = time\)

\(t_{1/2} = half-life\)

\(A_{0} = initial amount/concentration\)

Given:

\(t_{1/2}\) of \(^{239}_{94}Pu\) = 24,000 years

t = 1000 years

\(A_{0} = 1\) since we assume that Pu has not started to decay yet and we are starting from t=0

Plug into equation to find what fraction of \(^{239}_{94}Pu\) present today will be present in 1000 years.

# of half-lives = \(\frac{t}{t_{1/2}}\) =\(\frac{1000 years}{24000 years}\)

= 0.041666

\(A = 1 \times 2^{-0.041666}\) = 0.973

\(0.973 \times 100\% = {97.3}\%\)

20.3.3

What is the difference between a galvanic cell and an electrolytic cell? Which would you use to generate electricity?

Although both converts energy in using electrodes, solid metals that are used to provide an electrical connection between two parts of the system, a galvanic cell drives a spontaneous reaction whereas an electrolytic cell drives a non-spontaneous reaction. Furthermore, a galvanic cell converts the energy released by the spontaneous reaction to make electricity while an electrolytic cell converts electrical energy into chemical energy. The anode and cathode for a galvanic cell and electrolytic cell is also different. For a galvanic cell, the anode will be negative while the cathode is positive. For an electrolytic cell, the anode will be positive while the cathode will be negative. If it is positive, that means oxidation takes place there and if it is negative that means reduction is taking place.

Therefore, you would use a galvanic cell to generate electricity.

20.5.14

Explain why the sum of the potentials for the half-reactions Sn²⁺(aq) + 2e⁻ → Sn(s) and Sn⁴⁺(aq) + 2e⁻ → Sn²⁺(aq) does not equal the potential for the reaction Sn⁴⁺(aq) + 4e⁻→ Sn(s). What is the net cell potential? Compare the values of ΔG° for the sum of the potentials and the actual net cell potential.

You cannot sum the cell potentials for 1/2 reactions because the sum of the cell potentials equals 2 times the cell potential of the third equation. This equation relating the sum of the two potentials and the net cell potential can be derived from relating the equations of each reaction's \(\Delta G^{\circ}\) values. The sum of the first two reaction's \(\Delta G^{\circ}\) values should equal the \(\Delta G^{\circ}\) value using the actual net cell potential.

\(-E_{2+\rightarrow (s)}^{\circ}\left ( 2 \right )F=\Delta G_{2+\rightarrow (s)}^{\circ}\)

\(-E_{4+\rightarrow 2+}^{\circ}\left ( 2 \right )F=\Delta G_{4+\rightarrow 2+}^{\circ}\)

\(-E_{4+\rightarrow (s)}^{\circ}\left ( 4 \right )F=\Delta G_{4+\rightarrow (s)}^{\circ}\)

\(\Delta G_{2+\rightarrow (s)}^{\circ}+\Delta G_{4+\rightarrow 2+}^{\circ}=\Delta G_{4+\rightarrow (s)}^{\circ}\)

\(-E_{2+\rightarrow (s)}^{\circ}\left ( 2 \right )F+-E_{4+\rightarrow 2+}^{\circ}\left ( 2 \right )F=-E_{4+\rightarrow (s)}^{\circ}\left ( 4 \right )F\)

\((E_{2+\rightarrow (s)}^{\circ}+E_{4+\rightarrow 2+}^{\circ})= 2(E_{4+\rightarrow (s)}^{\circ})\)

Net Cell Potential:

\(E_{4+\rightarrow (s)}^{\circ}=\frac{-0.14+0.15}{2}\)

\(E_{4+\rightarrow (s)}^{\circ}=.005V\)

0.005V does not equal 0.01V.

\(\Delta G^{\circ} = -nFE\) where n = # of electrons, F = Faraday's constant (96485C), and E = cell potential.

The \(\Delta G^{\circ}\) of the summed potentials is \(\Delta G^{\circ} = (-4 mol\, e^{-})(96485\frac{C}{mol})(0.01V) = -4kJ\) while \(\Delta G^{\circ}\) of the actual net potential is \(\Delta G^{\circ} = (-4 mol\, e^{-})(96485\frac{C}{mol})(0.005V) = -2kJ\). This shows that the bigger the cell potential, the more negative \(\Delta G^{\circ}\) is which means it is also more spontaneous.

24.6.9

The ionic radii of V²⁺, Fe²⁺, and Zn²⁺ are all roughly the same (approximately 76 pm). Given their positions in the periodic table, explain why their ionic radii are so similar.

Note below is the trend of ionic radii on the periodic table.

This phenomenon is called the shielding effect in which an electron and the nucleus of the atom experience a decrease in attraction. The ionic radii of V²⁺ , Fe²⁺ , and Zn²⁺ are all similar because they are in the same d-block of the periodic table so they don't change significantly across a period. In addition for V²⁺, Fe²⁺, and Zn²⁺ , the 4s electrons are pushed further away from the nucleus as electrons fill the 3d orbital because of the shielding effect. Then the 4s electrons are removed from being pushed away and the shielding effect is not as effective as the shielding decreases. This results in keeping the ionic radii of the metals similar to each other.

14.4.7

Benzoyl peroxide is a medication used to treat acne. Its rate of thermal decomposition at several concentrations was determined experimentally, and the data were tabulated as follows:

What is the reaction order with respect to benzoyl peroxide (\(C_{14)H_{10}O_{4}\))? What is the rate law for this reaction?

\(rate=k[Benzoyl Peroxide]^{x}\)

\(2.22\times 10^{-4}=k[1]^{x}\)

\(1.12\times 10^{-4}=k[.5]^{x}\)

*From looking at the data, as the concentration doubles and the rate doubles, it is a first order reaction. Therefore, \(rate = k[Benzoyl Peroxide]\)

In knowing that this is a first order reaction, the rate constant (k) can be found.

\(2.22\times 10^{-4}=k[1]^{1}\)

\(k=2.2\times 10^{-4}s^{-1}\)