Extra Credit 48

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Q17.7.2

What mass of each product is produced in each of the electrolytic cells of Q17.7.1 if a total charge of 3.33 × 10⁵ C passes through each cell? Assume the voltage is sufficient to perform the reduction.

CaCl₂
LiH
AlCl₃
CrBr₃

Given: the charge passes through each cell

Asked for: mass of each product that is produced

Strategy:

A. Calculate the moles of electrons passed through the cell using Faraday's Law.

Q = n×F

Q = total charge

n = moles of electrons

F = Faraday's constant

B. Determine the decomposition reaction of the molecule to find the products that are produced in each of the electrolytic cells. Use the anode and cathode reactions of those products then determine the amount of electrons transferred.

C. Calculate the mass of each product given the moles of electrons passed through the cell, moles of electrons transferred in the reaction per mole of each product and the molar mass of the products.

S17.7.2

A. Q = n×F

3.33×10⁵ col = n×96485 C/mol

n = 3.451 mol e⁻

1. CaCl₂

Decomposition	CaCl₂ → Ca + Cl₂
Cathode	Cl₂+ 2e⁻⇌ 2Cl⁻
Anode	Ca ⇌ Ca²⁺+ 2e⁻

\(3.451 mol e^-\times \dfrac{1molCl_2}{2mole^-}\times \dfrac{70.9g}{1molCl_2}=122.3g \text{ of } Cl_2\)

\(3.451 mol e^-\times \dfrac{1molCa}{2mole^-}\times \dfrac{40.08g}{1molCa}=69.2g \text{ of } Ca\)

2. LiH

Decomposition	2LiH → 2Li + H₂
Cathode	2H⁺+ 2e⁻⇌ H₂
Anode	2Li ⇌ 2Li⁺+ 2e⁻

\(3.451 mol e^-\times \dfrac{1molH_2}{2mole^-}\times \dfrac{2.0158g}{1molH_2}=3.5g \text{ of } H_2\)

\(3.451 mol e^-\times \dfrac{2molLi}{2mole^-}\times \dfrac{6.94g}{1molLi}=23.9g \text{ of } Li\)

3. AlCl₃

Decomposition	2AlCl₃ → 2Al + 3Cl₂
Cathode	3Cl₂ + 6e⁻ ⇌ 6Cl⁻
Anode	2Al ⇌ 2Al³⁺+ 6e⁻

\(3.451 mol e^-\times \dfrac{3mol Cl_2}{6mole^-}\times \dfrac{70.906g}{1mol Cl_2}=122.4g \text{ of } Cl_2\)

\(3.451 mol e^-\times \dfrac{2molAl}{6mole^-}\times \dfrac{26.982g}{1mol Al}=31.0g \text{ of } Al\)

4. CrBr₃

Decomposition	2CrBr₃→ 2Cr + 3Br₂
Cathode	3Br₂ + 6e⁻ ⇌ 6Br^-
Anode	2Cr ⇌ 2Cr³⁺ + 6e⁻

\(3.451 mol e^-\times \dfrac{3molBr_2}{6mole^-}\times \dfrac{159.808g}{1molBr_2}=275.7g \text{ of } Br_2\)

\(3.451 mol e^-\times \dfrac{2molCr}{6mole^-}\times \dfrac{51.996g}{1molCr}=59.8g \text{ of } Cr\)

(edited)

Q12.3.11

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:

[C₂H₅OH] (M)	4.4 × 10⁻²	3.3 × 10⁻²	2.2 × 10⁻²
Rate (mol/L/h)	2.0 × 10⁻²	2.0 × 10⁻²	2.0 × 10⁻²

Determine the rate equation, the rate constant, and the overall order for this reaction.

Given: the concentrations of [C₂H₅OH] and the rates of the reactions

Asked for: rate equation, the rate constant, and the overall order for this reaction

Strategy:

A. Determine the overall order for this reaction by looking at how concentration changes with the rate.

B. Determine the rate equation given the overall rate order.

C. Determine the rate constant given the rate equation, concentrations of [C₂H₅OH] and the rates of the reactions.

S12.3.11

A. The rate stays constant as concentration changes. Rate and concentration are independent of each other, therefore it's zero order.

B. The rate equation with x as the reaction order of these reactions is k[C₂H₅OH]^x. Because it is zero order, rate = k.

C. Substitute the rate from the chart above into the rate equation. The rate constant is then found to be k = 2.0 × 10⁻² mol/L/h.

(edited)

Q12.6.3

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:

step 1:	Cl + CO → COCl
step 2:	COCl + Cl₂→ COCl₂ + Cl

A. Write the overall reaction equation.

B. Identify any reaction intermediates.

C. Identify any intermediates.

Given: steps of the reaction

Asked For: overall reaction equation, reaction intermediates and intermediates

Strategy:

A. Add the steps together to and cancel molecules that are present on both sides to determine the overall reaction equation.

B. Determine any molecules that appear as a product in one step and as a reactant in the next step. These molecules were canceled out and should not appear in the overall reaction as determined in A.

C. Determine all the molecules that were canceled out in A along with the ones from B.

S12.6.3

A. Cl + CO + ~~COCl~~ + Cl₂→ ~~COCl~~ + COCl₂+ Cl

Overall reaction equation: CO + Cl₂→ COCl₂

B. COCl is a reaction intermediate because it is produced in the first step then used up in the second step.

C. Cl and COCl are intermediates because they are canceled out when determining the overall reaction.

(edited)

Q21.4.15

²³⁹Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the ²³⁹Pu present today will be present in 1000 y?

Given: half-life of atom

Asked for: fraction of atom left after a certain amount of years

Strategy:

A. Find the value of k, the radioactive decay constant.

B. Determine fraction of the atom left after 1000 years, given the radioactive decay constant.

S21.4.15

A. Because radioactive decay is first order, the equation to determine k is,

\(k=\dfrac{ln2}{t}\)

t = the half-life of the atom

\(k=\dfrac{ln2}{24000y}\)

k = 2.888x10^-5y^-1

B. With the value of k, use the first order integrated rate law to determine the fraction left of ²³⁹Pu.

ln(x) = -kt x = fraction of atom left; t = certain amount of years

ln(x) = -(2.888x10^-5y^-1) x 1000y

ln(x) = -0.02888

e^ln(x) = e^-0.02888

x = 0.9715 = 97.15%

(edited)

Q20.3.3

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

S20.3.3

Galvanic Cells VS Electrolytic Cells
Galvanic Cell	Electrolytic Cell
Converts chemical energy into electrical energy The redox reaction involved is *spontaneous* and is responsible for the production of electrical energy The two half-cells are set up in different containers. They are connected via a salt bridge or porous partition The anode is the negative electrode and the cathode is the positive electrode The electrons are supplied by the species being oxidized. They move from the anode to the cathode in the external circuit.	Converts electrical energy into chemical energy The redox reaction involved in non-spontaneous and energy needs to be supplied to initiate the reaction Both electrodes are places in the same container in the solution of molten electrolyte The anode is the positive electrode and the cathode is the negative electrode The external battery supplies the electrons. They enter through the cathode and come out through the anode

(https://chem.libretexts.org/Core/Ana...trolytic_cells )

As stated by the table above, the Galvanic Cell generates electricity by converting chemical energy into electrical energy.

(edited)

Q20.5.14

Calculate the pH of this cell constructed with the following half reactions when the potential is 0 at 25 °C

MnO₄^-+ 8H⁺+ 5e⁻→ Mn²⁺+ 4H₂O

Au³⁺+ 3e⁻→ Au(s)

Under this condition, the concentrations of other species in the cell are:

0.36 M: MnO₄^-
0.004 M: Au³⁺
0.001 M: Mn²⁺

Given: half reactions when the potential is 0 at 25 °C and the concentrations of the species in the cell

Asked for: pH of the cell

Strategy:

A. Determine cathode and anode reaction.

B. Balance the electrons in the two equations and add equations together cancelling out the electrons.

C. Use the Nernst equation to calculate the concentration of H⁺.

D. Determine the pH of the cell given the concentration of H⁺.

S20.5.14

A. The higher reduction potential of the two reactions is the cathode

MnO₄^-+ 8H⁺+ 5e⁻→ Mn²⁺+ 4H₂O reduction potential = +1.51

Au³⁺+ 3e⁻→ Au(s) reduction potential = +1.40

The MnO₄^- reaction equation in the cathode and the Au³⁺ reaction equation in the anode.

B. 3 x (MnO₄^-+ 8H⁺+ 5e⁻→ Mn²⁺+ 4H₂O)

5 x (Au(s)→ Au³⁺+ 3e⁻)

---------------------------------------------------------

3MnO₄^-+ 24H⁺+ 15e⁻→ 3Mn²⁺+ 12H₂O

5Au(s)→ 5Au³⁺+ 15e⁻

---------------------------------------------------------

3MnO₄^-+ 24H⁺+ 5Au(s) → 3Mn²⁺+ 12H₂O + 5Au³⁺

C. The Nernst equation is \(E_{cell}=E_{cell}^o-\dfrac{RT}{nF}lnQ\)

E_cell= 0

\(E_{cell}^o=E_{cathode}-E_{anode}\) 1.51V - 1.40V = 0.11V

R = 8.3145 J x mol^- x k^-

T = 25°C+273=298 K

n = number of electrons transferred = 15

F = 96485 C/mol

\(Q=\dfrac{\text [products]}{\text [reactants]}\)

\(0=0.11V-\dfrac{298\times 8.3145}{15\times 96485}ln(\dfrac{[0.001]^3\times [0.004]^5}{[0.36]^3\times [H^+]^{24}}\))

[H⁺]= 0.01043

D. pH = −log(H⁺)

=−log(0.01043)

= 1.98

(edited)

Q24.6.9

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.

S24.6.9

V²⁺: [Ar] 3d³

Fe²⁺:[Ar] 3d⁶

Zn²⁺: [Ar] 3d¹⁰

V²⁺, Fe²⁺, and Zn²⁺ are all transition metals that are in the 4^thperiod and same d-block. The size of the atoms in this d-block are determined by the 4s electrons because the 4s electrons are further away from the nucleus due to the shielding effect.

When the 4s electrons are removed like for V²⁺, Fe²⁺, and Zn²⁺, there is less shielding and so the ionic radii of these metals are almost the same.

(edited)

Q14.4.7

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:

Experiment	[Benzoyl Peroxide]₀ (M)	Initial Rate (M/s)
1	1.00	2.22 × 10⁻⁴
2	0.70	1.64 × 10⁻⁴
3	0.50	1.12 × 10⁻⁴
4	0.25	0.59 × 10⁻⁴

What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction?

Given: initial concentrations and rates

Asked for: reaction order and rate law

Strategy:

A. Determine the overall order for this reaction by looking at the relationship between concentration and rate from one experiment to the next.

B. Determine rate law given the overall order of the reaction.

S14.4.7

rate=k[A]^x

A. Can can look at the relationship between any of the experiments. Comepare experiment 1 and 3:

2.22*10^-4M/s=k*1M^x

1.12*10^-4M/s=k*0.5M^x

x=0.98 which is close to 1

k=2.22*10^-⁴s^-1

B. Without the overall rate order determined in A the rate law is

Including that the reaction is first order is

rate = 2.22*10^-⁴s^-1[Benzoyl Peroxide]¹

(edited)

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