Extra Credit 3

Edits by Reece Schweibold in blue! Also added the questions to the set of answers as part of edit.

Q17.1.3

For each of the following balanced half-reactions, determine whether an oxidation or reduction is occurring.

1. \(\ce{Fe^3+ + 3e- ⟶ Fe}\)
2. \(\ce{Cr ⟶ Cr^3+ + 3e-}\)
3. \(\ce{MnO4^2- ⟶ MnO4- + e-}\)
4. \(\ce{Li+ + e- ⟶ Li}\)

S17.1.3

• Oxidation and Reduction:

Remember:

1. Oxidation is the loss of electrons (electrons are on reactants side of the equation)
2. Reduction is the gain of electrons (electrons are on the product side of the equation)
3. All equations are balanced, so don't worry about balancing

a) Fe³⁺+ 3e^- → Fe

• Since electrons are on the reactants side, its being reduced

b) Cr→ Cr³⁺+3e^-

• Since electrons are on the product side, its being oxidized

c) MnO^2-₄ ⇒ MnO_{4 +}e^-

• Since electrons are on the product side, its being oxidized

d) Li⁺→Li+e^-

• Since electrons are on the reactants side, its being reduced

Q19.1.1

Write the electron configurations for each of the following elements:

1. Sc
2. Ti
3. Cr
4. Fe
5. Ru

S19.1.1

• Remember:

S can hold a max of 2 electrons

P can hold a max of 6 electrons

D can hold a max of 10 electrons

F can hold a max of 14 electrons

• Order of Filling:

- S orbitals are filled first ("s" orbital of the row the element falls on is filled)

- D orbitals are filled second ("d" orbital minus 1 is filled after "s")

i.e: Cr falls on row 4. 4s and 3d are filled

- F orbitals are filled last

- S electrons are removed first

a) Sc

• Sc has 21 electrons
• The Noble Gas associated with Sc is Ar
• Remaining electrons are Number of Electrons Sc - Number of Electrons Ar. In this case, we have 3
• Sc is in the fourth row, meaning we will fill the 4s orbital 3d orbital
• Remember the order and rules of filling orbitals (listed above)
• [Ar]4s²3d¹

b) Ti

• Ti has 22 electrons
• Noble Gas associated with Ti is Ar
• Remaining electrons are Number of Electrons Ti- Number of Electrons Ar. In this case, we have 4
• Ti appears in the fourth row, meaning we fill 4s and 3d
• Remember the order and rules of filling orbitals (listed above)
• [Ar]4s²3d²

c) Cr

• Cr has 24 electrons
• Noble Gas associated with Cr is Ar
• Remaining electrons are Number of Electrons Cr - Number of Electrons Ar. In this case, we have 6
• Cr appears in the fourth row, meaning we fill 4s and 3d
• Remember the order and rules of filling orbitals (listed above)
• [Ar]4s²3d⁴

d) Fe

• Fe has 26 electrons
• Noble Gas associated with Fe is Ar
• Remaining electrons are Number of Electrons Fe - Number of Electrons Ar. In this case, we have 8
• Fe appears in the fourth row, meaning we fill 4s and 3d
• Remember the order and rules of filling orbitals (listed above)
• [Ar]4s²3d⁶

e) Ru

• Ru has 44 electrons
• Noble Gas associated with Ru is Kr
• Remaining electrons are Number of Electrons Ru - Number of Electrons Kr. In this case, we have 8
• Ru appears in the fifth row, meaning we fill 5s and 4d
• Remember the order and rules of filling orbitals (listed above)
• [Ar]5s²4d⁶

Q19.2.3

Give the coordination number for each metal ion in the following compounds:

1. [Co(CO₃)₃]³⁻ (note that CO₃²⁻ is bidentate in this complex)
2. [Cu(NH₃)₄]²⁺
3. [Co(NH₃)₄Br₂]₂(SO₄)₃
4. [Pt(NH₃)₄][PtCl₄]
5. [Cr(en)₃](NO₃)₃
6. [Pd(NH₃)₂Br₂] (square planar)
7. K₃[Cu(Cl)₅]
8. [Zn(NH₃)₂Cl₂]

S19.2.3

• Coordination Number is the number of "bites" on the central metal
• Ligands outside brackets ARE NOT considered
• The MOST common polydentate ligands are "en" with 2, EDTA with 6, and "ox" with 2
• The remaining ligands are assumed to be monodentate (unless otherwise stated)

a) [Co(CO₃)₃]^3-

1. Count the number of monodentate ligands INSIDE the brackets.
2. Count the number of bidentate ligands INSIDE the brackets. 3 CO₃ molecules.
3. Multiply 2 (number of bites) by 3 (number of CO₃ molecules)
4. Coordination Number = 6

b) [Cu(NH₃)₄]²⁺

1. Count the number of monodentate ligands INSIDE the brackets. 4 NH₃ molecules.
2. Count the number of bidentate ligands INSIDE the brackets.
3. Multiply 1 (number of bites) by 4 (number of NH₃) molecules.
4. Coordination Number = 4

c) [Co(NH₃)₄Br₂]₂(SO₄)₃

1. Count the number of monodentate ligands INSIDE the brackets. NH₃ and Br

• The subscript outside the brackets, 2, tells us to multiply the number of NH₃, Br, and Co by 2

2. Count the number of bidentate ligands INSIDE the brackets.

3. Multiply 1 (number of bites of NH₃) by 8 (4 * 2) + 1 (number of bites of Br) by 4 (2 * 2)

4. Coordination Number = 8 + 4 = 12

d) [Pt(NH₃)₄][PtCl₄]

The first compound [Pt(NH₃)₄]:

1. Count the number of monodentate ligands INSIDE the brackets. NH₃

2. Count the number of bidentate ligands INSIDE the brackets.

3. Multiply 1 (number of bites of NH₃) by 4

4. Coordination Number = 4

The second compound [PtCl₄]:

1. Count the number of monodentate ligands INSIDE the brackets. 4 Cl^-

2. Count the number of bidentate ligands INSIDE the brackets.

3. Multiply 1 (number of bites of Cl) by 4 Cl molecules

4. Coordination Number = 4

e) [Cr(en)₃](NO₃)₃

1. Count the number of monodentate ligands INSIDE the brackets.

2. Count the number of bidentate ligands INSIDE the brackets. (en)

3. Multiply 2 (number of bites of en) by 3 (number of en molecules)

4. Coordination Number = 6

f) [Pd(NH₃)₂Br₂]

1. Count the number of monodentate ligands INSIDE the brackets. 2 NH₃ and 2 Br

2. Count the number of bidentate ligands INSIDE the brackets.

3. Multiply 1 (number of bites of NH₃) by 2 (number of NH₃ molecules) + Multiply 1 (number of bites of Br) by 2 (number of Br molecules)

4. Coordination Number = 4

g) K₃[Cu(Cl)₅]

1. Count the number of monodentate ligands INSIDE the brackets. 5 Cl

2. Count the number of bidentate ligands INSIDE the brackets.

3. Multiply 1 (number of bites of Cl) by 5 (number of Cl molecules)

4. Coordination Number = 5

h) [Zn(NH₃)₂Cl₂]

1. Count the number of monodentate ligands INSIDE the brackets. 2 NH₃ and 2 Cl

2. Count the number of bidentate ligands INSIDE the brackets.

3. Multiply 1 (number of bites of Cl) by 2 (number of Cl molecules) + Multiply 1 (number of bites of NH₃) by 2 (number of NH₃ molecules)

4. Coordination Number = 4

Q12.3.15

Nitrogen(II) oxide reacts with chlorine according to the equation:

\(\displaystyle 2NO(g)+Cl2(g)\rightarrow2NOCl\)

The following initial rates of reaction have been observed for certain reactant concentrations:

What is the rate equation that describes the rate’s dependence on the concentrations of NO and Cl₂? What is the rate constant? What are the orders with respect to each reactant?

S12.3.15

• To write the rate law, we must first determine the order of each compound

Determine the order:

1. Use the 2^nd row compound or element to determine the 1^st compound's order [Cl₂]
2. Find when the concentration of the second compound is equal [Cl₂] = .50M in row 1 and 2
3. Use the same rows used above, but for the 1^st compound [NO] = 1.0M and .5M in row 1
4. Use equation: [Concentration_1x]^x[Concentration_1y]^y/[Concentration_2x]^x[Concentration_2y]^y=Rate₁/Rate₂.

• You can place 1M of [NO] in either Concentration_1x or Concentration_2x. Just make sure you place ALL contents from the same row in its corresponding spot. (i.e. Row 1 material in each location marked with 1)
• [1.0M]^x[.50M]^y/[.5M]^x[.5M]^y = [4.56]₂/[1.14]₁

5. The concentrations of the "y" compound should cancel

[1.0M]^2x/[.5M]^1x = [4.56]₂/[1.14]₁

6. Determine an exponent "x", that would make x = rate

[2.0M]^x= 4

7. The exponent that makes the concentration equal the rate is the order of the first compound

x = 2 ( [2M]²=4

• Common Exponents are: 1/2, 0, 1, 2
• Repeat the process, solving for y, instead of x

8. [1.0M]^x[1.0M]^y/[1.0M]^x[.5M]^y = [9.12]₂/4.56]₁

9. [1.0M]^y/[.5M]^y = 2

10. [2M]^y= 2

11. y = 1

12. Rate is written as [Compound A]^x{Compound B]^y

• rate of reaction = k[NO]²[Cl₂]¹

13. To find the rate constant, plug in values of [NO],[Cl₂], and rate of reaction from the same row

• Using values from the first row:
• 1.14 = k[.50]²[.5M]^y
• Solve for k
• 1.14/[.5M]²[.5M]^y= k
• k = 9.12

14. The orders with respect to each reactant is the exponent of each reactant

• [NO] = 2
• [Cl₂] = 1

Q12.6.7

Write the rate equation for each of the following elementary reactions:

1. \(O_{3}\rightarrow O_{2} + O\)

2. \(O_{3}+Cl\rightarrow O_{2}+ClO\)

3. \(ClO +O\rightarrow Cl+O_{2}\)

4. \(O_{3}+NO\rightarrow NO_{2}+O_{2}\)

5. \(NO_{2}+O\rightarrow NO+O_{2}\)

S12.6.7

Rate Equations for elementary steps involve the reactants at 1^st order. Rate reactions don't depend on products.

a) rate = k[O₃]

b) rate = k[O₃][Cl]

c) rate = k[ClO][O]

d) rate = k[O₃][NO]

e) rate = k[NO₂][O]

The rate equation involves the rate constant k and the reactants. The exponents for these equations are determined experimentally.

Q21.4.19

Technetium-99 is often used for assessing heart, liver, and lung damage because certain technetium compounds are absorbed by damaged tissues. It has a half-life of 6.0 h. Calculate the rate constant for the decay of \(\displaystyle _{43}^{99}\textrm{Tc}\)

S21.4.19:

• Remember, the equation for half-life is N = N₀e^-kt
• N is the final concentration (.5 in this case)
• N₀ is initial concentration
• k is rate constant
• t is time (in any unit)

1. (.5) = 1e^-k(6)
2. To bring down the -kt and eliminate the e, we take the natural log (ln) of both sides

• ln (.5) = ln(e^-k(6))
• ln (.5) = -k(6)

3. Solve for k

• k = ln (.5)/-6
• k = .12 h^-1

- Its always in first order, so its 1/unit of time

Q20.3.7

Copper(II) sulfate forms a bright blue solution in water. If a piece of zinc metal is placed in a beaker of aqueous CuSO₄ solution, the blue color fades with time, the zinc strip begins to erode, and a black solid forms around the zinc strip. What is happening? Write half-reactions to show the chemical changes that are occurring. What will happen if a piece of copper metal is placed in a colorless aqueous solution of \(\displaystyle ZnCl_2\)?

S20.3.7

(answer left blank in phase I)

Copper(II) sulfate forms a bright blue solution in the presence of water due to the formation of the blue hexaaquacopper(II) ion, \(\displaystyle\left [CuH_2O)_6 \right ]^{2+}\). If a piece of zinc metal were to be placed in a beaker of aqueous \(\displaystyle CuSO_4\) solution, the blue color fades because this chemical reaction is a displacement reaction. Since zinc is more reactive than copper, it will displace the less reactive copper from its compound. The solution becomes lighter in color as the zinc ion displaces the copper ions. Since there will be less copper ions in the solution, the blue color will fade. This is exemplified by the half reactions of \(\displaystyle Zn\)(s) and \(\displaystyle Cu\)(aq):

oxidation reaction: \(\displaystyle Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}\)

reduction reaction: \(\displaystyle Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)\)

overall reaction: \(\displaystyle Zn(s)+CuSO_4(aq)\rightarrow ZnSO_4(s)+Cu(s)\)

Thus zinc will consolidate around the metal rod the form a black solid.

From looking at these half reactions, we can also tell that this chemical reaction is a redox reaction. In other words, there will be a transfer of electrons from a coupled oxidation and reduction reaction in which the oxidation reaction accounts for the loss of electrons and the reduction reaction accounts for the gaining of electrons. In this reaction, the zinc strip begins to erode because it is being oxidized. In addition the \(\displaystyle CuSO_4\) will become reduced and gain electrons.

If a piece of copper metal were to be placed in a colorless aqueous solution of \(\displaystyle ZnCl_2\), there would be no change in the solution. You can determine this by looking at the standard reduction potentials for the \(\displaystyle Zn^{2+}\) and the \(\displaystyle Cu^{2+}\) ions. According to the potentials, the reduction of \(\displaystyle Cu^{2+}\) has a greater standard reduction potential than the reduction of \(\displaystyle Zn^{2+}\)(aq) and it is unlikely for the copper to become oxidized because it is less willing to give up its electons.

Q20.5.18

In acidic solution, permanganate (MnO₄⁻) oxidizes Cl⁻ to chlorine gas, and MnO₄⁻ is reduced to Mn²⁺(aq).

1. Write the balanced chemical equation for this reaction.
2. Determine E°_cell.
3. Calculate the equilibrium constant.

S20.5.18

a. Write half reactions

Begin by writing the compounds. The compound that oxidized is the element that is reduced.

• MnO₄^- → Mn²⁺
• Cl^- → Cl₂

Next, start by balancing elements, excluding Hydrogen and Water, by multiplying by a coefficient needed to make the left side equal the right. In the first case, the Mn is balanced.

• 1 Mn on reactants and 1 Mn on the products

Notice in the second reaction, there are 2 Cl molecules on the products. We need 2 on the reactants. So we Multiply the left by 2

2Cl^- → Cl₂

Next, balance Oxygen molecules using H₂O and a coefficient to balance Oxygen.

• In the first equation, notice that the reactants has 4 Oxygen molecules, while the reactants has 0. We need to add 4 Oxygen molecules by adding 4H₂O to the products.

MnO₄^- ⇒ Mn²⁺ + 4H₂O

• In the second equation, neither the reactants nor the products have any oxygen molecules. Its balanced.

2Cl^- → Cl₂

Next, we need to balance the number of Hydrogen molecules by adding H⁺

• In the first equation, the products have 8 Hydrogen molecules (4 * 2), while the reactants have 0. So add 8 H⁺ molecules to the reactants.

MnO₄^- + 8H⁺ ⇒ Mn²⁺ + 4H₂O

• In the second equation, neither the reactants nor the products have any H. So, its balanced. NO₃^- + H⁺→

2Cl^- → Cl₂

Next, we need to balance the overall charges. To do so, we add electrons (e^-)

• In the first equation, the reactants have a +7 charge (-1 from the MnO₄^- and 8⁺ from the H⁺). The products have a 2+ overall charge ( +2 from Mn²⁺ and 0 from H₂O). To balance the reactants charge and products side, we need to add 5e^- to the reactants, giving an overall charge of 2+.

MnO₄^- + 8H⁺ + 5e^- ⇒ Mn²⁺ + 4H₂O

• In the second equation, the reactants has a charge of -2 (-1 from each Cl^-). The products side has a charge of 0. To balance the reactants and products charge, we must add two e^- to the products.

2Cl^- → Cl₂ + 2e^-

Finally we need to get the same number of e^- in both equations in order for them to cancel from the overall equation when you add the two equations.

• The first equation has 5 e^- on the reactants.
• The second equation has 2 e^- on the product side.
• We need 10 e^- in both equations to cancel the electrons. So we multiply the ENTIRE first equation by 2 and the ENTIRE second equation by 5

2(MnO₄^- + 8H⁺ + 5e^- ⇒ Mn²⁺ + 4H₂O)

5(Cl^- → Cl + e^-) = 5Cl^- → 5Cl + 5e^-

Finally, we add the two balanced equations (make sure the electrons cancel out; they shouldn't appear in the overall equation):

5Cl^- → 5Cl + 5e^-

+ 2MnO₄^- + 16H⁺ + 10e^- ⇒ 2Mn²⁺ + 8H₂O

Which yields:

5Cl^- + 2MnO₄^- + 16H⁺ ⇒ 2Mn²⁺ + 8H₂O + 5Cl

B) Find E^°_cell:

To find E^°_cell, we go to the Standard Reductions Table:

https://chem.libretexts.org/Core/Ana...tion_Potential

Remember:

1. We use the equation E^°_cell = E°_cathode - E^°_anode, where cathode is the the element or compound being reduced (electrons on the reactants side) and anode is the element or compound being oxidized (electrons on the product side). To do so, look at the half reactions above.
2. Notice that Cl^- → Cl₂ + 2e^- is the anode, since the electrons are on the product side. Refer to the table, to find that the reduction is +1.358V. (DO NOT SCALE this value by 5 or flip the sign)
3. Notice that MnO₄^- + 8H⁺ + 5e^- ⇒ Mn²⁺ + 4H₂O is the cathode since the electrons are on the reactants side. Refer to the table, to find that the reduction is +1.51V. (DO NOT SCALE this value by 2 or flip the sign)
4. Now plug in the values. Remember, the subtraction in the equation will take care of the sign flipping.
5. E^°_cell = 1.51 - 1.358 = +.152V

C) Find the equilibrium constant:

1. Since we have the value of E^°_cell that we calculated above, and dont have any other information, we want to use the equation E^°_cell = RT/nF * ln(K)
2. Isolate K:( E^°_cell * nF)/RT = ln(K)
3. Raise both sides to the e power, to eliminate the ln.
4. Now, we have e^{(E°_cell * nF)/RT} = K
5. Plug in the values of E^°_cell, which was .152V, R = 8.3145, T = 298, F =96,485 , and n = 10. R and F are constants, F is the temperature, which we assume to be 298^°, since no other temperature was given, and n is the number of electrons in the overall balanced equations, which was 10.
6. Plugging these values in, we get 59.19