Extra Credit 3

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17.1.3

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

Fe³⁺+3e⁻⟶ Fe
Cr ⟶ Cr³⁺+3e⁻
MnO₄^2- ⟶ MnO₄^-+e⁻
Li⁺+e⁻⟶ Li

Answer

1. Reduction -- The charge of the reactant, Fe³⁺, is +3. The charge of the product, Fe, is 0. Reduction is when the element gains electrons, or becomes more negative. In this reaction Fe gains 3 electrons to go from a charge of +3 to 0.

2. Oxidation -- Reactant (Cr) charge: 0 Product (Cr³⁺) charge: +3 Oxidation is the loss of electrons, or when an atom becomes more positive. Cr had to lose electrons to go from a charge of 0 to a charge of +3.

3. Oxidation -- Reactant (MnO₄^2-) charge:-2 Product (MnO₄^-) charge:-1 Oxidation=loss of electrons. MnO₄had to lose electrons to go from a charge of -2 to -1.

4. Reduction -- Reactant (Li⁺) charge: +1 Product (Li) charge: 0 Li had to gain electrons to go from +1 to 0 in charge, therefore reduction occurred.

Remember, OIL RIG or Oxidation Is Loss and Reduction Is Gain. In a reduction reaction the element or compound will Gain electrons, and in an oxidation reaction it will Loose electrons.

19.1.1

Write the electron configurations for each of the following elements:

Answer

Procedure: When making an electron configuration for an element, you must start by looking at the periodic table and finding the noble gas in the period above the assigned element. This will put put in brackets and written first. Then you look at the period the assigned element is in. If the element is in the s category (the first two groups), then the next part of the electron configuration will be either s¹ or s² depending on if the element is in the or second column. The number that comes before s corresponds to which row the element is in. Because we are focusing on transition metals (the d block), the next step of this process is figuring out what number precedes d. This is based on which row the element is in, and then subtracting one. For example if the element is in the fifth row, then the number that precedes d in the electron configuration is 4. Finally you must find the superscript of d. This can be done by counting from the third group to wherever the assigned element is located.

1. Sc: [Ar]4s²3d¹ (or 1s²2s²2p⁶3s²3p⁶4s²3d¹) -- When looking at the periodic table, Ar is the last element before Sc to have all possible electron shells completely full, it is the noble gas before Sc, so in this abbreviated notation we put Ar in brackets. Then we see that Sc is in the 4th period and that it's 4s shell is full, too, so we write 4s². Finally we look at the 4th period transition elements (the 3ds) and see that Sc is the first element, and has one electron in the 3d shell, so we write 3d¹

2. Ti: [Ar]4s²3d²(1s²2s²2p⁶3s²3p⁶4s²3d²) -- This is very similar to part one, the only difference being the number of electrons in the 3d shell. Ti has two electrons in the 3d shell (as it is in group 4 and row 4), so we say that it is 3d².

3. Cr: [Ar]4s²3d⁴ (1s²2s²2p⁶3s²3p⁶4s²3d⁴) -- Follow the above procedure.

4. Fe: [Ar]4s²3d⁶ (1s²2s²2p⁶3s²3p⁶4s²3d⁶) -- Follow the above procedure.

5. Ru: [Kr]5s¹4d⁷ (1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹4d⁷) -- The above procedure only half works in this case because Ru has an anomalous configuration. One electron that we assume would go in the 5s orbital actually goes into the 4d orbital. Intuitively, this goes against Aufbau's principle, but it is due to the field splitting that occurs in the 4d orbital, changing the energy level of the orbital into two different levels (e_g and t_2g). Thus allowing one of the electrons from what is generally considered a lower energy orbital (5s) to slip into a "higher" energy (4d) orbital. This can also be done in order to have more stability. Another example would be Cu, which would have an electron configuration of [Ar]4s¹3d¹⁰or Cr with [Ar]4s¹3d⁵. This is done so that it will have a full or half full d orbital shell. Source: https://eic.rsc.org/feature/the-trou...000133.article

19.2.3

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

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

Answer

The first step is to locate the metal, this can easily be done by finding the first element within the brackets. The next step is to identify the ligands in the compound, figuring out if they are monodentate, bidentate, etc.. This will tell you how many times each ligand binds to the metal. The ligands that must be counted are the ones inside of the brackets, that will directly touch the metal in its structure. Then you must add up all ligand bonds to the metal and that is your coordination number.

1. [Co(CO₃)₃]³⁻-- Metal: Co Ligand(s): 3 CO₃s (bidentate) 3x2=6 Coordination Number: 6

2. [Cu(NH₃)₄]²⁺ -- Metal: Cu Ligand(s): 4 NH₃s (monodentate) 4x1=4 Coordination Number:4

3. [Co(NH₃)₄Br₂]₂(SO₄)₃ -- Metal: Co Ligand(s): 4 NH₃s (monodentate), 2 Br₂s (monodentate), SO₄ is not considered when finding c.n. as it is not directly bonded to the metal Coordination Number: 6

4. [Pt(NH₃)₄][PtCl₄] -- Metal: Pt (AND A SECOND COMPLEX--also Pt) Ligand(s): Complex 1=4 NH₃s (monodentate), complex 2=4 Cl s (monodentate) (4x1),(4x1)=4 Coordination Number: C1=4, C2=4

5. [Cr(en)₃](NO₃)₃ -- Metal: Cr Ligand(s): 3 en s (bidentate) Coordination Number: 6

6. [Pd(NH₃)₂Br₂] (square planar) -- Metal: Pd Ligand(s): 2 NH₃ (mondentate), 2 Br s (monodentate) Coordination Number: 4

7. K₃[Cu(Cl)₅] -- Metal: Cu Ligand(s): 5 Cl s (monodentate), K is outside the complex and therefore not counted Coordination Number: 5

8. [Zn(NH₃)₂Cl₂] -- Metal: Zn Ligand(s): 2 NH₃ (dentate), 2 Cl s (monodentate) Coordination Number: 4

12.3.15

Nitrogen(II) oxide reacts with chlorine according to the equation: 2NO(g)+Cl₂(g)⟶2NOCl(g)

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

Reaction	[NO] (mol/L¹)	[Cl₂] (mol/L)	Rate (mol/L/h)
a	0.50	0.50	1.14
b	1.00	0.50	4.56
c	1.00	1.00	9.12

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?

Answer

Rate equation: Rate Law:

First, you must solve for m, the order for the reactant [Cl₂]. You do this by using two reactions in which the initial values for [Cl₂] differ and [NO] are the same. In this case those reactions are b and c. This allows you to cancel out the initial values of [NO], in order to solve for m. The same can be done for [NO] to solve for n.

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Because m=1 for [Cl], this reactant is first order. n=2 for [NO], so it is second order. Adding these together, the overall reaction is third order. Therefore, the units for k will be L²*mol^-2*hr^-1. Now that you have m and n, you can plug any of the reaction values into the rate equation to solve for k.

TIP: If there was a reaction with three different parts, make sure to pick two of the reactants to be of the same value so that they will cancel out and leave just one order possibility. In order to avoid errors when calculating the rate constant, only use the information from one reaction, in this case, only use all of the information from a.

12.6.7

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

O₃(sunlight)→O₂+O
O₃+Cl ⟶ O₂+ClO
ClO+O ⟶ Cl+O₂
O₃+NO ⟶ NO₂+O₂
NO₂+O ⟶ NO+O₂

Answer

Procedure: Use the rate law Rate=k[A]^m[B]ⁿ for 2A + B --> A₂B . k is the reaction constant and depends on the reaction. Then you multiply k by the reactant concentrations, raised to whatever order that reactant is in any given equation. In this case, the order is given in the equation, as the question states that each reaction is an elementary reaction. This means that order can be found by looking at the balanced equation. If this were not an elementary reaction, the rate law could only be determined experimentally.

1. rate₁=k[O₃], m=1 because O₃'s coefficient is 1

2. rate₂=k[O₃][Cl]

3. rate₃=k[ClO][O]

4. rate₄=k[O₃][NO]

5.rate₅=k[NO₂][O]

Another Example: If this reaction was given: 2NO₂+O ⟶ NO+O_2,the rate of the reaction would be:

rate=K[NO₂]²[O]

this would also make the it a third order equation, since the addition of both coefficients (2+1 = 3)

21.4.19

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 $\ce{^{99}_{43}Rh^2+}$

Answer

You already know t_1/2, so in order to solve for k, you can use the first order equation (although the order is not stated, neither are any concentrations-we do know that without any initial concentrations only the first order half life equation is applicable):

\[t_{1/2}=6 hours\]

\[t_{1/2}=\dfrac{ln2}{k}\]

\[{k}=\dfrac{ln2}{t_{1/2}}\]

\[{k}=\dfrac{ln2}{(6 hours)}\]

\[{k}= 0.1155 \,hr^-\]

find k from 0.5 life.JPG

20.3.7

Copper(I) 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 ZnCl₂?

Answer

This is an example of an electrochemical reaction. Looking at the standard reduction potential table, we can see that Copper(I) is the oxidizing agent and being reduced, while Zinc is the reducing agent and being oxidized. Half reactions:

\[ Cu^{1+}(aq) + e^- \rightarrow Cu(s) \]

\[ Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-\]

\[Zn_{(s)} \; | \; Zn^{2+}_{(aq)} \;|| \; Cu^{1+}_{(aq)} \;|\; Cu_{(s)}\]

This means that the electrons are flowing from the zinc metal to the copper(I) ions in the solution (reducing them to copper metal), and this copper simultaneously goes onto the zinc (s) surface to plate it in a red/brown colored covering. Zinc is also eroding to form aqueous Zn²⁺.

If copper metal is placed in a solution of ZnCl₂, then the Zn could potentially plate the copper in an electrochemical reaction by the same process. However, the reaction would be non-spontaneous according to the equation:

\[E^o_{cell}=E^o_{cathode}-E^o_{Anode}\]

where the ZnCl₂ solution is the cathode (where reduction occurs) and the copper metal is the anode (where oxidation occurs).

E^o_anode= 0.52V

E^o_cathode= -0.763V

\[E^o_{cell}=E^o_{cathode}-E^o_{anode}=-0.763\; V -(0.52\; V)=-1.28 \; V\]

The change in energy is negative, therefore we can conclude that the reaction is non-spontaneous and is unlikely to occur at all.

20.5.18

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

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

Answer

1. First, determine the oxidation and reduction half reactions. Then balance the elements in each reaction (that aren't O or H). Next balance O by adding water molecules. After that balance the H by adding H⁺ ions. Then you can balance the overall charge of the half reaction by adding electrons. Finally, you may need to multiply each half reaction equation so that the number of electrons are equal in each. Then you can combine the two equations.

Reduction: