Extra Credit 3

Q 17.1.3:

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

a. Fe³⁺ + 3e^- → Fe

Solution: Oxidation reactions occur when there is a loss of electrons and reduction reactions occur when there is a gain of electrons. This can be memorized through the mnemonic, OIL RIG (oxidation is loss, reduction is gain). Because the reactant, Fe³⁺ is gaining 3 electrons, that indicates that a reduction reaction is occurring.

b. Cr → Cr³⁺ + 3e^-

Solution: A simple way to tell if an oxidation or reduction reaction is happening is to look at what side of the equations the electrons are on (but note that this is only easily applicable to equations dealing with a single molecule that is simply changing oxidation states or charges). If the electrons are on the left, they are being gained and it is a reduction reaction. If the electrons are on the right, they are being lost and it is an oxidation reaction. Through this, we can see that electrons are being lost in this example and this is an oxidation reaction.

c. MnO₄^2- → MnO₄^- + e^-

Solution: Here we see the permanganate ion is experiencing a change in its overall charge. The charge goes from 2- to 1-. This means that electrons are being lost because the charge is increasing and becoming more positive. As we read earlier, a loss of electrons means an oxidation reaction is taking place here.

d. Li⁺ + e^- → Li

Solution: In this equation, we see that the oxidation state of lithium decreases from +1 to 0. A decrease in the oxidation state of an element or molecule signifies that electrons were gained (the negative charge of the electron accounts for the charge/oxidation state decrease of the lithium), and if electrons were gained, this must be a reduction reaction.

Q 19.1.1:

Write the electron configurations for each of the following elements.

Background Information on electron configurations: The electron configuration of an element is a short-hand way of finding out how many electrons are in each orbital of the element. The shell is denoted by the number coefficient in front (i.e. 1, 2, 3,...), the subshell is denoted by the lowercase letter following the shell number (i.e. s, d, p, f,...), and the orbital numbers vary from subshell to subshell, each orbital holding two electrons. The s subshell holds 1 orbital, the d subshell holds 3 orbitals, the p subshell holds 5 orbitals, and the f subshell holds 7 orbitals. The first shell can hold the s subshell, the second shell can hold the s and p subshells, the third shell holds s, p, and d subshells, and the fourth shell (and every shell beyond this) holds s, p, d, and f subshells.

a. Sc

Solution: The first step is to find the number of electrons for Sc. By looking at a periodic table, we find that Sc has 21 electrons. When doing electron configurations, the lower energy orbitals are always filled first (which are those of s, then d, then p, then f). Next, fill the orbitals with 2 electrons each until 21 electrons have been accounted for in the correct order (be sure to pay attention to which shells contain which subshells). The configuration for Sc is 1s²2s²2p⁶3s²3p⁶4s²3d¹. If you add up the exponents (each telling how many total electrons are in the orbitals), you find that we have correctly accounted for all 21 electrons of Sc. Usually, we abbreviate long electron configurations like this by putting the noble gas directly before it in brackets and following it with the next orbitals and their electrons. The noble gas before Sc is Ar (this means Ar has the same electron configuration as Sc up until 3p⁶), so we would write: [Ar]4s²3d¹.

b. Ti

Solution: For this problem we will use the same process as stated above. First, we see that Ti has 22 electrons and the prior noble gas is still Ar. Following the same procedure as above, we find that the full configuration for Ti would be 1s²2s²2p⁶3s²3p⁶4s²3d², but if we used the noble gas abbreviating system, we would shorten the electron configuration to be [Ar]4s²3d².

c. Cr

Solution: The electron configuration of Cr is similar but has a slight exception to the normal rule we use to fill the orbitals. Usually, we completely fill one subshell's orbitals before moving up to the next (i.e., s should be completely filled before starting to fill p). In Cr, there are 24 electrons and the prior noble gas is Ar, so the configuration should look like this, based on the way we have been finding the others: 1s²2s²2p⁶3s²3p⁶4s²3d⁴. However, in the transition metals (which Cr is a part of), it is more stable to keep one electron in the s orbital and put 5 electrons in the d orbitals. Due to this slight variation in stability, the actual configuration would look like this: 1s²2s²2p⁶3s²3p⁶4s¹3d⁵, and the shortened version would be [Ar]4s¹3d⁵.

d. Fe

Solution: Fe follows the normal rule for electron configurations, so we would use the earlier procedure. There are 26 electrons in Fe and the prior noble gas is Ar, still. Following the earlier procedure, we would get an electron configuration of 1s²2s²2p⁶3s²3p⁶4s²3d⁶, and the abbreviated configuration would be [Ar]4s²3d⁶.

e. Ru

Solution: For Ru, it is similar to the Cr example, but it is on the next row of the periodic table so the noble gas will change in the configuration. Ru has 44 electrons and the prior noble gas is Kr. This configuration should be 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d⁶, however this is in the same situation as Cr, in which it is more stable to take one electron from the s subshell and add it to the next d orbital. The actual configuration is 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹4d⁷, and the shorthand notation of it is [Kr]5s¹4d⁷.

Q 19.2.3:

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

a. [Co(CO₃)₃]^3- (note that Co₃^2- is bidentate in this complex)

Solution: The coordination number for a metal ion is defined as the total number of points of attachment to the central atom (this is usually the number of ligands if they are all monodentate, but often there are polydentate ligands that will bite/attach in more than one place, so we are counting the total number of "bites" or attachments). There are three carbonato ligands (CO₃) attached to the central Co. Usually, carbonato ligands are monodentate (only attaching once), but the instructions noted that in this specific complex the ligands should be considered bidentate, meaning they "bite twice". This gives a coordination number of 6 (if 3 ligands attach twice, two times three = 6).

b. [Cu(NH₃)₄]²⁺

Solution: The central Cu has four ammine ligands (NH₃) attached to it, and they are monodentate, so the coordination number is 4.

c. [Co(NH₃)₄Br₂]₂(SO₄)₃

Solution: The central Co has four ammine ligands and two bromo ligands (Br) attached to it. Both ammine and bromo ligands are monodentate, so adding the four ammine and two bromo gives a total of 6 attachments for the coordination number. There are three more sulfate ions (SO₄) outside the complex ion, but because they are not directly attached to the central metal, they are not counted in the coordination number.

d. [Pt(NH₃)₄][PtCl₄]

Solution: There are two different complex ions in this system. The first Pt has four ammine (monodentate) ligands attached, so the coordination number of the first complex ion is 4. The second complex ion has four chloro (Cl) ligands attached, which are also monodentate, meaning that the coordination number for this second complex ion is also 4.

e. [Cr(en)₃](NO₃)₃

Solution: The central Cr has three ethylenediamine (en) ligands attached to it. Though we may have assumed that this had a coordination number of 3, that would be incorrect. Each ethylenediamine is a bidentate ligand, meaning it attaches to the central atom two times. If there are three ligands, each attaching twice, multiplying 2 by 3 gives a total coordination number of 6. Like a similar problem above, this has more ligands outside the complex ion, in this case they are three nitrates. Again, these are not directly bound to the central metal and should not be counted in the coordination number.

f. [Pd(NH₃)₂Br₂] (square planar)

Solution: The central Pd has two ammine ligands (monodentate) and two bromo ligands (monodentate) attached to it, so adding these together gives this a coordination number of 4.

g. K₃[Cu(Cl)₅]

Solution: The central Cu has five chloro ligands attached to it, which are all monodentate. This gives a total coordination number of 5.

h. [Zn(NH₃)₂Cl₂]

Solution: The central Zn has two ammine ligands and two chloro ligands attached to it. As we have learned from previous questions, these ligands are all monodentate, so the total coordination number is the sum of the two ligands, which equals 4.

Q 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:

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?

Solution: The rate law will take the form:

rate = k[NO]^m[Cl₂]ⁿ

where m and n represent the orders of their respective components. The variable, k, represents the rate coefficient for the reaction, and this will differ for each reaction.

• To determine m, we must find two rows in which [NO] varies and [Cl₂] stays constant. Then, write the ratios, dividing one rate equation by the other. We can use the first and second row to do this as it satisfies the concentration rules we are looking for. This ratio will leave us with 1.14/4.56=(k[0.5]^m[0.5]ⁿ)/(k[1]^m[0.5]ⁿ). After simplifying, we have 1/4=([0.5]^m) so m=2
• To determine n, we must find two rows in which [Cl₂] varies and [NO] stays constant. We can use the second and third rows for this. The ratio will be 4.56/9.12=(k[1]^m[0.5]ⁿ)/(k[1]^m[1.0]ⁿ). After simplifying, we have 1/2=([0.5]ⁿ) so n=1

Overall, our rate equation is now

rate=k[NO]²[Cl₂]

To find the rate constant, k, plug data in from one row/experiment and solve for k:

1.14mol/L·h=k[0.5mol/L]²[0.5mol/L]

This gives a k value of 9.12 L²mol^-2h^-1.

NO is second order and Cl₂ is first order for this reaction.

Q 12.6.7:

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

a. O₃ → O₂ + O

Solution: Usually, we cannot simply use the balanced equation to tell us the orders of a reaction. However, these are exceptions as the instructions say that all of these reactions are elementary. When a reaction is elementary, we can use the balanced equation to create the rate law. We simply use the reactants as the different components, and their coefficients would give their respective orders. For this reaction, the only reactant is O₃ and it has a coefficient of one, so our rate law would look like this: rate=k[O₃]

b. O₃ + Cl → O₂ + ClO

Solution: This is another elementary reaction, and none of the coefficients are greater than one, so each reactant will be in the first order. The two reactants are O₃ and Cl, so our rate law would look like this: rate=k[O₃][Cl]

c. ClO + O → Cl + O₂

Solution: This elementary reaction has two reactants, ClO and O. Both of them will be in the first order based on their coefficients. The rate law will look like this: rate=k[ClO][O]

d. O₃ + NO → NO₂ + O₂

Solution: Again, this is an elementary reaction. The reactants are O₃ and NO, both being first order. The rate law should then look like this: rate=k[O₃][NO]

e. NO₂ + O → NO + O₂

Solution: This last reaction is still elementary so we can proceed with the same steps as the rest of the reactions. Both reactants, NO₂ and O, are in the first order, so the rate law should look like this: rate=k[NO₂][O]

Q 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 Technetium-99.

Solution: For this problem, we will assume that this is a first-order reaction, even though it is not explicitly stated. We can assume this because first-order reactions are the only ones (out of zero, first, and second order) that we can calculate the rate constant from half-life without needed any concentrations present. This tells us that the half-life of a first-order reaction is independent of concentration. The equation to find the half-life of a first-order reaction is

t_1/2 = (ln 2)/k

where k is the reaction constant, which is what we are looking for. To solve for k, we can do some algebraic manipulation of the equation to isolate it. By multiplying both sides by k and then dividing both sides by t_1/2 we end up with k = (ln 2)/t_1/2 . From here, we can plug in the information given in the problem, which is that the half-life (or t_1/2) is 6 hours. Now, we have k = (ln 2)/6 h = 0.12 h^-1

Q 20.3.7:

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 ZnCl₂?

Solution: What is happening in this reaction is that the zinc metal placed in the copper sulfate solution will begin to dissolve and form an aqueous zinc solution (this is representative of the zinc strip eroding). While this happens, the aqueous copper will start to form solid copper (this is seen in the black solid that started forming around the zinc strip). The half-reactions that occur are solid zinc becoming aqueous ( Zn(s) → Zn²⁺(aq) + 2e^- ) and aqueous copper forming into a solid ( Cu²⁺(aq) + 2e^- → Cu(s) ). If a piece of copper metal was placed in an aqueous solution of ZnCl₂, the most likely outcome is that the aqueous zinc will start to form into solid zinc, and the solid copper will dissolve into an aqueous CuCl₂ solution.

Q 20.5.18:

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

a. Write the balanced chemical equation for this reaction.

Solution: To balance this redox reaction, there a several steps we must go through:

1. First, write the half reactions. The instructions tell us that chlorine was oxidized and permanganate was reduced, so the oxidation half-reaction is Cl^-(aq) → Cl₂(g) and the reduction half reaction is MnO₄^-(aq) → Mn²⁺(aq)
2. Though it is not crucial to this problem, you can also identify the oxidation states of each component. Cl^- is -1, Cl₂ is 0, Mn²⁺ is +2, the Mn of MnO₄^- is +7, and each O is -2
3. Next, balance all elements that are not oxygen or hydrogen. The only thing that needs this is the oxidation half-reaction, which will become 2Cl^-(aq) → Cl₂(g)
4. After that, balance oxygen with H₂O and then balance hydrogen with H⁺. The only reaction that needs this is the reduction half-reaction. The left side has 4 oxygens and the right has none, so add 4 waters to the right side. Now the right side has 8 hydrogens and the left has none, so add 8H⁺ to the left to balance this. The end result of this step will be 8H⁺+ MnO₄^-(aq) → Mn²⁺(aq) + 4H₂O(l)
5. Now we must balance the charges for both half-reactions. The oxidation half-reaction has an overall charge of -2 on the left and 0 on the right, so we must add 2 electrons to the right, leaving us with 2Cl^-(aq) → Cl₂(g) + 2e^-. The reduction half reaction has a charge of +7 on the left and +2 on the right, so we must add 5 electrons to the left side, leaving us with 5e^- + 8H⁺+ MnO₄^-(aq) → Mn²⁺(aq) + 4H₂O(l)
6. After this, we have to balance the electrons for both equations. We want them both to include the same number of electrons so they can later cancel each other out. The first has 2 electrons and the second has 5 electrons. This means we have to multiply the whole first equation by 5 and the second by 2. This will leave us with 10 electrons in each equation. The resulting half-reactions will be 10Cl^-(aq) → 5Cl₂(g) + 10e^- for the first equation and 10e^- + 16H⁺+ 2MnO₄^-(aq) → 2Mn²⁺(aq) + 8H₂O(l) for the second equation.
7. Now all that is left is to combine the two half-reactions and cancel out any repeating terms. The parts cancelling out should just be the 10 electrons on either side. Overall, our balanced equation for this reaction should look like this:

16H⁺+ 2MnO₄^-(aq) + 10Cl^-(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 5Cl₂(g)

b. Determine E°_cell.

Solution: To find E°_cell we must follow the equation: E°_cell = E°_cathode - E°_anode

• Each E value represents a Standard Reduction Potential (SRP), or the tendency for a chemical species to be reduced. The cathode is where reduction occurs and the anode is where oxidation occurs, so we must find those SRP values for our specific chemical reaction. The SRP for our oxidation reaction is +1.35827 V and the SRP for our reduction reaction is +1.507 V.
• Now we will plug these values into the equation to find E°_cell:

E°_cell = 1.507 - 1.35827 = 0.14873 V

c. Calculate the equilibrium constant.

Solution: To find the equilibrium constant, K, we must use an equation that relates our known value of E°_cell to the K we wish to find. The equation that will accomplish this is E°_cell = [(0.0592)/n]logK where n is the number of electrons transferred (in our case, this is 10. This can be found from the balanced equation). Now we plug in our values and solve for K:

0.14873 = (0.00592)logK

logK = 25.12331081

K = 1.328x10²⁵