Extra Credit 3

Q17.1.3

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

a) \(Fe^{3+}+3e^{-}\rightarrow Fe\)

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

c) \(MnO_{4} ^{2-}\rightarrow MnO_{4}^{-}+e^{-}\)

d) \(Li^{+}+e^{-}\rightarrow Li\)

S17.1.3

To determine if a reaction is being oxidized or reduced, you must determine the oxidation state of the elements on either side of the equation. If the oxidation state is increased, then the reaction was oxidized. If the oxidation state decreased, then the reaction was reduced. Alternatively, you can examine the number of electrons for the elements on either side of the reaction. If the number of electrons is reduced, then the reaction is an oxidation reaction. If the number of electrons is increased, then the reaction is a reduction reaction. We can remember this by the common chemistry mnemonic: OIL RIG. Oxidation is losing [electrons], Reduction is gaining [electrons. The electrons on either side of the half-reactions keep the charges balanced.

a) In this reaction, the oxidation states are very easy to find. The oxidation state of Fe³⁺ is (+3). The oxidation state of Fe is (0) because it is in elemental form. The oxidation state decreases from 3 to 0, so reduction is occurring.

b) This is similar to part A. Cr is in its elemental form, so its oxidation state is (0). Cr³⁺ is an ion, and its oxidation state is (+3). In this case, the oxidation state increases from 0 to 3, so oxidation is occurring.

c) For this reaction, you must find the oxidation states of Manganese in the permanganate compounds.

The first compound is MnO₄^2- . You can set up a simple equation to find the oxidation state of Manganese.

\(X + 4(-2) = -2\)

X represents the oxidation state of Mn. It is added to 4(-2) because there are 4 oxygen molecules, and oxygen always has an oxidation state of (-2). The equation is equal to -2 because the overall compound has a charge of -2. Then, you solve for X. X is equal to 6, so the oxidation state of Manganese in MnO₄^2- is (+6).

You repeat this process for MnO₄^- .

\(X + 4(-2) = -1\)

This time, X = +7, so the oxidation state of Mn in MnO₄^- is (+7).

The oxidation state for Manganese increased from 6 to 7, so oxidation occurred.

d) This problem is similar to the first 2. Li⁺ has an oxidation state of (+1), and Li has an oxidation state of (0). The oxidation state decreased from 1 to 0, so the reaction is going through reduction.

My solution matches! Great. I don't know why this guy did your phase 2 (I Mattew Grabert was assigned to do this phase 2). My edits are in Blue.

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

We are going to use the noble gas configurations to write the electron configurations of each of these elements. This means locating the noble gas on the row before the element in question and using the noble gas' electron configuration as a jumping off point. We can refresh our memory on the essential concept of electron configurations.

1) Sc

Scandium is #21 on the periodic table. The noble gas on the row before it is #18, Argon (Ar).

So, we must determine the number of electrons in Scandium that come after Argon, and their configuration.

Scandium is in the third column. This means it is in the D-orbital block. By examining the periodic table, we see that there are 2 electrons that come before Scandium, in the S block.

To write the electron configuration, we start with [Ar], and then denote the orbital and how many electrons are in each orbital.

For Scandium, the electrons that come after Argon are in the 4s orbital and the 3d orbital. There are 2 electrons in the 4s orbital and 1 electron in the 3d orbital.

Then, we put all this information together into the electron configuration.

Sc: [Ar]4s²d¹

2) Ti

Titanium is #22 on the periodic table, right after Scandium. We go through the same process to write the electron configuration.

Argon is the closest noble gas. Titanium has electrons in the 4s orbital and the 3d orbital. There are 2 electrons in the 4s orbital, and 2 electrons in the 3d orbital.

Putting this information together, we get this:

Ti: [Ar]4s²3d²

3) Cr

Chromium is #24 on the periodic table. Argon is the closest noble gas. By observing the periodic table, we see that there are 2 electrons in the 4s orbital, and 4 electrons in the 3d orbital.

It would be easy to assume that the electron configuration is written as:

Cr: [Ar]4s²3d⁴

However, this is incorrect. The structure will always want to produce the lowest energy possible for the system. This system leaves the s orbital full, but the d orbital has one empty orbital. So, to decrease the energy level, one electron from the s orbital is promoted to the d orbital, so that all of the outer orbitals are half full. Therefore, the electron configuration is as follows:

Cr: [Ar]4s¹3d⁵

4) Fe

Iron is #26 on the periodic table. Argon is the closest noble gas. According to the periodic table, there are 2 electrons in the 4s orbital, and 6 electrons in the 3d orbital. Following our usual rules, the electron configuration is:

Fe: [Ar]4s²3d⁶

5) Ru

Ruthenium is #44 on the periodic table. The closest noble gas is Krypton. From the periodic table, we can see that the 5s orbital has 2 electrons, and the 4d orbital has 6. We would expect the electron configuration to be:

Ru: [Kr]5s²4d⁶

However, Ruthenium has an anomalous configuration. Instead of the usual configuration, Ruthenium has a configuration somewhat similar to our chromium example. One electron from the s orbital is promoted to the d orbital. Now, our electron configuration looks like this:

Ru: [Kr]5s¹4d⁷

Well done. My solution matches.

Q19.2.3

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

1. [Co(CO₃)₃]^3- (note that CO₃^2- 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

You can determine a compound's coordination number based on how many ligands are bound to the central atom.

1) In this compound, Cobalt is the central atom, and it has 3 CO₃^2- molecules attached to it. However, CO₃^2- is a bidentate ligand, which means it binds to the central atom in two places rather than one. This means that the coordination number of [Co(CO₃)₃]^3- is 6. A coordination number of 6 means that the structure is most likely octahedral.

2) In this compound, Copper is the central atom. 4 ammonia molecules are attached to it. This means the coordination number is 4, and the structure is likely tetrahedral.

3) For this compound, we can ignore the (SO₄)₃ because it is not bound to the central atom. The central atom is cobalt, and it has 4 ammonia molecules and 2 bromine molecules bound to it. The coordination number is 6.

4) There are two compounds here, indicated by the brackets. The central atom for both is platinum. One of them has 4 ammonia molecules attached, and the other has 4 chlorine atoms attached. Both complexes have a coordination number of 4.

5) We can ignore (NO₃)₃ for this compound. The central atom is Chromium. There are 3 ethylenediamine molecules attached to the chromium. Ethylenediamine is a bidentate ligand, so the coordination number is 6.

6) Palladium is the central atom. 2 ammonia molecules and 2 bromine atoms are bound to the palladium atom. The coordination number is 4.

7) We can ignore the K₃ structure. Copper is the central atom, and there are 5 chlorine molecules attached to it. The coordination number is 5, so the structure is either trigonal bipyramidal or square pyramidal.

8) In this compound, zinc is the central atom. There are 2 ammonia molecules and 2 chlorine atoms attached. This means that the coordination number is 4.

Awesome. I agree. I edited a few grammatical issues, otherwise, great explanation.

Q12.3.15

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

\(2NO(g) + Cl_{2}(g) \rightarrow 2NOCl(g)\)

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?

S12.3.15

For the general equation,

\(aA + bB \rightarrow cC + dD\)

The rate can be written as

\(rate = k[A]^{m}[B]^{n}\) where k is the rate constant, and m and n are the reaction orders.

For our equation

\(2NO(g) + Cl_{2}(g) \rightarrow 2NOCl(g)\)

the \(rate = k[NO]^{m}[Cl_{2}]^{n}\)

Now, we need to find the reaction orders. Reaction orders can only be found through experimental values. We can compare two reactions where one of the reactants has the same concentration for both trials, and solve for the reaction order.

\(\frac{rate_{1}}{rate_{2}}=\frac{[NO]_{1}^{m}[Cl_{2}]_{1}^{n}}{[NO]_{2}^{m}[Cl_{2}]_{2}^{n}}\)

We can use the data in the table provided. If we plug in the values for rows 1 and 2, we see that the values for the concentration of Cl will cancel, leaving just the rates and the concentrations of NO.

\(\frac{1.14}{4.56}=\frac{[0.5]^{m}}{[1.0]^{m}}\)

We can now solve for m, and we find that m =2. This means that the reaction order for [NO] is 2.

Now we must find the value of n. To do so, we can use the same equation but with the values from rows 2 and 3. This time, the concentration of NO will cancel out.

\(\frac{4.56}{9.12}=\frac{[0.5]^{n}}{[1.0]^{n}}\)

When we solve for n, we find that n = 1. This means that the reaction order for [Cl₂] is 1.

We are one step closer to finishing our rate equation.

\(rate = k[NO]^{2}[Cl_{2}]\)

Finally, we can solve for the rate constant. To do this, we can use one of the trials of the experiment, and plug in the values for the rate, and concentrations of reactants, then solve for k.

\(1.14 mol/L/h = k[0.5 mol/L]^{2}[0.5mol/L]\)

\(k=9.12L^{2}mol^{-2}h^{-1}\)

So, our final rate equation is:

\(rate = (9.12 L^{2} mol^{-2}h^{-1})[NO]^{2}[Cl_{2}]\)

*A common mistake is forgetting units. Make sure to track your units throughout the process of determining your rate constant. Be careful because the units will change relative to the reaction order.

My solution matches.

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

The rate law of a reaction can be found using a rate constant (which is found experimentally), and the initial concentrations of reactants.

A general solution for the equation

\(aA + bB \rightarrow cC + dD\)

is \(rate = k[A]^{m}[B]^{n}\) where m and n are reaction orders.

However, reaction orders are found experimentally, and since we do not have experimental data for these reactions, we can disregard that part of the equation.

To find the rate laws, all we have to do is plug the reactants into the rate formula. This is only due to the case that these are elementary reactions. Further reading on elementary reactions can be found on Libre Texts.

1) \(rate = k[O_{3}]\)

2) \(rate = k[O_{3}][Cl]\)

3) \(rate = k[ClO][O]\)

4) \(rate = k[O_{3}][NO]\)

5) \(rate = k[NO_{2}][O]\)

My solution matches.

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 technetium-99.

S21.4.19

To solve for the rate constant for the decay of technetium-99, we can use first order kinetics.

First order kinetics tells us that

\(t_{1/2} = \frac{ln(2)}{k}\)

We can rearrange this to be

\(k = \frac{ln(2)}{t_{1/2}}\)

When we plug in 6.0 hours for the value of the half life, we find that k = .12 h^-1.

My solution matches. Good job being concise and clear.

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 \(CuSO_{4}\) solution, the blue color fades with time, the zinc strip begins to erode, and 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 \(ZnCl_{2}\)?

S20.3.7

When you place a strip of zinc metal into a beaker of aqueous \(CuSO_{4}\) solution, the copper in the solution becomes displaced by the zinc. The solution changes color as it changes from a \(CuSO_{4}\) solution to a \(ZnSO_{4}\) solution. The black solid forming around the zinc strip is copper consolidating around the metal rod. Electrons leave zinc and combine with copper.

The reaction observed in this process is

\(CuSO_{4}(aq) +Zn\rightarrow ZnSO_{4}(aq)+Cu(s)\)

To find the half reactions, you first have to find the oxidation states of Cu and Zn, then balance the equations with electrons.

\(Cu^{2+}(aq) + 2e^{-}\rightarrow Cu(s)\)

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

If you place a strip of zinc into a colorless aqueous solution of \(ZnCl_{2}\), then a similar type of reaction will occur. The copper will displace the zinc in solution, making the solution turn blue. Zinc will consolidate around the metal rod.

\(ZnCl_{2}(aq)+Cu\rightarrow CuCl_{2}(aq)+Zn(s)\)

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

\(Cu(s)\rightarrow Cu^{2+}(aq)+2e^{-}\)

My solution matches.

Q20.5.18

In acidic solution, permanganate \(MnO_{4}^{-}\) oxidizes \(Cl^{-}\) to chlorine gas, and \(MnO_{4}^{-}\) is reduced to \(Mn^{2+}(aq)\).

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

S20.5.18

1. We know this reaction is a redox reaction, because we are told that \(MnO_{4}^{-}\) oxidizes \(Cl^{-}\). We can start balancing the equation by first writing out the equation with the given information.

\(MnO_{4}^{-}(aq)+Cl^{-}(aq)\rightarrow Mn^{2+}(aq)+Cl_{2}(g)\)

Then write the half reactions, and balance with electrons, \(H_{2}O\), and \(H^{+}\) as necessary.

\(MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\rightarrow Mn^{2+}(aq)+4H_{2}O(l)\)

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

Now, you balance the number of electrons in both reactions, and cancel any overlapping compounds on either side of the equation.

The end result is the balanced reaction.

\(2MnO_{4}^{-}(aq)+10Cl^{-}(aq)+16H^{+}(aq)\rightarrow 2Mn^{2+}(aq)+5Cl_{2}(g)+8H_{2}O(l)\)

2. To determine the E°cell of the reaction, you must find the standard potentials for the reduction half reaction and the oxidation half reaction.

\(E^{o}_{cell}= E_{reduction}^{o}+E_{oxidation}^{o}\)

We know from the balanced equation that the oxidation reaction is

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

and the reduction reaction is

\(MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\rightarrow Mn^{2+}(aq)+4H_{2}O(l)\)

By using the activity series, we can find the reduction potentials.

We find that the reduction potential for \(Cl_{2}\) is 1.36 volts. Because the reduction potential is given for the equation, \(Cl_{2} +2e^{-}\rightarrow 2Cl^{-}\), we must flip the sign of the value.

The value of the reduction for \(MnO_{4}^{-}\) is 1.49 volts.

When we plug these values into the equation, we get

\(E_{cell}^{o}=(1.49 volts)+(-1.36volts)=.13volts\)

3. Using the value we just found for the \(E_{cell}^{o}\), we can find the value of the equilibrium constant by using the Nernst Equation.

\(E_{cell}^{o}=(\frac{RT}{nF})ln(k)\)

R = 8.31 J/molK, the universal gas constant

F = 96485 C/mol e^-, Faraday's constant

T = temperature in Kelvin

n = number of moles of electrons transferred

We can rearrange the equation to read

\(ln(k)=\frac{nFE^{o}}{RT}\)

Then, we can plug in our values and solve for k. (Assume the temperature is 298K.)

\(ln(k)=\frac{(10)(96485)(.13)}{(8.31)(298)}\)

k= 9.94.

My solution matches.