Extra Credit 23

Q19.7

Write and balance the net reaction using the two following half reactions. Find the Voltage for this reaction

\(\mathrm{BrO^-(aq) + H_2O(l) + 2e^- \rightarrow Br^-(aq) +2OH^-(aq)}\)

\(\mathrm{Mg(s) + 2OH^-(aq) \rightarrow Mg(OH)_2(s) +2e^-}\)

Q19.7 Ans

The half reactions for this problem have already been balanced with \(\mathrm{OH^-}\) ions and electrons. Therefore, we only have to find the net reaction.

A. First step would be to add the two half reactions and cancel out the common terms.

\(\mathrm{[BrO^-(aq) + H_2O(l) + 2e^- \rightarrow Br^-(aq) +2OH^-(aq)]+[Mg(s) + 2OH^-(aq) \rightarrow Mg(OH)_2(s) +2e^-]}\)

=

\(\mathrm{BrO^-(aq) + H_2O(l) + 2e^-+Mg(s) + 2OH^-(aq) \rightarrow Br^-(aq) +2OH^-(aq)+ Mg(OH)_2(s) +2e^-}\)

We can cancel out the 2 electrons on each side and 2 \(\mathrm{OH^-}\) thus resulting in

\(\mathrm{BrO^-(aq) + H_2O(l)+Mg(s) \rightarrow Br^-(aq) ++ Mg(OH)_2(s)}\)

as the final net reaction

B. To find the voltage we would have to determine which half reaction is the anode and the cathode and its Standard Reduction Potential values from the table.

Anode: \(\mathrm{Mg(s) + 2OH^-(aq) \rightarrow Mg(OH)_2(s) +2e^-}\) \(\mathrm{E^°_{red,anode}=-2.687}\)

because it is getting oxidized thus losing electrons.

Cathode: \(\mathrm{BrO^-(aq) + H_2O(l) + 2e^- \rightarrow Br^-(aq) +2OH^-(aq)}\) \(\mathrm{E^°_{red,cathode}=0.766}\)

because it is getting reduced thus gaining electrons.

Using the Formula: \(\mathrm{E^°_{cell}=E^°_{red,cathode}-E^°_{red,anode}}\)

\(\mathrm{E^°_{cell}=0.766-(-2.687)=3.453}\)

Q19.63B

1. \(\ce{Mg^2+}\)
2. \(\ce{Sn^4+}\)
3. \(\ce{Fe^3+}\)
4. \(\ce{Ni^2+}\)

Q19.63B Ans.

First and foremost, for all parts of the this problem we would have to convert the 75 mins into seconds.

\(\mathrm{(75mins)(\frac{60seconds}{1min})=4500seconds}\)

Then we can use

\(\mathrm{n_e=\frac{It}{f}}\)

where I=current and t= time in seconds and f is faraday's constant= 96500 C/mole

Using the formula we can calculate the moles of electrons transferred then convert that number to moles of each element in each individual problem.

\(\mathrm{\frac{(2.3)(4500)}{96500}=.11}\)moles of electrons transferred

A. \(\ce{Mg^{2+}}\)

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

So we now know for 2 electrons transferred results in 1 mole of Mg formed and we know that the Molar mass of Mg= to 24.31

\(\mathrm{Mass\:Mg = 0.11\:mol\:e^- \times \dfrac{1\:mol\:Mg^{2+}}{2\:mol \:e^-}\times\frac{24.31g}{1mol} = 1.34\:g\:Mg}\)

B. \(\ce{Sn^4+}\)

\(\mathrm{Sn^{4+}(aq)+4e^-\rightarrow Sn(s)}\)

So we now know for every 4 electrons transferred results in 1 mole of Sn formed and we know that the Molar mass of Sn=118.71

\(\mathrm{Mass\:Sn = 0.11\:mol\:e^- \times\dfrac{1\: mol\: Sn^{4+}}{4\: mol\: e^-}\times\frac{118.71g}{1 mol}= 3.26\: g\: Sn }\)

C. \(\ce{Fe^3+}\)

\(\mathrm{Fe^{3+}(aq)+3e^-\rightarrow Fe(s)}\)

So we now know for every 3 electrons transferred results in 1 mole of Fe formed and we know that the Molar mass of Fe= 55.85

\(\mathrm{Mass\:Fe = 0.11\: mol\: e^- \times\dfrac{1\: mol\: Fe^{3+}}{3\: mol\: e^-}\times\frac{55.85g}{1mol}= 2.05\: g\: Fe}\)

D. \(\ce{Ni^2+}\)

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

So we now know for every 2 electrons transferred results in 1 mole of Ni formed and we know that the Molar mass of Ni= 58.69

\(\mathrm{Mass\: Ni = 0.11\: mol\: e^- \times\dfrac{1\: mol\: Ni^{2+}}{2\: mol\: e^-}\times\frac{58.69g}{1 mol}= 3.23\: g\: Ni}\)

Q21.4C

1. How many unpaired e^- would you find in the octahedral complex \(\mathrm{[ZnCl_6]^4-}\)?
2. How many unpaired electrons would you expect to find in the tetrahedral complex \(\mathrm{[CuI_4]^2-}\)? Would you expect more unpaired electrons in the octahedral complex \(\ce{[Mn(NH_3)_8]^3+}\)?

Q21.4C Ans

a1. The first step would be determine the oxidation state of Zn. We know that the total charge of the ion is -4, the charge of Cl is -1, and that there are 6 Cl .

total charge=(charge of Zn)(# of Zn atoms)+(charge of Cl)(# of Cl atoms)

-4= (charge of Zn)+(6)(-1)

-4+6=(charge of Zn)

Charge of Zn=+2

a2. The second step would be to determine the electron configuration of \(\ce{Zn^{2+}}\)

\(\mathrm{[Ar]4s^{2}{3d^{10}}\rightarrow}\)\(\mathrm{\underline{[Ar]3d^{10}}}\)

because \(\ce{Zn^{2+}}\) contains in total 28 electrons and there are 10 electrons in the 3d subshell/orbitals because the 2 electrons are lost due the charge comes from the 4s orbital (This is the case of for all 2+ transition state metals as the s orbital electrons are always lost first). As you can see the outer shell, 3d, has 10 electrons which the most the 3d shell can fit. Therefore, this octahedral complex does not have any unpaired electrons. Furthermore, we can assume that the all the other shells are completely filled due to the Aufbau principle. Also, for this complex ion, the case of whether it is low spin or high spin due to the its ligand would not matter because all 5 d orbitals are completely filled completely.

b1. Here again would to determine the oxidation state of Cu. We know the total charge is -2, the charge of I is -1, and that there are 4 I.

total charge=(charge of Cu)(# of Cu particles)+(Charge of I)(# of I atoms)

-2=(charge of Cu)+(-1)(4)

-2+4=charge of Cu

Charge of Cu=+2

b2. Then we would have to figure out the electron configuration of \(\ce{Cu^{2+}}\)

\(\mathrm{[Ar]4s^{1}{3d^{10}}\rightarrow\underline{[Ar]3d^{9}}}\)

because \(\ce{Cu^{2+}}\) contains in total 27 electrons and there are 9 electrons in the 3d subshell/orbitals since 2 electrons are lost due to the 2+ charge; one electron comes from the 4s orbital and one of the 3d orbital (This is the case of for all 2+ transition state metals as the s orbital electrons are always lost first). This means at most that there is one unpaired electron in the 3d orbital. Again, we can assume that all of the other shells are completely filled due to the Aufbau principle. Also, the case of the low spin or high spin due to the ligand does not matter because 4 of the 5 orbitals are going to be filled completely leaving one with only one electron.

Now that we have determined that \(\mathrm{[CuI_4]^2-}\) has one unpaired electron

b3. Determining the charge of Mn in \(\mathrm{[Mn(NH_3)_8]^3+}\)

total charge=(charge of Mn)(# of Mn)+(charge of \(\mathrm{NH_3}\))(# of \(\mathrm{NH_3}\))

+3=(charge of Mn)(1)+(0)(8)

Charge of Mn=3

b4. Now that the charge is determine we can determine the electron configuration \(\mathrm{Mn^{3+}}\)

\(\mathrm{[Ar]4s^{2}{3d^{5}}\rightarrow\underline{[Ar]3d^{4}}}\)

Because \(\ce{Mn^{3+}}\) contains 22 electrons and there are 4 electrons in the 3d orbital/subshell since a total of 3 electrons are lost due to the 3+ charge; 2 electrons from the 4s orbital and 1 electron from the 3d orbital (This is the case of for all 2+ transition state metals as the s orbital electrons are always lost first). Again, we can assume that all of the other shells are completely filled due to the Aufbau principle. This means at most there is 4 unpaired electrons in this octahedral complex because we have not determined whether this complex is low or high spin. Looking at the ligand \(\mathrm{NH_3}\), it is a strong field ligand therefore it induces a low spin thus resulting in only 2 unpaired electrons because it has to fill up the \(\ce{t_{2g}}\) level first.

Finally, the octahedral complex has 2 unpaired electrons and the square planar complex has only 1 unpaired electron, so the octahedral complex has more.

Q24.12B

For the reaction A + B \(\ce{\rightarrow}\)C+ D the following initial rate of reaction were found. What is the rate law for this reaction?

Q24.12B Ans

In order to find the rate law for this problem we have determine the rate constant, k, the exponents of each of the reactants x and y.

Rate=k\(\mathrm{[A]^x[B]^y}\)

In order to determine k we have to determine x and y

Let's first determine x

Lets pick two experiments we can set a ratio for ideally one with the same [B] values so it can cancel out.

Picked experiment 1 and 2 and ratio set up should look like this

\(\mathrm{\frac{0.0042}{0.013}}\)=\(\mathrm{\frac{(k)[0.6]^x[1.8]^y}{(k)[1.8]^x[1.8]^y}}\)

So as as stated before, the \(\mathrm{1.8^y}\) cancels out and the k's cancel out leaving us with

\(\mathrm{\frac{0.0042}{0.013}}\)=\(\mathrm{\frac{[0.6]^x}{[1.8]^x}}\)

Simplify.

\(\mathrm{(\approx{\frac{1}{3}})}\)=\(\mathrm{(\frac{1}{3})^x}\)

x=1

Then lets determine y

Lets pick two experiments we can set a ratio for ideally one with the same [A] values so it can cancel out.

Picked experiments 3 and 1 and the ration set up should like this

\(\mathrm{\frac{0.017}{0.0042}}\)=\(\mathrm{\frac{(k)[0.6]^x[3.6]^y}{(k)[0.6]^x[1.8]^y}}\)

like before the k and \(\mathrm{[0.6]^x}\) cancels out resulting in

\(\mathrm{\frac{0.017}{0.0042}}\)=\(\mathrm{\frac{[3.6]^y}{[1.8]^y}}\)

Simplify.

\(\mathrm{(\approx{4})=2^y}\)

y=2

this results in the

Rate=k\(\mathrm{[A]^1[B]^2}\)

to determine k lets pick an experiment to get the values of A,B and rate from.

Chose experiment 1. Plugging in the values.

0.0042=(k)\(\mathrm{[0.6][1.8]^2}\)

k=\(\mathrm{\frac{0.0042}{[0.6][1.8]^2}}\)

k=0.00216

Which get us our rate law

Rate=0.00216\(\mathrm{[A]^1[B]^2}\)

Q24.59B

1. Do catalysts take part in the reaction they catalyze? Do catalysts always speed up a reaction? Explain.
2. What is the function of a catalyst?

Q24.59B Ans

1. Catalysts do not take part in reactions they catalyze because they don't necessarily change the concentrations of the final products of the reaction at all, but they do affect at least one mechanism of the reaction. Catalysts always speeds up a reaction by lowering the activation energy of a reaction. An exception would be negative catalysts called inhibitors . Negative catalysts slow down reaction rates by increasing the activation energy.

2. The function of a catalyst is to lower the activation energy by altering one or more mechanisms of a reaction thus creating a different pathway with a lower activation energy of the final products, which consequently increases the reaction rate of the whole reaction.

Q25.27E

Approximately how old is a wooden object given that the object has a disintegration rate of 11.0 dis/minxg? It is given that the half-life of this particular wood is 5730 years.

Q25.27E

In this problem we are asked to how old an object is therefore we have use to the half-life given to us.

The first step of the problem would be to determine the the decay constant λ from the half life. Since half-life and rate of the decay are constant we can assume it a first order reaction. Therefore we can use

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

Solving for λ

λ=\(\mathrm{\frac{ln(2)}{t_(1/2)}}\)

\(\mathrm{\lambda = \frac{ln(2)}{5730\:years} = 1.21\times10^{-4}}\)

Now that we found λ we can use the equation \(\mathrm{N_t=(N_0)e^{-(\lambda)t}}\) where \(\ce{N_0}\), the initial disintergration rate of wood, as 15 dis/min which is stated in the one of the problems above and \(\ce{N_t}\) as 11 dis/min, which is stated in the problem.

Using \(\mathrm{N_t}\)=\(\mathrm{(N_0)e^{-(\lambda)t}}\) we have to solve for t

Divide \(\ce{N_0}\) from both sides

\(\mathrm{\frac{N_t}{N_0}}\)=\(\mathrm{e^{-(\lambda)t}}\)

Take the ln of both sides

\(\mathrm{ln({\frac{N_t}{N_0})}}\)=-λt

Now plug in the values

\(\mathrm{ln({\frac{11 dis/min}{15 dis/min})}}\)=\(\mathrm{-1.21\times10^{-4}t}\)

-0.30155=\(\mathrm{-1.21\times10^{-4}t}\)

Now solve for t

\(\mathrm{\frac{-0.30155}{-1.21\times10^{-4}}}\)=t

t=2563.3 years

Q21.1.2

Describe the differences between nonionizing and ionizing radiation in terms of the intensity of energy emitted and the effect each has on an atom or molecule after collision. Which nuclear decay reactions are more likely to produce ionizing radiation? nonionizing radiation?

Q21.1.2 Ans

Nonionizing radiation is low in terms of the intensity of energy emitted and ionizing radiation is very high in terms of the intensity of energy emitted. As a result, nonionizing radiation does not destroy matter and is harmless, and the kinetic energy of nonionizing radiation makes the atoms or molecules move more rapidly once absorbed but doesn't cause structural or chemical damage. On the other hand, ionizing radiation is so high in energy that it causes electrons to become excited thus creating positive charged ions, which is reactive and can decompose or go through chemical changes. Alpha, beta(both), and gamma decay are all ionizing radiation because of their ability to penetrate matter and strip away electrons from the atoms they hit. There are no nuclear decay reactions that are nonionizing.

Q24.3.1

Write the name of the following complexes

1. [CoCl₃(NH₃)₃]
2. [Co(ONO)₃(NH₃)₃]
3. [Fe(ox)₂(H₂O)₂]^-
4. Ag₂[HgI₄]

Q24.3.1 Ans

A. [CoCl₃(NH₃)₃]

1. Let's start with the ligands in this complex ion, which are NH₃ and Cl. They are monodentate ligands with the names of "amine" and "chloro", respectively. There are three atoms of each so there will be a prefix of "tri" for both ligands.

2. Alphabetically, amine is before chloro so that will be order within the name. That is to say, amine will come before chloro.

2. Putting the three previous things together we get "triamminetrichloro". The name of the metal is cobalt so that will come last and to figure out its charge:

In order to figure out the charge:

total charge=(charge of cobalt)+(charge of Cl)(# of Cl)+(charge of NH₃)(# of NH₃)

0=charge of cobalt+(-1)(3)+0

Charge of cobalt= +3

So putting this all together we get

triamminetrichlorocobalt(III)

B. [Co(ONO)₃(NH₃)₃]

1. Let's start with the ligands in this complex ion, which are ONO and NH₃. These are both monodentate ligands with the names "nitrito" (linkage isomer) and "amine", respectively. There 3 atoms atoms of each ligand so have to use "tri" as the prefix for both

2. Alphabetically, amine is before nitrito so that will be the order of name of the complex.

2. Putting all this together we get "triamminetrinitrito." The name of the metal is cobalt. Now, we have to figure out the the charge of the cobalt in the complex ion

total charge=(charge of cobalt)+(charge of ONO)(# of ONO)+(charge of NH₃)(# of NH₃)

0=(charge of cobalt)+(-1)(3)+0

Charge of cobalt=+3

so putting this all together we get

triamminetrinitritocobalt(III)

c. [Fe(ox)₂(H₂O)₂]^-

1. Lets start with the two ligands ox and H₂O. H₂O is monodentate ligand with name "aqua" and ox is a bidentate ligand with name "oxolate." Since there two atoms of each ligand we use the prefix "di." Alphabetically, aqua is before oxolato.

2. Putting this all together we get "diaquadioxolate."

2. Now time for the metal Fe which is called iron, but since it is a anion in a complex molecules, we have to use its formal name "ferrate."Also we have to put ion at the end because it is anion. We have to figure out the charge of iron/ ferrate.

total charge=(charge of iron)+(charge of ox)(# of ox)+(charge of H₂O)(# of H₂O)

-1=charge of iron+(-2)(2)+0

-1=charge of iron-4

charge of iron=+3

Putting everthing together we get

diaquadioxalatoferrate(III) ion

d. Ag₂[HgI₄]

1. So lets start with the complex ion. The ion, [HgI₄], has the ligand I. The ligand is a monodentate ligand and has the name "iodo". There are 4 atoms of these ligands therefore we have to the prefix tetra. Putting these two things together we get "tetraiodo." Now we should find the charge of the Hg to see if the ion is an anion.

total charge=((charge of Ag)(# of Ag))+((charge of Hg)(#of Hg)+(charge of I)(#of I))

0=(+1)(2)+(charge of Hg)+(-1)(4)

charge of Hg=+2

Furthermore since the charge of the complex ion is negative it is an anion therefore we have change mercury to mercurate. Putting this all together we get:

silver(I) tetraiodomercurate(II)

Corrected by : Arlo Chaudhuri