Extra Credit 1

Q17.1.1

If a 2.5 A current is run through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?

S17.1.1

This problem is asking for the amount of couloumbs per charge given how many amps and is being run through a circuit for 35 minutes. This problem is essentially gravimetric analysis, where the strategy is to use stoichiometry to cancel out units, and thus it's important to know what the units of the given information is.

Amps is given. Amps can also be expressed as \( \mathrm{A} = \mathrm{(\frac{Coulombs}{seconds})}\). So the information given can also be expressed as \( \mathrm{2.5 A} = \mathrm{(\frac{2.5 Coulombs}{seconds})}\)

Notice now that the Amps measurement is in seconds, where as the given time is expressed in minutes. Therefore it's important to know that minutes can also be expressed in seconds. \( \mathrm{1 minute} = \mathrm{60 seconds}\), so \( \mathrm{35 minutes} = \mathrm{(\frac{60 seconds}{1 minute})}\).

Looking at it without the numbers, when you multiply minutes by seconds dived by minutes as in : \( \mathrm{minutes} = \mathrm{(\frac{ seconds}{ minute})}\), the minutes will cancel out leaving only seconds.

\( \mathrm{35 minutes} * \mathrm{(\frac{60 seconds}{1 minute})}=\mathrm{2100 Seconds}\)

The last thing to know is that in the end, the final product needs to be coulombs. So multiply the given information in a way that cancels out the seconds and leaves only the units of coulombs. So start off with the given amp measurement and multiply in a way that the unwanted units cancel out or start off with the given time and do the same. There are multiple correct forms for this, where the order of the things being multiplied can vary, but the actual factors are the same. Here is one example:

\( \mathrm{35 minutes} * \mathrm{(\frac{60 seconds}{1 minute})}*\mathrm{(\frac{2.5 Coulombs}{seconds})}=5250 Coulombs\).

Seeing how only two sig figs are given, the answer cannot be presented as 5250 Coulombs and should be expressed as:

\( \mathrm{5.3*10^3 Coulombs}\)

Q17.7.4

A current of 2.345 A passes through the cell shown in Figure for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm? Assume the voltage is sufficient to perform the reduction. (Hint: Is hydrogen the only gas present above the water?)

S17.7.4

This is essentially gravimetrical analysis, where the goal is to use conversions to get unwanted units to cancel out, leaving only the wanted.

You're given a unit of time and this process can be compared to metal plating, only because this is a gas, one will need to convert moles to volume in the end. So convert from time in minutes to seconds. Convert the seconds to amps, which can then be converted to coulombs. Use the reaction for the oxidation of H₂.

\[45 minutes \times \frac{60 seconds}{1 minute} \times \frac{2.345 C}{1 second} \times \frac{1 mole e^-}{96500Coulombs} \times \frac{2 mole H_2}{4 mole e^-} = 0.0328 \quad moleH_2\]

Now that the mole quanity of H₂ is known, use the formula: \(\mathrm{pv}=\mathrm{nrt}\) to get =(0.0328 mol)(0.08206 L atm mol-1 K-1)(298.15 K)(95/100)/0.9684 atm=0.79 L

Q19.2.1

Indicate the coordination number for the central metal atom in each of the following coordination compounds:

1. [Pt(H₂O)₂Br₂]
2. [Pt(NH₃)(py)(Cl)(Br)] (py = pyridine, C₅H₅N)
3. [Zn(NH₃)₂Cl₂]
4. [Zn(NH₃)(py)(Cl)(Br)]
5. [Ni(H₂O)₄Cl₂]
6. [Fe(en)₂(CN)₂]⁺ (en = ethylenediamine, C₂H₈N₂)

S19.2.1

The coordination number is the number of coordinate covalent bonds that are made given the complex. The number of bonds the metal center can form is found through the number of ligands attached to the metal center of the complex. Monodentate ligands can only form one coodinate covalent bond and bidentate ligands can for two bonds. There are ligands, such as EDTA, which can form six coordinate covalent bonds. So finding the coordination number is essentially counting the number of ligands and multiplying the number of polydentate ligands by their degree of "bite". For instance a bidentate ligand will make two bonds. Therefore multiply the number of bidentate ligands by two. So if you have 6 monodentate ligands, the coordination number is 6. If you have 2 bidentate ligands, you have a total of four bonding spots and thus you have a coordination number of 4.

1. The 2 aqua and the 2 bromo ligands form a total of 4 coordinate covalent bonds and has a result coordination number (CN) of 4.
2. The ammine, pyridine, chloro and bromo each form one coordinate covalent bond that gives a total of 4 and hence CN=4
3. two ammine and two chloro ligands give a total of 4 coordinate covalent bonds and a CN = 4.
4. one ammine, a pyrimidine, a chloro and a bromo ligand give a total of 4 covalent bonds, resulting in CN = 4.
5. 4 aqua ligands and 2 chloro ligands form a total of 6 coordinate covalent bonds and a CN =6
6. ethylenediamine is a bidentate ligand that forms two coordinate covalent bonds. Since there are two ethylenediamine ligands, there are four bonds and along with two cyano ligands it forma a total of 6 bonds. hence has a CN=6

Q12.3.13

Nitrosyl chloride, NOCl, decomposes to NO and Cl₂.

\( \mathrm{2NOCl(g)}\rightarrow\mathrm{2NO(g)}+\mathrm{Cl_2(g)}\)

Determine the rate equation, the rate constant, and the overall order for this reaction from the following data:

S12.3.13

The rate will have the form: \( \mathrm{Rate}=\mathrm{k}*\mathrm{[A^x]}*\mathrm{[B^y]} \), where k is the rate constant, A and B represent reactants, and x and y represent the order of each reactant. To solve for the rate, the reactants are usually given and thus the only unknown that actually needs to be solved for are the orders of the reactants. K is usually left unspecified.

To solve for the reactant's order when given a table of experiments, follow the format of :

\(\mathrm{{(\frac{[A]_{from experiment "n"}}{[A]_{from experiment "n+1"}}})^Z}\) = \( \mathrm{(\frac{Rate_n}{Rate_{n+1}})}\), where n represents the experiment number.

So applying the format to the problem: \(\mathrm{{(\frac{[NOCl]_{from experiment "1"}}{[NOCl]_{from experiment "2"}}})^x}\) = \( \mathrm{(\frac{Rate_1}{Rate_{2}})}\)

Filling in the equation with the given information: \(\mathrm{{(\frac{.10}{.2}})^x}\) = \( \mathrm{(\frac{8.1*10^{-10}}{3.2*10^{-9}})}\)

The answer would then be x=1.98. However, this number can be rounded up since only two sig figs are given. So the order for the reactant would be 2 and since there are no other reactants, the rate can be written as:

\( \mathrm{Rate}=\mathrm{k}*\mathrm{[NOCl]^2} \)

Now, the value of K also needs to be determined. To do so, simply take the derrived rate formula and plug in all the values except for K, which one cannot because that's what's being solved for. Any of the experiments can be chosen, but the values must remain consistent for a single experiment.

Here is one example:

1. \( \mathrm{Rate}=\mathrm{k}*\mathrm{[NOCl^2]} \)
2. \( \mathrm{8.1*10^{-10}}=\mathrm{k}*\mathrm{[.10]^2} \)
3. k=8.0 × 10⁻⁸

Now, the k value needs units. Because the final rate has to have the units of \(\frac{M}{unit of time}\), the k will need units where once the concentration, which has units \(\mathrm{M^{total order}}\), is multipled to the k value, the end result are the units for rate. Total order is found by adding all the individual orders of reactants together. Since there is only one reactant and it's order is 2, the total order is also 2. So k value will follow the format of \( \frac{1}{M^{total order-1}*unit of time^{-1}}\). Thus because this is a second order reaction, the units for k are \(\frac{1}{M^{1}*seconds^{-1}}\)

All together, k= \( \mathrm{8.1*10^{-10}}\frac{1}{M^{1}*seconds^{-1}}\)

Q12.6.5

What is the rate equation for the elementary termolecular reaction:

\( \mathrm{A}+\mathrm{2B}\rightarrow\mathrm{products}\)?

\( \mathrm{3A}\rightarrow\mathrm{Products}\)?

S12.6.5

The stoichiometry can be used to predict the rate in elementary steps only. Because these are stated to be elementary steps, the coefficients are equivalent to the order. The rates will then be equal to the multiplication of all the reactants raised to their coefficient and the rate constant.

Thus for : \( \mathrm{A}+\mathrm{2B}\rightarrow\mathrm{products}\)?

Since the reactant "B" has a coefficient of 2, it's order will be 2. The reactant "A" has the coefficient of 1 in front of it. Thus putting everything together:

\(\mathrm{rate}\)=\(\mathrm{k}*\mathrm{[A]}*\mathrm{[B]^2}\)

And for: \( \mathrm{3A}\rightarrow\mathrm{Products}\)?

Since the reactant "A" has the coefficient of 3 in front of it, it will have the order of 3. There are no other reactants, so putting everything together:

\(\mathrm{rate}\)=\(\mathrm{k}*\mathrm{[A]^3}\)

Q21.4.17

If 1.000 g of \(\ce{^{226}_{88}Ra}\) produces 0.0001 mL of the gas \(\ce{^{226}_{86}Sr}\) at STP (standard temperature and pressure) in 24 h, what is the half-life of ²²⁶Ra in years?

S21.4.17

First, find the moles of radon produced via:

PV=nRT

n(Rn)=PV/RT

=(1 atm)(0.0001 mL*1 L/10^3 mL)/(0.08206 L atm mol^(-1) K^(-1))/(273.15 K)

=4.4614*10^(-9) mol

Because moles of (Rn) produced=moles of (Rn) decayed,

mass Ra lost=4.4614*10^(-9) mol*226g/mol

=1.00827*10^(-6) g (then convert to grams) using the given mass number

Then, use graviemtrical analysis to convert the units of mass Ra remaining after 24 h to years.

24 hr=1-(1.00827*10^(-6)g)=0.99999899g

ln(C0/C)= λt=ln(1/0.99999899)=4.3785*10^(-7)

λ=4.2015*10^(-8) h^(-1)

t(1/2)=0.693/4.2015*10^(-8)*(1 d/24 h)*(1 y/365 d)

=1.6494*10^7 h*(1 d/24 h)*(1 y/365 d)

=1883 y

Q20.3.5

One criterion for a good salt bridge is that it contains ions that have similar rates of diffusion in aqueous solution, as K⁺ and Cl⁻ ions do. What would happen if the diffusion rates of the anions and cations differed significantly?

Q20.3.5

The resistance in salt bridge would be increased and the electron flow through the circuit would be limited.

Q20.5.16

For each reaction, calculate E°_cell and then determine ΔG°. Indicate whether each reaction is spontaneous.

1. \(\mathrm{2Na(s)}+\mathrm{2H_2O(l)}\rightarrow\mathrm{2NaOH(aq)}+{H_2(g)}\)
2. \(\mathrm{K_2S_2O_6(aq}+\mathrm{I_2(s)}\rightarrow\mathrm{2KI(aq)}+{2K_2SO_4(aq)}\)
3. \(\mathrm{Sn(s)}+\mathrm{CuSO_4(aq)}\rightarrow\mathrm{Cu(s)}+{SnSO_4(aq)}\)

Q20.5.16

\(\mathrm{E^o_{cell}}\), is the difference in the standard reduction potentials of the cathode and anode. So simply subtract the standard reduction potential of the cathode and anode. Remember the cathode is the one being reduced and the anode is the one being oxidized. Then to figure out ΔG°, plug the the E°_cell value into the following equation: \(\mathrm{dG^o}\) =\(\mathrm{-n}\) *\(\mathrm{F}\)*\(\mathrm{E^o_{cell}}\). If this value is positive, the reaction will not be spontaneous and if the value is negative, then, the reaction is spontaneous.

1. For:\(\mathrm{2Na(s)}+\mathrm{2H_2O(l)}\rightarrow\mathrm{2NaOH(aq)}+{H_2(g)}\),

First identify what is being reduced/oxidized. Because free elements always have an oxidation state of "0", it's easy to see how it changed. Thus, because Na(s) starts off as a free element and then becomes NaOH, with an oxidation state of +1, so it has become oxidized. The same logic can be applied with Hydrogen, where it's a free element in the product. This means that while it starts off with an oxidation state of +1, it later becomes oxidized so that it has an oxidation state of 0.

The next step would be to write out the half reactions. This isn't needed for the calculating E°_cell but is needed later on to calculate ΔG°. However in this case, it would be less work to just think about each reaction individually and see which involves a greater amount of electrons being trasnfered. The table of standard reduction potentials can be used in this case because it will contain all the already balanced half reactions. If one were to look at the table, one would see that to oxidize Na(s), 1 electron is being transferred and to get the reduction of H₂(g), 2 electrons are needed. Thus 2 electrons are being trasnferred.

Now, it's time to caluclate E°cell. Look up the corresponding values of E^o on the table of standard reduction potentials for reactions that match.

\(\mathrm{E°cell}= \mathrm{E°cathode}-\mathrm{E°anode}=\mathrm{-0.83 v}-\mathrm{2.71 v}=\mathrm{3.54 v}\)

Now, to get ΔG°, multiply the caluclated E°cell value, the negative number of electrons being transferred and faraday's constant.

\(\mathrm{dG^o}\) =\(\mathrm{-2}\) *\(\mathrm{96486}\)*\(\mathrm{3.54}\) =\(\mathrm{-683120.88}\)

Notice how the value calculated is negative. The reaction is spontaneous.

2. For:\(\mathrm{K_2S_2O_6(aq}+\mathrm{I_2(s)}\rightarrow\mathrm{2KI(aq)}+{2K_2SO_4(aq)}\),

First identify what is being reduced/oxidized. Because free elements always have an oxidation state of "0", it's easy to see how it changed. Thus, because I₂(s) starts off as a free element and then becomes KI, with an oxidation state of -1, so it has become reduced. This means that Sulfur will be oxidized.

The next step would be to write out the half reactions. This isn't needed for the calculating E°_cell but is needed later on to calculate ΔG°. However in this case, it would be less work to just think about each reaction individually and see which involves a greater amount of electrons being trasnfered. The table of standard reduction potentials can be used in this case because it will contain all the already balanced half reactions. If one were to look at the table, one would see that to reduce I₂(s), 2 electrons are needed. Thus 2 electrons are being trasnferred.

Now, it's time to caluclate E°cell. Look up the corresponding values of E^o on the table of standard reduction potentials for reactions that match.

\(\mathrm{E°cell}= \mathrm{E°cathode}-\mathrm{E°anode}=\mathrm{0.54 v}-\mathrm{-2.92 v}=\mathrm{3.46 v}\)

Now, to get ΔG°, multiply the caluclated E°cell value, the negative number of electrons being transferred and faraday's constant.

\(\mathrm{dG^o}\) =\(\mathrm{-2}\) *\(\mathrm{96486}\)*\(\mathrm{3.46}\) =\(\mathrm{-667683.12}\)

Notice how the value calculated is negative. The reaction is spontaneous.

3. For : \(\mathrm{Sn(s)}+\mathrm{CuSO_4(aq)}\rightarrow\mathrm{Cu(s)}+{SnSO_4(aq)}\)

First identify what is being reduced/oxidized. Because free elements always have an oxidation state of "0", it's easy to see how it changed. Thus, because Sn(s) starts off as a free element and then becomes SnSO₄, with an oxidation state of +5, so it has become oxidized. This means that copper will be reduced.

The next step would be to write out the half reactions. This isn't needed for the calculating E°_cell but is needed later on to calculate ΔG°. However in this case, it would be less work to just think about each reaction individually and see which involves a greater amount of electrons being transfered. The table of standard reduction potentials can be used in this case because it will contain all the already balanced half reactions. If one were to look at the table, one would see that to reduce Cu(s), 2 electrons are needed. Thus 2 electrons are being trasnferred.

Now, it's time to caluclate E°cell. Look up the corresponding values of E^o on the table of standard reduction potentials for reactions that match.

\(\mathrm{E°cell}= \mathrm{E°cathode}-\mathrm{E°anode}=\mathrm{0.54 v}-\mathrm{-2.92 v}=\mathrm{.48 v}\)

Now, to get ΔG°, multiply the caluclated E°cell value, the negative number of electrons being transferred and faraday's constant.

\(\mathrm{dG^o}\) =\(\mathrm{-2}\) *\(\mathrm{96486}\)*\(\mathrm{.48}\) =\(\mathrm{-92626.56}\)

Notice how the value calculated is negative. The reaction is spontaneous.