Extra Credit 23

Q17.3.2

For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.

a. \(\ce{Mn}(s) +\ce{Ni^2+}(aq)→\ce{Mn^2+}(aq)+\ce{Ni}(s)\)

b. \(\ce{3Cu^2+}(aq) +\ce{2Al}(s)→\ce{2Al^3+}(aq)+\ce{2Cu}(s)\)

c.\(\ce{Na}(s) +\ce{LiNO3}(aq)→\ce{NaNO3}(aq)+\ce{Li}(s)\)

d. \(\ce{Ca(NO3)2}(aq) +\ce{Ba}(s)→\ce{Ba(NO3)2}(aq)+\ce{Ca}(s)\)

Answer 17.3.2

For all the reactions, we are given that the reactions take place at standard conditions: 25°C and 1 M

a. \(\ce{Mn}(s) +\ce{Ni^2+}(aq)→\ce{Mn^2+}(aq)+\ce{Ni}(s)\)

The two half reactions would be:

Oxidation: \(\ce{Mn}(s) →\ce{Mn^2+}(aq)+\ce{2e^-}\)

Reduction: \(\ce{Ni^2+}(aq)+\ce{2e^-}→\ce{Ni}(s)\)

We have to write in terms of standard reduction potentials:

\(\ce{Mn^2+}(aq)+\ce{2e^-} →\ce{Mn}(s)\) \(\ce{E^°}\) = -1.185 V = \(\ce{E_{anode}^°}\)

\(\ce{Ni^2+}(aq)+\ce{2e^-}→\ce{Ni}(s)\) \(\ce{E^°}\) = -0.257 V = \(\ce{E_{cathode}^°}\)

Cathode is where reduction occurs. Anode is where oxidation occurs.

We know that:

\(\ce{E_{cell}^°}\) = \(\ce{E_{cathode}^°}\) - \(\ce{E_{anode}^°}\)

So,

\(\ce{E_{cell}^°}\) = -0.257 - (-1.185) = +0.928V (spontaneous)

b. \(\ce{3Cu^2+}(aq) +\ce{2Al}(s)→\ce{2Al^3+}(aq)+\ce{2Cu}(s)\)

The two half reactions:

Reduction: \(\ce{Cu^2+}(aq)+\ce{2e^-} →\ce{Cu}(s)\)

Oxidation: \(\ce{Al}(s)→\ce{Al^3+}(aq)+\ce{3e^-}\)

We have to write in terms of standard reduction potentials:

\(\ce{Cu^2+}(aq)+\ce{2e^-} →\ce{Cu}(s)\) \(\ce{E^°}\) = 0.337 V = \(\ce{E_{cathode}^°}\)

\(\ce{Al^3+}(aq)+\ce{3e^-}→\ce{Al}(s)\) \(\ce{E^°}\) = -1.662 V = \(\ce{E_{anode}^°}\)

NOTE: The coefficients do not affect the potentials' values.

Cathode is where reduction occurs. Anode is where oxidation occurs.

We know that:

\(\ce{E_{cell}^°}\) = \(\ce{E_{cathode}^°}\) - \(\ce{E_{anode}^°}\)

So,

\(\ce{E_{cell}^°}\) = 0.337 - (-1.662) = +1.999V (spontaneous)

c. \(\ce{Na}(s) +\ce{LiNO3}(aq)→\ce{NaNO3}(aq)+\ce{Li}(s)\)

The two half reactions:

Oxidation: \(\ce{Na}(s) →\ce{Na^+}(aq)+\ce{e^-}\)

Reduction: \(\ce{Li^+}(aq)+\ce{e^-}→\ce{Li}(s)\)

We have to write in terms of standard reduction potentials:

\(\ce{Na^+}(aq)+\ce{e^-}→\ce{Na}(s)\) \(\ce{E^°}\) = -2.71 V = \(\ce{E_{anode}^°}\)

\(\ce{Li^+}(aq)+\ce{e^-}→\ce{Li}(s)\) \(\ce{E^°}\) = -3.04 V = \(\ce{E_{cathode}^°}\)

Cathode is where reduction occurs. Anode is where oxidation occurs.

We know that:

\(\ce{E_{cell}^°}\) = \(\ce{E_{cathode}^°}\) - \(\ce{E_{anode}^°}\)

So,

\(\ce{E_{cell}^°}\) = -3.04 - (-2.71) = -0.33V (non-spontaneous)

d. \(\ce{Ca(NO3)2}(aq) +\ce{Ba}(s)→\ce{Ba(NO3)2}(aq)+\ce{Ca}(s)\)

The two half reactions:

Reduction: \(\ce{Ca^2+}(aq)+\ce{2e^-} →\ce{Ca}(s)\)

Oxidation: \(\ce{Ba}(s)→\ce{Ba^2+}(aq)+\ce{2e^-}\)

We have to write in terms of standard reduction potentials:

\(\ce{Ca^2+}(aq)+\ce{2e^-} →\ce{Ca}(s)\) \(\ce{E^°}\) = -2.868 V = \(\ce{E_{cathode}^°}\)

\(\ce{Ba^2+}(aq)+\ce{2e^-}→\ce{Ba}(s)\) \(\ce{E^°}\) = -2.912 V = \(\ce{E_{anode}^°}\)

Cathode is where reduction occurs. Anode is where oxidation occurs.

We know that:

\(\ce{E_{cell}^°}\) = \(\ce{E_{cathode}^°}\) - \(\ce{E_{anode}^°}\)

So,

\(\ce{E_{cell}^°}\) = -2.868 - (-2.912) = +0.044V (spontaneous)

Q19.1.21

Predict the products of each of the following reactions and then balance the chemical equations.

1. Fe is heated in an atmosphere of steam.
2. NaOH is added to a solution of Fe(NO₃)₃.
3. FeSO₄ is added to an acidic solution of KMnO₄.
4. Fe is added to a dilute solution of H₂SO₄.
5. A solution of Fe(NO₃)₂ and HNO₃ is allowed to stand in air.
6. FeCO₃ is added to a solution of HClO₄.
7. Fe is heated in air.

Answer 19.1.21

a. Steam is water (\(\ce{H_{2}O}\))

We can write out the reaction as:

\(\ce{Fe}\) + \(\ce{H_{2}O}\) → ?

This is a single replacement reaction, so \(\ce{Fe}\) replaces \(\ce{H_{2}}\). So, one of the products is \(\ce{Fe_{3}O_{4}}\) since it is a combination of iron(II) oxide, \(\ce{FeO}\), and iron(III) oxide, \(\ce{Fe_{2}O_{3}}\).

The \(\ce{Fe}\) is heated in an atmosphere of steam. \(\ce{H_{2}}\) becomes neutrally charged and becomes another product.

After balancing the coefficients, the final reaction is:

\(\ce{3Fe}(s)\) + \(\ce{4H_{2}O}(g)\) → \(\ce{Fe_{3}O_{4}}(s)\) + \(\ce{4H_{2}}(g)\)

b. \(\ce{NaOH}\) added to a solution of \(\ce{Fe(NO_{3})_{3}}\) is a double replacement and precipitation reaction.

We can write out the reaction as:

\(\ce{NaOH}\) + \(\ce{Fe(NO_{3})_{3}}\) → ?

The \(\ce{Na}\) and \(\ce{Fe}\) switch to form \(\ce{Fe(OH)_{3}}(s)\) and \(\ce{NaNO_{3}}(aq)\).

\(\ce{Fe(OH)_{3}}\) is solid because it is insoluble according to solubility rules.

After balancing the coefficients in the reaction, the final reaction is:

\(\ce{Fe(NO_{3})_{3}}(aq)\) + \(\ce{3NaOH}(aq)\) → \(\ce{Fe(OH)_{3}}(s)\) + \(\ce{NaNO_{3}}(aq)\)

c. For instance, the acid used to make the acidic solution is \(\ce{H_{2}SO_{4}}\), then the reaction is:

\(\ce{FeSO_{4}}\) + \(\ce{KMnO_{4}}\) + \(\ce{H_{2}SO_{4}}\) → \(\ce{Fe_{2}(SO_{4})_{3}}\) + \(\ce{MnSO_{4}}\) + \(\ce{H_{2}O}\) + \(\ce{K_{2}SO_{4}}\)

Next, the net ionic reaction has to be written to get rid of the spectator ions in the reaction, this is written as:

\(\ce{Fe^{2+}}\) + \(\ce{MnO_{4}^{-}}\) + \(\ce{H^{+}}\) → \(\ce{Fe^{3+}}\) + \(\ce{Mn^{2+}}\) + \(\ce{H_{2}O}\)

As seen in the net ionic equation above, \(\ce{Fe^{2+}}\) is oxidized to \(\ce{Fe^{3+}}\) and \(\ce{MnO_{4}^{-}}\) is reduced to \(\ce{Mn^{2+}}\). These can be written as two half reactions:

\(\ce{Fe^{2+}}\) → \(\ce{Fe^{3+}}\)

\(\ce{MnO_{4}^{-}}\) → \(\ce{Mn^{2+}}\)

To balance the oxidation half reaction, one electron as to be added to the \(\ce{Fe^{3+}}\), this is shown as:

\(\ce{Fe^{2+}}\) → \(\ce{Fe^{3+}}\) + \(\ce{e^{-}}\)

The reduction half reaction also has to be balanced, but with \(\ce{H^{+}}\) ions and \(\ce{H_{2}O}\), this is shown as:

\(\ce{MnO_{4}^{-}}\) + \(\ce{8H^{+}}\) → \(\ce{Mn^{2+}}\) + \(\ce{4H_{2}O}\)

After the charge of the \(\ce{Mn}\) atoms are balanced, the overall charge has to be balanced on both sides because on the reactants side, the charge is \(\ce{7+}\), and the charge on the products side is \(\ce{2+}\). The overall charge can be balanced by adding electrons, this is shown as:

\(\ce{MnO_{4}^{-}}\) + \(\ce{8H^{+}}\) + \(\ce{5e^{-}}\) → \(\ce{Mn^{2+}}\) + \(\ce{4H_{2}O}\)

Now since both half reactions are balanced, the electrons in both half reactions have to be equal, and then the half reactions are added together. After this is done, the reaction looks like this:

\(\ce{MnO_{4}^{-}}\) + \(\ce{8H^{+}}\) + \(\ce{5Fe^{2+}}\) + \(\ce{5e^{-}}\) → \(\ce{Mn^{2+}}\) + \(\ce{4H_{2}O}\) + \(\ce{5Fe^{3+}}\) + \(\ce{5e^{-}}\)

The \(\ce{5e^{-}}\) on both sides cancel out and the final balanced reaction is:

\(\ce{MnO_{4}^{-}}\) + \(\ce{8H^{+}}\) + \(\ce{5Fe^{2+}}\) →\(\ce{Mn^{2+}}\) + \(\ce{4H_{2}O}\) + \(\ce{5Fe^{3+}}\)

d. \(\ce{Fe}\) added to a dilute solution of \(\ce{H_{2}SO_{4}}\) is a single replacement reaction.

The \(\ce{Fe}\) is added to a dilute solution so the \(\ce{H_{2}SO_{4}}\) is written as separate ions.

We can write out the reaction as:

\(\ce{Fe}(s)\) + \(\ce{2H^+}(aq)\) + \(\ce{(SO_{4})^{2-}}(aq)\) → ?

The Fe replaces the \(\ce{H^+}\) ion, and becomes an \(\ce{Fe^{2+}}\) ion.

\(\ce{H_{2}O}\) is also a product because the solution is dilute.

Furthermore, the \(\ce{FeSO_{4}}\) also has to be separated into ions as a result of the \(\ce{Fe}\) being added to a dilute solution.

After balancing all of the coefficients, the final reaction is:

\(\ce{Fe}(s)\) + \(\ce{(2H_{3}O)^+}(aq)\) + \(\ce{(SO_{4})^{2-}}(aq)\) → \(\ce{Fe^{2+}}(aq)\) + \(\ce{SO_{4}^{2-}}(aq)\) + \(\ce{H_{2}}(g)\) + \(\ce{2H_{2}O}(l)\)

Note: \(\ce{H^+}\) can also be written as the the hydronium ion, \(\ce{(H_{3}O)^{+}}\).

e. We initially can initially write out:

\(\ce{4Fe(NO_{3})_{2}}\) + \(\ce{4HNO_{3}}\) + \(\ce{O_{2}}\) → ?

We write the oxygen term in the reactants because it is stated that the solution is allowed to stand in air.

We just have to analyze the possible products that can be formed and we can see that the hydrogen from nitric acid can combine with oxygen gas to form water and then combining everything together, we get the final reaction to be:

\(\ce{4Fe(NO_{3})_{2}}(aq)\) + \(\ce{4HNO_{3}}(aq)\) + \(\ce{O_{2}}(g)\) → \(\ce{2H_{2}O}(l)\) + \(\ce{4Fe(NO_{3})_{3}}(aq)\)

f. When \(\ce{FeCO_{3}}\) is added to \(\ce{HClO_{4}}\), a double replacement reaction occurs.

The \(\ce{Fe^{2+}}\) ion switches spots with the \(\ce{H^+}\) ion to form \(\ce{Fe(ClO_{4})_{2}}\) as a product.

When the \(\ce{H^+}\) ion is added to the \(\ce{(CO_{3})^{2-}}\) ion, \(\ce{H_{2}CO_{3}}\) is formed.

After balancing the coefficients, the final reaction is:

\(\ce{FeCO_{3}}(s)\) + \(\ce{HClO_{4}}(aq)\) → \(\ce{Fe(ClO_{4})_{2}}(aq)\) + \(\ce{H_{2}O}(l)\) + \(\ce{CO_{2}}(g)\)

g. Air is composed of oxygen gas, which is a diatomic molecule, so it is \(\ce{O_{2}}\).

Adding \(\ce{Fe}\) to \(\ce{O_{2}}\) will cause a synthesis reaction to occur forming \(\ce{Fe_{2}O_{3}}\).

After balancing coefficients, the final reaction is:

\(\ce{3Fe}(s)\) + \(\ce{2O_{2}}(g)\) → \(\ce{Fe_{2}O_{3}}(s)\)

Q19.3.13

[CuCl₄]²⁻ is green. [Cu(H₂O)₆]²⁺is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting?

Answer 19.3.13

We are given that [CuCl₄]²⁻ is green and [Cu(H₂O)₆]²⁺ is blue.

First of all, the observed color is complementary color of one that is absorbed. What that means is that the complexes absorb the color opposite to the color they resemble. Using the color wheel above, we can see that the opposite if green is red and the opposite of blue is orange. So, [CuCl₄]²⁻ absorbs red light wavelengths and [Cu(H₂O)₆]²⁺ absorbs orange light wavelengths. Therefore, [Cu(H₂O)₆]²⁺ absorbs higher energy photons.

To find out which complex has the highest crystal-field splitting, we have to use the following:

\(\ce{E_{photon}}\) = \(\ce{hv}\) = \(\ce\frac{\ce{hc}}{\ce{\lambda}}\) = Δ_o

where Planck's Constant, \(\ce{h}\), is \(\ce{6.6262 \times \ce{10^{-34}}}\)\(\ce{J \cdot s}\) and the speed of light in a vacuum, \(\ce{c}\), is \(\ce{3.00 \times \ce{10^8} }\) \(\ce\frac{\ce{m}}{\ce{s}}\).

For the complex [Cu(H₂O)₆]²⁺:

This complex absorbs orange light so its maximum absorption is at approximately 620 nm. This complex is an octahedral complex.

Using the above equation, we get:

Δ_o = \(\ce{E_{photon}}\) = \(\ce{hv}\) = \(\ce\frac{\ce{hc}}{\ce{\lambda}}\)⇒ (\(\ce\frac{\ce{6.6262} \times \ce{10^{-34}} \times \ce{3.00} \times \ce{10^8}}{\ce{620} \times \ce{10^{-8}}}\)) =\(\ce{3.206 \times \ce{10^{-20}} }\) \(\ce\frac{\ce{J}}{\ce{photon}}\)

For the complex [CuCl₄]²⁻:

This complex absorbs orange light so its maximum absorption is at approximately 800 nm. This complex is a tetrahedral complex.

Using the above equation, we get:

Δ_o = \(\ce{E_{photon}}\) = \(\ce{hv}\) = \(\ce\frac{\ce{hc}}{\ce{\lambda}}\)⇒ (\(\ce\frac{\ce{6.6262} \times \ce{10^{-34}} \times \ce{3.00} \times \ce{10^8}}{\ce{800} \times \ce{10^{-8}}}\)) =\(\ce{2.485 \times \ce{10^{-20}} }\) \(\ce\frac{\ce{J}}{\ce{photon}}\)

Because it's a tetrahedral complex:

Δ_t = 0.44Δ_o

Which means that:

Δ_t = (0.44)(\(\ce{2.485 \times \ce{10^{-20}} }\)) = \(\ce{1.0934 \times \ce{10^{-20}} }\) \(\ce\frac{\ce{J}}{\ce{photon}}\)

Therefore, [Cu(H₂O)₆]²⁺ has a larger crystal field splitting.

Q12.4.13

Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?

Answer 12.4.13

For technetium-99:

Radioactive decay is a first-order process and can be described in the terms of the following equation:

\(\ce\ln({\frac{N}{N_{0}})}\) = \(\ce{-kt}\)

where \(\ce{N_{0}}\) is the initial amount available and \(\ce{N}\) is the amount at some time, \(\ce{t}\) .

In order to find the percentage that would remain after some time, \(\ce{t}\), we would use the above equation.

In this case, \(\ce{t}\) = \(\ce{48 h}\) because we want to find out how much remains after 2 days and we want the units to be consistent so the value is in terms of hours \(\ce{h}\) .

In order to plug in the values and solve for the ratio of \(\ce\frac{\ce{N}}{\ce{N_{0}}}\), we need to find out what " \(\ce{k}\) " is.

We can solve for " \(\ce{k}\) " by using the following formula:

\(\ce{t_{1/2}}\) = \(\ce\frac{0.693}{k}\)

Solving for " \(\ce{k}\) " and plugging in \(\ce{t_{1/2}}\) = \(\ce{6}\) h, we get " \(\ce{k}\) " to be:

\(\ce{k}\) = \(\ce\frac{0.693}{t_{1/2}}\) ⇒\(\ce{k}\) =\(\ce\frac{0.693}{6}\) = 0.1155

Using the integrated rate law and plugging in the known values, we can solve for the ratio of \(\ce\frac{\ce{N}}{\ce{N_{0}}}\):

\(\ce\ln({\frac{N}{N_{0}})}\) =\(\ce{-kt}\)

\(\ce\ln({\frac{N}{N_{0}})}\) =\(\ce{(-0.1155)(48)}\)

\(\ce\ln({\frac{N}{N_{0}})}\) =\(\ce{-5.544}\)

\(\ce\frac{\ce{N}}{\ce{N_{0}}}\) = \(\ce{e^{-5.544}}\)

\(\ce\frac{\ce{N}}{\ce{N_{0}}}\) = \(\ce{3.911 \times \ce{10^{-3}}}\)

Multiplying this value by 100, we get:

0.3911% of technetium-99 would remain after 2 days.

For thallium-201:

Radioactive decay is a first-order process and can be described in the terms of the following equation:

\(\ce\ln({\frac{N}{N_{0}})}\) =\(\ce{-kt}\)

where \(\ce{N_{0}}\) is the initial amount available and \(\ce{N}\) is the amount at some time, \(\ce{t}\) .

In order to find the percentage that would remain after some time, \(\ce{t}\), we would use the above equation.

In this case, \(\ce{t}\) = \(\ce{48 h}\) because we want to find out how much remains after 2 days and we want the units to be consistent so the value is in terms of hours \(\ce{h}\) .

In order to plug in the values and solve for the ratio of \(\ce\frac{\ce{N}}{\ce{N_{0}}}\), we need to find out what " \(\ce{k}\) " is.

We can solve for " \(\ce{k}\) " by using the following formula:

\(\ce{t_{1/2}}\) = \(\ce\frac{0.693}{k}\)

Solving for " \(\ce{k}\) " and plugging in \(\ce{t_{1/2}}\) = \(\ce{73}\) h, we get " \(\ce{k}\) " to be:

\(\ce{k}\) = \(\ce\frac{0.693}{t_{1/2}}\) ⇒\(\ce{k}\) =\(\ce\frac{0.693}{73}\) = 0.009493

Using the integrated rate law and plugging in the known values, we can solve for the ratio of \(\ce\frac{\ce{N}}{\ce{N_{0}}}\):

\(\ce\ln({\frac{N}{N_{0}})}\) =\(\ce{-kt}\)

\(\ce\ln({\frac{N}{N_{0}})}\) =\(\ce{(-0.009493)(48)}\)

\(\ce\ln({\frac{N}{N_{0}})}\) =\(\ce{-0.4557}\)

\(\ce\frac{\ce{N}}{\ce{N_{0}}}\) = \(\ce{e^{-0.4557}}\)

\(\ce\frac{\ce{N}}{\ce{N_{0}}}\) = \(\ce{0.634}\)

Multiplying this value by 100, we get:

63.40% of thallium-201 would remain after 2 days.

Q21.2.8

The mass of the atom \(\ce{^{23}_{11}Na}\) is 22.9898 amu.

1. Calculate its binding energy per atom in millions of electron volts.
2. Calculate its binding energy per nucleon.

Answer 21.2.8

a. We are given that the mass of the atom is 22.9898 amu.

We first need to calculate the mass defect of the particle. To do that, by looking it up, we get that the masses of a free proton and neutron are 1.007825 amu and 1.008665 amu respectively. From the problem above, we see that there are 11 protons and 23 - 11 = 12 neutrons.

Mass defect is equal to the mass of the free nucleus minus the mass of the nucleus.

We can calculate the mass of the free nucleus by summing the total mass of the protons and neutrons:

Mass of free nucleus = (\(\ce{11 \times \ce{1.007825}}\)) + (\(\ce{12 \times \ce{1.008665}}\)) = \(\ce{23.190055 amu}\)

We can calculate mass defect by:

Mass defect = (\(\ce{23.190055}\)) - (\(\ce{22.9898}\)) = \(\ce{0.200255 amu}\)

From this we can calculate the binding energy by using the formula:

\(\ce{E}\) = \(\ce{mc^2}\)

Plugging in all the values, we get:

\(\ce{E}\) = (0.200255 amu)(\(\ce\frac{\ce{1.6605} \times \ce{10^{-27} kg}}{\ce{1 amu}}\)) \(\ce{(3.00 \times \ce{10^8} )^2}\) = \(\ce{2.993 \times \ce{10^{-11}}}\) \(\ce\frac{\ce{J}}{\ce{nucleus}}\)

"per nucleus" is same as "per atom" so we don't need to change the units in the denominator. However, this time, instead of Joules, we want the units to be in terms in millions of electron volts, MeV.

1 MeV = \(\ce{1.602 \times \ce{10^{-13}}}\)\(\ce{J}\)

To convert our value in terms of MeV, we do:

(\(\ce{2.993 \times \ce{10^{-11}}}\) \(\ce\frac{\ce{J}}{\ce{nucleus}}\))(\(\ce{1.602 \times \ce{10^{-13}}}\)\(\ce{J}\)) = \(\ce{186.8}\) \(\ce\frac{\ce{MeV}}{\ce{nucleus}}\) = \(\ce{186.8}\) \(\ce\frac{\ce{MeV}}{\ce{atom}}\)

b. We are given that the mass of the atom is 22.9898 amu.

We first need to calculate the mass defect of the particle. To do that, by looking it up, we get that the masses of a free proton and neutron are 1.007825 amu and 1.008665 amu respectively. From the problem above, we see that there are 11 protons and 23 - 11 = 12 neutrons.

Mass defect is equal to the mass of the free nucleus minus the mass of the nucleus.

We can calculate the mass of the free nucleus by summing the total mass of the protons and neutrons:

Mass of free nucleus = (\(\ce{11 \times \ce{1.007825}}\)) + (\(\ce{12 \times \ce{1.008665}}\)) = \(\ce{23.190055 amu}\)

We can calculate mass defect by:

Mass defect = (\(\ce{23.190055}\)) - (\(\ce{22.9898}\)) = \(\ce{0.200255 amu}\)

From this we can calculate the binding energy by using the formula:

\(\ce{E}\) = \(\ce{mc^2}\)

Plugging in all the values, we get:

\(\ce{E}\) = (0.200255 amu)(\(\ce\frac{(\ce{1.6605} \times \ce{10^-27 kg}}{\ce{1 amu}}\)) \(\ce{(3.00 \times \ce{10^8} )^2}\) = \(\ce{2.993 \times \ce{10^{-11}}}\) \(\ce\frac{\ce{J}}{\ce{nucleus}}\)

This binding energy is in terms of \(\ce\frac{\ce{J}}{\ce{nucleus}}\) but we want it in terms of \(\ce\frac{\ce{J}}{\ce{nucleon}}\) . The number of nucleons defines an isotope's mass number. Given in the problem, we see that the isotope's mass number is 23 which is the number of nucleons.

To convert our value to "per nucleon," we do:

(\(\ce{2.993 \times \ce{10^{-11}}}\) \(\ce\frac{\ce{J}}{\ce{nucleus}}\))(\(\ce\frac{\ce{1 nucleus}}{\ce{ 23 nucleons}}\)) =\(\ce{1.301 \times \ce{10^{-12}}}\)\(\ce\frac{\ce{J}}{\ce{nucleon}}\)

Q21.6.3

Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine-131 is injected into the body. In larger doses, I-133 is also used as a means of treating cancer of the thyroid. I-131 has a half-life of 8.70 days and decays by β⁻ emission.

1. Write an equation for the decay.
2. How long will it take for 95.0% of a dose of I-131 to decay?

Answer 21.6.3

a. In the problem, we are given that the type of decay is β⁻ emission which can be represented by: \(\ce{^0}_{-1}e\)

By looking at the periodic table, we see that Iodine has an atomic number of 53.

We can start out by writing:

\(\ce{^{133}_{53}I}\) → \(\ce{?}\) + \(\ce{^{0}_{-1}e}\)

To complete the equation, we need to find out what " \(\ce{?}\) " is. All equations need to balance to conform to the two conservation laws: the mass number and the electrical charge.

Using the conservation laws to find the unknown,?, in a nuclear reaction equation, we get \(\ce{?}\) = \(\ce{^{A}_{Z}ElementalSymbol}\) in which A is the sum of the total number of protons, Z, and neutrons. We find that A = 133-0 = 133; Z= 53-(-1) = 54. Looking up the periodic table, we find that Xenon(Xe) has an atomic number of 54. Putting everything together, we get the equation for the decay to be:

\(\ce{^{133}_{53}I}\) → \(\ce{^{133}_{54}Xe}\) + \(\ce{^{0}_{-1}e}\)

b.

We are given that the half-life \(\ce{t_{1/2}}\) is 8.70 days.

First thing we have to do is find " \(\ce{k}\) ." We can do this by using the following equation:

\(\ce{t_{1/2}}\) = \(\ce\frac{0.693}{k}\)

Solving for " \(\ce{k}\) " and plugging in \(\ce{t_{1/2}}\) = \(\ce{8.70}\) days, we get " \(\ce{k}\) " to be:

\(\ce{k}\) = \(\ce\frac{0.693}{t_{1/2}}\) ⇒\(\ce{k}\) =\(\ce\frac{0.693}{8.70}\) = 0.07966

Radioactive decay is a first-order process and can be described in the terms of the integrated rate law:

\(\ce\ln({\frac{N}{N_{0}})}\) =\(\ce{-kt}\)

where \(\ce{N_{0}}\) is the initial amount available and \(\ce{N}\) is the amount at some time, \(\ce{t}\) .

If 95.0% has decayed, that means that there is 100%-95.0%= 5.0% of the dose left. For instance, lets say we start out with 100 grams, that is, \(\ce{A_{0}}\) =100. If 95.0% of this amount has decayed, that means there's only 5 grams left ( \(\ce{({0.95})(100)}\) = \(\ce{5}\) ).

Plugging these values into the integrated rate law above, we can find out the time, \(\ce{t}\) , it will take 95.0% of the dose to decay:

\(\ce\ln({\frac{5}{100}})\) =\(\ce{-0.07966t}\) ⇒ \(\ce{t}\) = 37.6 days.

Q20.4.9

All reference electrodes must conform to certain requirements. List the requirements and explain their significance.

Answer 20.4.9

An ideal reference electrode should be reversible and obey the Nernst equation, exhibit a potential that is constant with time, return to its original potential after being subjected to small currents, and exhibit little lag in response( also known as hysteresis) with temperature fluctuations. These properties are significant in maintaining the function of the reference electrode which is to maintain a constant electrical potential against any deviations. A reference electrode is an electrode that has a known half-cell potential, is constant, and is altogether insensitive to the composition of the solution.

Q20.8.1

Do you expect a bent nail to corrode more or less rapidly than a straight nail? Why?

Answer 20.8.1

A bent nail will corrode more rapidly than a straight nail. Bending the nail creates stress in the structure. The atoms will dislocate and the layers will slip and stress due to the bend. Oxidation occurs at stress points in a metal, because the point(s) of stress becomes the anode for the reaction: For instance if the nail is iron, then the anode for the reaction of Iron with oxygen will yield the half reaction of: \(\ce{Fe}(s)→\ce{Fe^2+}(aq)+\ce{2e^-}\) .The reduction half reaction will be \(\ce{O2}(g)+\ce{4H^+}(aq)+{4e^-}→\ce{2H2O}\) . Here, iron is oxidized to \(\ce{Fe^2+}\) at the anodic site on the surface of the iron, which is often a defect. Oxygen is reduced to water at a different site on the surface of the iron, which acts as a cathode. Rust is formed due to corrosion. Rust is formed by the oxidation of iron by atmospheric oxygen. Overall, tensile stress makes a nail more anodic and if the bend is sharp, there will be more corrosion there.