Extra Credit 25

Q17.3.4

Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell made from a half-cell consisting of a silver electrode in 1 M silver nitrate solution and a half-cell consisting of a zinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?

Remember: \(E_{cell}^o=E^o_{cathode}-E^o_{anode}\)

1. First find the which reaction happens on each electrode by using the standard reduction potentials (SRPs) for each half-reaction.

\(Zn^{2+}_{(aq)}+2e^- \rightarrow Zn_{(s)}\) \(E^o: -0.7618V\)

\(Ag^+_{(aq)}+e^- \rightarrow Ag_{(s)}\) \(E^o: 0.7996V\)

From this information, you can know that oxidation happens on the zinc electrode (thus the anode) while reduction happens on the silver electrode (the cathode). We can know this because a positive E^o value means the reaction is spontaneous and here, because the reduction of Ag reaction occurs spontaneously while the reduction of Zn is non-spontaneous (thus occurs as an oxidation reaction).

2. Balance the equation (most important part is getting the half reactions from the balanced equation to find correct SRPs):

Anode: \(Zn^{2+}_{(aq)}+2e^-\rightarrow Zn_{(s)}\) \(E^o: -0.7618V\)

Cathode: [multiply the silver equation by two to balance the electrons]: \((Ag^+_{(aq)}+e^-\rightarrow Ag_{(s)} ) \times 2\) \(E^o: 0.7996V \times (2) = 1.5992V\)

Balanced Equation: \(2Ag^+_{(aq)}+Zn_{(s)}\rightarrow 2Ag_{(s)}+Zn^{2+}\) correct

3. Use original equation to calculate the overall cell potential: \(E_{cell}^o=E^o_{cathode}-E^o_{anode} \Rightarrow 1.5992- (-0.7618) = 2.361V\) correct

the overall reaction is spontaneous because it equals + 2.361V

Q19.1.23

Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.

As CN− is added, the sodium cyanide dissociates to form sodium and cyanide ions (sodium is a spectator ion) while silver nitrate dissociates to form silver and nitrate ions (nitrate is a spectator ion). The cyanide combines with the silver ions to form an intermediate solid complex, AgCN:

\(Ag^+_{(aq)}+CN^−_{(aq)}⟶AgCN_{(s)}\) correct

However, as more CN− is added, extra cyanide continues bonding with the existing AgCN precipitate, forming an aqueous product, Ag(CN)₂:

\(AgCN_{(s)}+CN^−_{(aq)}⟶[Ag(CN)_2]^−_{(aq)}\) correct

_{Net Reaction:}

\(Ag^+_{(aq)}+2CN^−_{(aq)}⟶[Ag(CN)_2]^−_{(aq)}\) correct

Very clear steps that make sense.

Q12.4.16

Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge); the nuclear equation is: \({^{18}_{9}F+}⟶{_8^{18}O}+{^0_1e^+}\)

Physicians use ¹⁸F to study the brain by injecting a quantity of fluoro-substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.

1. What is the rate constant for the decomposition of fluorine-18?

Use equation: \(ln\frac{[A]_0}{[A]}=kt\)

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

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

By plugging in our value for the half-life of oxygen-18 (109.7 min), you get

\(k\approx0.006317 min^{-1}\)

2. If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h?

a) First convert 5.59h into minutes by multiplying by 60, giving: 335.4 minutes.

b) Next, using the original half life expression: \(ln\frac{[A]_0}{[A]}=kt\)

c) Plug in t=335.4, \([A]_0=100\) (for 100%), and your k constant, \(k=0.006317 min^{-1}\):

\(ln\frac{100}{[A]}=0.006317 minutes^{-1} \times335.4minutes\)

d) Your minutes will cancel out, and multiplication gives us,

\(ln\frac{100}{[A]}=2.119\)

e) To get rid of the natural log, use e:

\(e^{ln\frac{100}{[A]}}=e^{2.119}\)

f) This gives us the expression:

\(\frac{100}{[A]}=8.3249\quad or \quad\frac{100}{8.3249}=[A]\)

Thus, [A] = 12.01 or 12.01%

3. How long does it take for 99.99% of the 18F to decay?

a) Using a similar process, we can find this answer. Going back to our original equation:

\(ln\frac{[A]_0}{[A]}=kt\)

b) We can plug in A₀=100, A=100-99.99=0.01, and use our decay constant k=0.006317 min^-1

\(ln\frac{100}{0.01}=0.006317min^{-1}t\)

c) Which yields us:

\(9.21=0.006317t\)

So, t = 1458 minutes or by dividing by 60, gives us 24.3 hours.

Very clear steps that makes sense.

Q21.2.10

Which of the following nuclei lie within the band of stability?

1. argon-40
2. oxygen-16
3. ¹²²Ba
4. ⁵⁸Ni
5. ²⁰⁵Tl
6. ²¹⁰Tl
7. ²²⁶Ra
8. magnesium-24

Steps to find stability:

• Find atomic number (Z) to give you the number of protons.
• Subtract atomic number from given mass number to get the number of neutrons (N)
• If mass number (total number of nucleons is even, element is likely stable).
• If there is a magic # of protons or neutrons, the nucleus is particularly stable: 2, 8, 28, 50, 82, 114 (protons), 126 (neutrons), 184 (neutrons)
• Calculate N/Z ratio and use belt of stability diagram to determine best way to get from an unstable to a stable nucleus.

1. argon-40:

Z = 18

N = 40-18 = 22

#nucleons is even thus the element argon-40 is stable

2. oxygen-16:

Z = 8

N = 16-8 = 8

# nucleons is even, and both Z and N are magic numbers so oxygen-16 is highly stable.

3. ¹²²Ba

Z = 56

N = 122-56 = 66

# nucleons is even, however, looking at the belt of stability you can see that this isotope actually lies just below the belt of stability, thus unstable.

4. ⁵⁸Ni

Z = 28

N = 58-28 = 30

# nucleons is even, and the proton number is a magic number, thus this element is highly stable.

5. ²⁰⁵Tl

Z = 81

N = 205-81 =124

# nucleons is odd, thus the element is unstable.

6. ²¹⁰Tl

Z = 81

N = 210-81 = 129

# nucleons is even, however, both N and Z are odd and Z is very close to a magic number and thus will decay to reach it; therefore unstable.

7. ²²⁶Ra

Z = 88

N = 226-88 = 138

# nucleons is even, however, looking at the belt of stability you see that this element is unstable because it is below the stable isotopes.

8. magnesium-24

Z = 12

N = 24-12 = 12

# nucleons is even, also Z<20 and the ratio of Z/N is 1:1 and thus stable.

Consistent and clear with each step.

Q21.7.2

Based on what is known about Radon-222’s primary decay method, why is inhalation so dangerous?

Radon is soluble in water and organic solvents. Atmospheric releases of radon-222 results in the formation of decay products that are radioisotopes of heavy metals and attach to airborne materials (dust) which makes it easy to inhale. Radon-222 is an alpha emitter and although alpha radiation is easily stopped by the skin, when ingested, our bloodstream and internal organs are highly vulnerable and these emitters shoot off helium nuclei which can damage the DNA of our cells. Alpha particles also have noticeably high "Relative biological effectiveness" meaning they are extremely damaging to our bodies. Radon can cause cell transformation, changes in chromosome structure and gene mutations.

Q20.4.12

Draw the cell diagram for a galvanic cell with an SHE and a zinc electrode that carries out this overall reaction: \(Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn_{2(aq)}^+ + H_{2(g)}\).

For cell notation there are a few rules:

• The anode (oxidation) half cell belongs on the left of the "||" symbol, cathode (reduction) on the right
• The reactants are stated before the products.
• The double vertical line represents a salt bridge or porous membrane separating the two oxidation and reduction half reactions. The single vertical line represents a phase change between two species in contact with one another (i.e. electrode (solid)| ion solution (liquid).
• Non-standard conditions are included in the parentheses with the phase notation.
• Spectator ions not included

With these rules, we can write:

\(Zn_{(s)}|Zn^{2+}_{(aq)}||H^+_{(aq)}|H_{2(g)}\)

easy to understand

Q20.4.14

Balance each reaction and calculate the standard reduction potential for each. Be sure to include the physical state of each product and reactant.

1. \(Cu^+_{(aq)} + Ag^+_{(aq)} → Cu_{(aq)}^{2+} + Ag_{(s)}\)

a) Separate the reaction into its oxidation and reduction half reactions:

oxidation: \(Cu^+_{(aq)}→ Cu^{2+}_{(aq)}\) -- we know this is oxidation because copper is losing an electron to become more positive

reduction: \(Ag^+_{(aq)}→ Ag_{(s)}\) -- we know this is reduction because silver is gaining an electron to become less positive, more negative

b) Make sure all elements are balanced in both half reactions, here they are.

c) Balance the charges of each side by adding electrons to the side with a more positive charge.

oxidation: \(Cu^+_{(aq)}→ Cu^{2+}_{(aq)}+e^-\)

reduction: \(Ag^+_{(aq)}+e^-→ Ag_{(s)}\)

d) Add the half reactions together, here the e^- cancel each other out, giving us the original equation, meaning it was given to us balanced.

\(\require{cancel}\cancel{e^-}+Cu^+_{(aq)} + Ag^+_{(aq)} → Cu_{(aq)}^{2+} + Ag_{(s)}+\cancel{e^-}\)

e) Using your earlier balanced half-reactions, you can find their standard reduction potentials from the table of standard reduction potentials:

oxidation: \(Cu^+_{(aq)}→ Cu^{2+}_{(aq)}\) E^o = 0.159V

reduction: \(Ag^+_{(aq)}→ Ag_{(s)}\) E^o = 0.7996V

f) \(E_{cell}^o=E^o_{cathode}-E^o_{anode}\Rightarrow\) 0.7996V-0.159V = 0.6406V

2. \(Sn_{(s)} + Fe^{3+}_{(aq)} → Sn^{2+}_{(aq)} + Fe^{2+}_{(aq)}\)

a) Separate the reaction into its oxidation and reduction half reactions:

oxidation: \(Sn_{(s)}→ Sn^{2+}_{(aq)}\) -- we know this is oxidation because tin is losing two electrons to become more positive

reduction: \(Fe^{3+}_{(aq)} →Fe^{2+}_{(aq)}\) -- we know this is reduction because iron is gaining an electron to become less positive, more negative

b) Make sure all elements are balanced in both half reactions, here they are.

c) Balance the charges of each side by adding electrons to the side with a more positive charge.

oxidation: \(Sn_{(s)}\rightarrow Sn^{2+}_{(aq)}+2e^-\)

reduction: \(Fe^{3+}_{(aq)}+e^-\rightarrow Fe^{2+}_{(aq)}\)

[Notice here that the two half-reactions do not have the same number of electrons. For them to cancel out in the next step, we have to multiply the reduction half-reaction by 2]

oxidation: \(Sn_{(s)}\rightarrow Sn^{2+}_{(aq)}+2e^-\)

reduction: \(2Fe^{3+}_{(aq)}+2e^-\rightarrow 2Fe^{2+}_{(aq)}\)

d) Add the half reactions together, here the e^- cancel each other out, giving us a new, balanced equation.

\(\require{cancel}\cancel{2e^-}+Sn_{(s)}+2Fe^{3+}_{(aq)}\rightarrow Sn^{2+}_{(aq)}+2Fe^{2+}_{(aq)}+\cancel{2e^-}\)

e) Using your earlier balanced half-reactions, you can find their standard reduction potentials from the table of standard reduction potentials:

oxidation: \(Sn_{(s)}\rightarrow Sn^{2+}_{(aq)}+2e^-\) E^o = -0.14V

reduction: \(2Fe^{3+}_{(aq)}+2e^-\rightarrow 2Fe^{2+}_{(aq)}\) E^o = 0.771V

f) \(E_{cell}^o=E^o_{cathode}-E^o_{anode}\Rightarrow\) 0.771V-(-0.14V) = 0.911V

3. \(Mg_{(s)} + Br_{2(l)} → 2Br^−_{(aq)} + Mg^{2+}_{(aq)}\)

a) Separate the reaction into its oxidation and reduction half reactions:

oxidation: \(Mg_{(s)}\rightarrow Mg^{2+}_{(aq)} \) -- we know this is oxidation because magnesium is losing two electrons to become more positive

reduction: \(Br_{2(l)}\rightarrow 2Br^−_{(aq)}\) -- we know this is reduction because bromine is gaining an electron to become less positive, more negative

b) Make sure all elements are balanced in both half reactions, here they are.

c) Balance the charges of each side by adding electrons to the side with a more positive charge.

oxidation: \(Mg_{(s)}\rightarrow Mg^{2+}_{(aq)}+2e^-\)

reduction: \(Br_{2(l)}+2e^-\rightarrow 2Br^−_{(aq)}\)

d) Add the half reactions together, here the e^- cancel each other out, giving us the original equation, meaning it was given to us balanced.

\(\require{cancel}\cancel{2e^-}+Mg_{(s)} + Br_{2(l)} → Mg^{2+}_{(aq)} +2Br^−_{(aq)}+\cancel{2e^-}\)

e) Using your earlier balanced half-reactions, you can find their standard reduction potentials from the table of standard reduction potentials:

oxidation: \(Mg_{(s)}\rightarrow Mg^{2+}_{(aq)}+2e^-\) E^o = -2.356V

reduction: \(Br_{2(l)}+2e^-\rightarrow 2Br^−_{(aq)}\) E^o= 1.087V

f) \(E_{cell}^o=E^o_{cathode}-E^o_{anode}\Rightarrow\) 1.087V-(-2.356V) = 3.443V

no mistakes

Q14.1.3

What is the relationship between each of the following factors and the reaction rate: reactant concentration, temperature of the reaction, physical properties of the reactants, physical and chemical properties of the solvent, and the presence of a catalyst?

Adding a catalyst, increasing the temperature of the reaction, and increasing the reactant concentration will all increase the rate of the reaction. Depending on the physical and chemical properties of the solvent, the reaction rate could increase, decrease or stay the same. High solubility will increase the reaction rate but increasing solvent viscosity will decrease the reaction rate. Solvent polarity can make the rate increase or decrease.

couldn't have said it better