Extra Credit 4

- (Edits made by the Phase II Student)

Q19.1C

For the half reaction, M2+(aq)+2e−→M(s), what range of E0 is possible if:

M displaces MnO−4(aq), but not AgCl(s). M is unable to react with HCl, but reacts with HNO3.

ANSWER:

State what you know about the reaction.

• M displaces MnO₄^-(aq), but not AgCl(s)
• M is unable to react with HCl, but reacts with HNO₃

Since M can displace MnO₄^-(aq) in this reaction, the standard reaction potential E₀ could be up to +.56, however, E₀ cannot be lower than +.2223 since M cannot displace AgCl(s).

Therefore,

+.2223V < E₀ < +.560

For more help on this topic visit: Electrochemistry - Libretexts.

Q19.31A

Calculate the theoretical cell voltage for the reaction between copper and zinc given that the overall reaction is:
Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)

ANSWER:

Start by separating the overall redox reaction into its two half reactions, then use a standard reduction potentials table to find each E° to identify the anode and cathode.

Zn(s) → Zn²⁺(aq)+2e^- and Cu(s)→ Cu²⁺(aq)+2e^-

ANODE: Zn(s) → Zn²⁺(aq)+2e^- (Oxidation) E^o=-(-.763V)

CATHODE: Cu²⁺(aq)+2e^- → Cu(s) (Reduction) E^o=-.340V

E^ocell= E(cathode) - E(anode)

E^ocell=.763V+.340V

E^ocell=+1.103V

To get more help on problems like these refer to: The Cell Potential - Libretexts.

Q20.13A

Draw mer and Fac isomers for the ZA3B3 complex. Assume A and B and different ligands and Z is a random metal.

ANSWER:

• The fac/mer isomerism occurs only in octahedral complexes when 3 identical ligands are either neighboring (fac) or across (mer) from eachother.
• Fac occurs when 3 identical ligands are bonded neighboring each other. In other words, these 3 ligands are each on different axes. A good way to remember this type of isomer would be to relate "fac" with facial, meaning on the same face/side of the complex when seen in three dimensions.
• Mer (think of meridian) occurs with 3 identical ligands, and 2 of the 3 are bonded on the same axis. Visualizing the complex in three dimensions helps us picture meridinal isomers have identical ligands along a meridian, in other words, on the same plane of the complex.

Scan_1.pdf

Q21.23A

[NiCl4]3−NiCl43is diamagnetic. Use the crystal field theory to speculate on its possible structure.

ANSWER:

Identify paramagnetic and diamagnetic.

Paramagnetic = unpaired electrons

Diamagnetic = paired electrons

Strong field tends to have more paired electrons and as a result are typically diamagnetic.

Weak field tends to have more unpaired electrons and as a result are typically paramagnetic.

Identify the charge of the metal.

Cl: (-1)x4=-4

Ni: +1

Overall charge: -4+1=-3

Identifying the number of electrons in D orbital.

Ni: [Ar]4s²3d⁸

Ni⁺¹: [Ar]4s3d⁸

Cl is a weak ligand which means it is high spin.

8 electrons in D orbital.

Scan.pdf

All electrons are paired, therefore, it is diamagnetic.

If you need more help distinguishing between paramagnetic and diamagnetic metals, refer to: Mangetic Properties - Libretexts.

Q24.33D

The reaction A+B→C+D. A is second order and first order with respect to B. The value of k
is .0351 M^-1min^-1. What is the rate of this reaction when [A]=0.120M and [B]=4.6M?

ANSWER:

The problem states that A is in second order and B is in first order, meaning the rate is:

Rate=K[A]²[B]

Now you substitute the values into the equation.

Rate=(.0351 M^-1mm^-1)(.12)²(4.6)

Rate=.0023 M/min.

For more help with Rate Law problems take a look at: The Rate Law - Libretexts

Q25.11C

Complete the equations of the nuclear reactions below

1. bombardment of ⁹Be with protons to produce ¹⁰Be and γ rays
2. bombardment of ¹¹B with ²H to produce ¹²C
3. bombardment of ¹⁶O with neutrons to produce ¹⁶N

ANSWER:

Identify the types of radioactive decay first.
Recall the different values of each particle: ⁴₂He, ⁰₁B, ⁰₀Y, ¹₀n, ¹₁H.

1. ⁹₄Be+¹₁H→¹⁰₅B+γ
2. ¹¹₅B+²₁H→¹²₆C+¹₀n
3. ¹⁶₈O+¹₀n→¹⁶₇N+¹₁H

Be sure to make sure each equation is balanced correctly, taking into account both the mass number (x) and the atomic number (y) of each element (z) involved in the nuclear reaction (^x_yZ).

For more help with radioactive decay problems, refer to: Nuclear Kinetics - Libretexts.

Q25.42D

Which member of the following pairs of nuclide would you expect most abundant in natural sources. Explain your answer

1. Y-89 or Y-91
2. Cs -132 or Cs- 130

ANSWER:

Recall that nuclides that are closest to their natural mass are the most stable forms and are more abundant in nature than nuclides that have masses far from their normal state masses.

A) Y has an atomic mass of 88.91, meaning the nuclide most abundant would be the one closest to its mass, in this case it is Y-89.

B)Cs has an atomic mass of 132.9, meaning the nuclide most abundant would be the one closest to its mass, which would be Cs-132.

Q21.2.33

Calculate the amount of energy released or required by the fusion of helium-4 to produce the unstable beryllium-8 (mass = 8.00530510 amu). Report your answer in kilojoules per mole. Do you expect this to be a spontaneous reaction?

ANSWER:

Write out the radioactive decay to identify the missing reactant.

¹²₆C -> ⁸₄Be+⁴₂He

We are solving for the energy change, but we must first calculate the change in masses.
Recall Einstein's equation: ΔE = (Δm)c² for this problem.

(ΔM)= (ΔE)/(C²)

(ΔM)= mass of products - mass of reactants.

(ΔM)=12.0079071-12.011

(ΔM)=-.0030929

We now have the change in mass, now we can plug in the values into the equation.

ΔE = (ΔM)(C²)

(ΔE)=(-3.093x10^-6kg)(2.998x10⁸m/s)²

(ΔE)=-2.779x10¹¹(kgxm²)/s²

The answer is given to us in J, we must convert to kJ.

(ΔE)=-2.78x10¹¹J x 1kJ/1000J

(ΔE)=-2.78x10⁸kJ

The reaction would not be spontaneous because ΔE is inversely proportional to ΔG, given the Nernst equation: ΔG = -nFE.
This means, ΔG would be positive making it a non-spontaneous reaction.

For more background regarding Einstein's equation, refer to: Wave-Particle Duality - Libretexts.