Extra Credit 38

Question 19.16A :

Find the weight of MnO₂ in grams from the following overall reaction knowing that the voltmeter reading at non standard state is 0.25V, the pH of the solution is 6 and the solution is diluted to 0.6L.

Pt(s)|Cr³⁺(0.2M),Cr₂O₇^2-(0.15M)||MnO₄(xM)|MnO₂(s)

Solution:

The question wants us to find the mass of MnO₂ in grams. This will be found using the molarity and molar mass of MnO_2. To find the molarity, we will use the Nernst equation. The first step is to write the oxidation and reduction half equations as follows:

Reduction Half:

2MnO₄^- (aq) + 8H⁺(aq) + 6e^- → 2MnO₂(s) + 2H₂O(l) Eº_red = +1.51V

Oxidation Half:

2Cr³⁺ (aq) + 7H₂O(l) → Cr₂O₇^2-(aq) + 14H⁺(aq) + 6e^- Eº_ox = +1.33V

The reduction potentials of both the oxidation and the reduction halves can be looked up in the table of standard reduction potentials. The next step is to find the Eº_cell.

Eº_cell = Eº_red - Eº_ox

= 1.51V - 1.33V

= 0.18V

Now, use the Nernst equation to find molarity of MnO₄^-.

\[E_{cell} = Eº_{cell} - \frac{0.0592}{6} + log\frac{[H^+]^6[Cr_2O_7^{2^-}]}{[Cr^{3+}][MnO_4^-]^2}\]

In this equation, [Cr₂O₇^2-] = 0.15 M and [Cr³⁺] = 0.2 M. The concentration of H⁺ can be found using the pH value.

\(pH = -log[H^+] = 6\)

⇒ \([H^+] = 10^{-6}\)

So,

\(0.25 = 0.18 - \frac{0.0592}{6} + log\frac{[H^+]^6[Cr_2O_7^{2-}]}{[Cr^{3+}][MnO_4^{-}]^2}\)

⇒ \(-10.135 = log\frac{[H^+]^6[Cr_2O_7^{2-}]}{[Cr^{3+}][MnO_4^{-}]^2}\)

⇒ \(7.3\times 10^{-11} = \frac{[H^+]^6[Cr_2O_7^{2-}]}{[Cr^{3+}][MnO_4^{-}]^2}\)

⇒ \(7.3\times 10^{-11}=\frac{[10^{-6}]^6[0.15]}{[0.2][MnO_4^{-}]^2}\)

⇒ \([MnO_4^{-}]^{-2} = 2.2\times 10^{-48}\)

⇒ \([MnO_4^{-}] = 1.48\times 10^{-24}\)

Now that we have the molarity of MnO₄^-, we can calculate the mass of MnO₂ using the stochiometric ratio between the two and the molar mass of MnO₂ as follows:

Mass of MnO₂ = \(0.6L \times \frac{1.48\times 10^{-24}}{1L of solution}\times \frac{2 mol MnO_2}{2 mol MnO_4^{-}}\times \frac{86.94g MnO_2}{1 mol MnO_2}\)

⇒ Mass of MnO₂ = \(7.73\times 10^{-23}\) grams.

Question 20.1 C:

Draw the orbital diagrams of the following atom and ions:

1. V
2. Cr³⁺
3. Co²⁺
4. Ti³⁺ (only theoretic)
5. Zn²⁺
6. Co⁴⁺ (only theoretic)

Solution:

See attached image file for solutions.

Question 21.11 B:

Which of these general structures for a complex ion would you expect to exhibit fac and mer isomerism? Explain.

1. tetrahedral
2. square-planar
3. octahedral

Solution:

1. Firstly, for a complex ion to exhibit fac and mer isomerism, it needs to have at least 6 ions. This is because fac isomerism occurs when 3 ions of the same kind form a plane of the octahedron. Mer isomerism occurs when 3 ligands of the same kind are in the same plane which also contains the central atom. Since the only structure from the given choices that can contain 6 ligands is octahedral, the right answer is octahedral. More information about fac and mer isomerism can be found here.

Question 24.19A

The reaction A→products is first order in A

1. If 4.2g A decomposes for 45 minutes, the undecomposed A is measured to be 1.05g. What is the half-life, t_1/2, of this reaction?
2. Starting with the same 4.2 g, what is the mass of undecomposed A after 75 minutes?

Solution:

1.) This is a simple problem about Radioactivity which requires us to use the equation

\[ln\frac{[A]}{[A_º]} = -kt\]

where A = 4.2 grams; A_º = 1.05 grams and t = 45 minutes. We can calculate k using this equation and then plug in k in the first order half life formula which is

\[t1/2= \frac{0.693}{k}\]

So, plugging in all the value that we have in the first equation,

\(ln\frac{1.05}{4.2} = -k \times 45\)

⇒ \(-1.386 = -k \times 45\)

⇒ \(-k = -\frac{1.386}{45}\)

⇒ k = 0.0308

Now we plug in this value of k in the half life formula above.

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

⇒ \(t_{1/2} = \frac{0.693}{0.0308}\)

⇒ t_1/2 = 22.5 minutes.

Hence, the half life of this reaction is 22.5 minutes.

2.) For this part also, we will use the same equation except that now the value of t is 75 minutes and we will solve for A as follows:

\(ln\frac{A}{4.2} = -0.0308 \times 75\)

⇒ \(ln\frac{A}{4.2} = -2.31\)

⇒\(\frac{A}{4.2} = e^{-2.31}\)

⇒ A = \(0.099\times 4.2\)

⇒ A = 0.417 grams

Therefore, the mass of undecomposed A after 75 minutes is 0.417 grams.

Question 25.1E

Write the Equation for each process

1. \({^{231}_{91}Pa^{}}\) alpha decay
2. \({^{249}_{96}Cm^{}}\) beta emission

Solution

1. \({^{231}_{91}Pa^{}}→{^{227}_{89}}Ac+{^4_{2}}He^{}\)

2. \({^{249}_{96}Cm^{}}→{^{249}_{97}}Bk+{^0_{-1}}e^{}\)

Question 25.37B

Calculate the energy in MeV, released in the below nuclear reaction

\[{^{6}_{3}Li^{}}+2({^{4}_{2}}He)→{^{13}_{6}C^{}}+{^1_{1}}H^{}\]

The nucleic mass

• \({^{6}_{3}Li^{}}\)=6.01514u,
• \({^{13}_{6}C^{}}\)=13.00335u,
• \({^1_{1}}H^{}\)=1.00783u, and
• \({^{4}_{2}}He\)=4.0026u.

Solution

We will find E using Einstein's equation \[E=\Delta mc^2\]

where \(\Delta m\) is the mass defect and c is the speed of light (\(2.98\times10^8\)). The mass defect can be easily calculated as follows:

\(\Delta m\) = \((6.01514+(2\times4.0026))-(13.0035+1.00783)=0.00901 u\)

Using \(E=\Delta mc^2\), we can now calculate the Energy

\(E = 0.00901\times(2.98\times10^8)^2\)

⇒ \(E = 8.00124\times10^{14} J\)

Lastly, we just have to convert Joules into MeV. The conversion factor for that is \(1J = 6.242\times10^{12} MeV\)

\(8.00124\times10^{14}J \times\frac{6.242\times10^{12}MeV}{1J}= 4.99\times10^{27} MeV\)

Question 21.2.6

Describe the energy barrier to nuclear fusion reactions and explain how it can be overcome.

Solution

Nuclear fusion is the opposite of nuclear fission where two lighter nuclei come together to form a heavier and more stable nuclei. Because the charges of both the nuclei are the same, there is a huge electrostatic energy barrier. For the nuclei to overcome this huge energy gap, a lot of energy is required. This is why these reactions occur at extremely high temperatures so that the nuclei have enough kinetic energy to overcome this energy gap. So, this electrostatic energy gap can be overcome by giving the nuclei enough kinetic energy.

For more information on Nuclear Fusion and Fission reaction, click here.

Question 20.3.4

What is the purpose of a salt bridge in a galvanic cell? Is it always necessary to use a salt bridge in a galvanic cell?

Solution

The purpose of the salt bridge is to maintain electrical neutrality in the overall galvanic cell. Without the salt bridge, electron flow from anode to cathode would result in a charge build-up near the 2 electrodes which would eventually stop the flow of the current. Therefore, it is always necessary to use a salt bridge in a galvanic cell.

For more information on Galvanic cell and salt bridges, click here.