Extra Credit 21

Q17.2.10

The mass of three different metal electrodes, each from a different galvanic cell, were determined before and after the current generated by the oxidation-reduction reaction in each cell was allowed to flow for a few minutes. The first metal electrode, given the label A, was found to have increased in mass; the second metal electrode, given the label B, did not change in mass; and the third metal electrode, given the label C, was found to have lost mass. Make an educated guess as to which electrodes were active and which were inert electrodes, and which were anode(s) and which were the cathode(s).

Solution: When an electrode goes through a redox reaction, electrons are either released or gained, depending on whether it is an anode or cathode. If the electrode is an active anode, it will release its electrons, and the oxidized substance will go into solution, causing the electrode itself to lose its mass as the reaction proceeds. In this case, since C lost its mass, it is an active anode. If the electrode is an active cathode, it will accept the electrons released by the anode. This causes the ions in the cathode compartment to plate onto the cathode, causing the electrode to gain mass. Electrode A would then be the active cathode. This leaves Electrode B to be the inert electrode because there are no electrons being released or gained causing the mass to remain constant.

Q19.3.11

Would you expect the Mg3[Cr(CN)6]2 to be diamagnetic or paramagnetic? Explain your reasoning.

Solution: When first determining if a complex ion is diamagnetic or paramagnetic, you have to take note of whether the ligand attached to the transition metal is a strong-field ligand or weak-field ligand by looking at the spectrochemical series. This will give us an idea of the difference in energy between t_2g and e_g. In this problem, the CN^- ligands attached to the transition metal are strong-field. This creates a big energy difference between the two levels, which makes it harder for electrons to overcome the energy difference. As a result, the electrons will pair up in the t_2g energy level, making this compound diamagnetic. Another way to look at magnetism is through the number of electrons.

Q12.4.11

The half-life of a reaction of compound A to give compounds D and E is 8.50 min when the initial concentration of A is 0.150 mol/L. How long will it take for the concentration to drop to 0.0300 mol/L if the reaction is (a) first order with respect to A or (b) second order with respect to A?

Solution

First, let's write down all the information we have from this problem:

t_1/2=8.50 minutes

[A₀]=0.150 mole/L

a)The problem tells us that this reaction is first order with respect to A. So, we can set up the first order integrated rate law that corresponds to first order reactions. The reason why we are using the integrated rate law, is because we are solving for a specific time value that will tell us how long it takes for the concentration to drop to a specific concentration. However, we need to find the rate constant first in order to solve for the time in our integrated rate law. We will use the first order half time reaction to find t.

t_1/2= 0.693/{k}

k=0.0815 1/s

Now, we can use the rate constant to solve for t.

ln[A]_t=-kt+ln[A]_o

-3.51=ln[0.0300]_t=-0.0815t+ln[.150]$$

-3.51=-0.0815t+-1.89712

t=19.8 minutes

b) Now, will follow the same steps as part a, except, we will use the second order integrated rate law and use the second order half time formula to find k in respect to A.

t_1/2=$$\frac{1}{(k)[A]_0}$$

= 8.50=$$\frac{1}{(k)[.150]} $$

k=0.784

$$\frac{1}{[A]_t}=0.7874t+\frac{1}{[0.150]_0}$$

=$$\frac{1}{[0.300]_t}=0.0815t+\frac{1}{[0.150]_0}$$

t=34 minutes

Q21.2.6

Calculate the density of the \frac{24}{12}Mg nucleus in g/mL, assuming that it has the typical nuclear diameter of 1 × 10–13 cm and is spherical in shape.

Solution: First, we have to find the mass of the nucleus in grams. From the given mass number of Magnesium, we know that there are 12 protons and 12 neutrons. We will be using conversion factors to convert the mass of the protons and neutrons in amu to grams.

First we will find the mass of protons and neutrons in amu using the standard masses of each in order to add them up allowing us to solve for the change in mass:

proton~ 12(1.007316 amu) =12.087792 amu

$$12.087792 amu (\frac{1.6727 x 10^{-24} g}{1 amu})= 2.02192497(10^{-23}) grams$$

neutron~12(1.008701 amu) = 12.1044 amu

$$12.1044 amu (\frac{1.6750 x 10^{-24} g}{1 amu})=2.02748901(10^{-23}) grams$$

When we add the masses together, we get= $$4.00941(10^{-23})$$ grams

The problem tells us that the diameter is $$1 × 10^{13}$$ cm. However, we have to cube this value, because it is a sphere, which will make the nuclear diameter of the sphere equal $$1 × 10^{-39} cm^3 $$. Since $$1cm^3=1 mL$$, we can divide the mass of the nucleus by the diameter of the sphere to get the density of the nucleus.

$$\frac{4.00941(10^{-23}) grams}{1 × 10^{-39} cm^3}$$ g/mL

Q21.6.1

How can a radioactive nuclide be used to show that the equilibrium:

$$AgCl(s)⇌Ag^+(aq)+Cl^−(aq)$$$

is a dynamic equilibrium?

The double arrows in the equation symbolize that the forward and reverse reaction are occurring at an equal rate, meaning that there is net charge of zero. When a nuclide, such as the positive +1 Ag ion, or -1 Cl ion, is added to the reaction, there will be a continuation of radioactivity because free electrons are being released. And, as long as the nuclides are in the dynamic equilibrium reaction, there will be a constant shift between the precipitate and the ionized nuclides. So in conclusion there is a dynamic equlibrium.

Q20.4.7

Identify the oxidants and the reductants in each redox reaction.

The oxidants, which are also known as oxidizing agents, are the elements that oxidize other elements and the reductants which are also called the reducing agents, reduce other elements.

1. $$Cr(s) + Ni^{2+}(aq) → Cr^{2+}(aq) + Ni(s)$$

oxidant- Ni^{2+} because the nickel accepts chromium's electrons.

reductant- Cr(s) because it donates 2 electrons to the nickel.

2.$$Cl_{2}(g) + Sn^{2+}(aq) → 2Cl^{−}(aq) + Sn^{4+}(aq)$$

oxidant-$$Cl_{2}(g)$$, because in accepts Sn electrons.

reductant- $$Sn^{2+}(aq)$$, because it donates 2 electrons to the chlorine

3.$$H_{3}AsO_{4}(aq) + 8H^{+}(aq) + 4Zn(s) → AsH_{3}(g) + 4H_{2}O(l) + 4Zn^{2+}(aq)$$

oxidant-$$H_{3}AsO_{4}(aq)$$, gets 2 electrons from Zinc

reductant-Zn(s), because it transfers 2 electrons to As

1. $$2NO_{2}(g) + 2OH^{−}(aq) → NO_2{−}(aq) + NO_{3}^{−}(aq) + H_{2}O(l)$$

oxidant-$$NO_{2}(g)$$ because it gains electrons, producing $$NO_{3}^{−}$$

reductant-$$NO_{2}(g)$$, because it gives electrons, producing $$NO_2{−}$$

Q20.7.5

This reaction is characteristic of a lead storage battery:

Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)

If you have a battery with an electrolyte that has a density of 1.15 g/cm3 and contains 30.0% sulfuric acid by mass, is the potential greater than or less than that of the standard cell?

Q20.7.5

In order to answer this question we must first look at the modifed version of the Nernst equation.$$E_{cell}=E^o-\frac{RT}{nF}lnQ$$

In this equation F and R are just constants. To solve this problem all that you need to do is solve for the concentration of your H₂SO_4. Since we are given density and percent mass we can use this to find our concentration.$$98.079g\frac{1cm^3}{1.15g}\frac{1L}{1000cm^3}\frac{30g}{100L}$$

Next take the molarity and take the natural log. From here you will see that your calculated value will be negative. This already tells you that your E_cell will be smaller than your E^o.

Q19.1.19

Predict the products of the following reactions and balance the equations.

1. Zn is added to a solution of Cr2(SO4)3 in acid.
2. FeCl2 is added to a solution containing an excess of Cr2O2−7Cr2O72 in hydrochloric acid.
3. Cr2+ is added to Cr2O2−7Cr2O72 in acid solution.
4. Mn is heated with CrO3.
5. CrO is added to 2HNO3 in water.
6. FeCl3 is added to an aqueous solution of NaOH.

1. Cr₂(SO4)₃(aq)+2Zn(s)+2H₃O⁺(aq)⟶2Zn²⁺(aq)+H₂(g)+2H₂O(l)+2C^r2+(aq)+3SO₂⁻⁴(aq)
2. 4TiCl₃(s)+CrO₂⁻⁴(aq)+8H⁺(aq)⟶4Ti⁴⁺(aq)+Cr(s)+4H₂O(l)+12Cl⁻(aq)
3. 3Cr₂⁺(aq)+CrO₂⁻⁴(aq)+8H₃O⁺(aq)⟶4Cr³⁺(aq)+12H₂O(l)
4. 8CrO₃(s)+9Mn(s)⟶4Cr₂O₃(s)+3Mn₃O₄(s)
5. CrO(s)+2H₃O⁺(aq)+2NO⁻³(aq)⟶Cr²⁺(aq)+2NO⁻³(aq)+3H₂O(l)
6. FeCl₃(s)+3NaOH(aq)⟶Fe(OH)₃(s)+3Na⁺(aq)+3Cl⁻(aq)