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).

The inert electrode is "B" because it does not gain or lose mass, which means that it was not involved in the reaction but possibly used to transfer electrons between the solutions, as is the case for carbon and platinum electrodes. "A" increases in mass and so it gains electrons which pull the metallic ions from the solution into a precipitate on the electrode. This means that "A" is the cathode being reduced. "C" is losing mass and therefore losing electrons and producing ions in the solution. "C" is the anode which is being oxidized. "A" has higher reduction potential than "C" therefore.

Q19.1.19

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

1. is added to a solution of in acid.
   • Oxidized half-reaction:
   • Reduction half reaction:
   • Overall reaction:
   • Chromium will precipitate out of the solution because it has a higher reduction potential than Zinc; the reaction is a single replacement.
2. is added to a solution containing an excess of in hydrochloric acid.
   • Dissociation reaction:
   • Oxidation half-reaction:
   • Reduction half-reaction:
   • Overall reaction:
   • The reduction potential for permanganate is larger so the reaction is still favorable even when the oxidation of is negative.
3. is added to in acid solution.
   • Reduction half-reaction:
   • Oxidation half-reaction:
   • Overall reaction:
   • The reaction is favorable with a high positive
4. is heated with .
   • Reduction half-reaction:
   • Oxidation half-reaction:
   • Overall reaction:
   • Heat creates a product with higher energy than both previous reactants.
5. is added to in water.
   • Strong acid dissociation:
   • Overall reaction:
   • This reaction works by exchange of electrons to yield Chromium ions.
6. is added to an aqueous solution of .
   • Overall reaction:
   • Iron hydroxide will precipitate because the two metals will exchange anions.

Q19.3.11

Would you expect the to be diamagnetic or paramagnetic? Explain your reasoning.

The complex will be paramagnetic because:

• Cyanide is a strong field ligand because its electrons are localized at specific locations. This gives the complex a low spin.
• The Chromium is attached to six ligands which means it has a octahedral structure .
• The oxidation number of the chromium is (+3) because the combined charge of the cyanide ligands will be (-6) (cyanide is always \(CN^{-1}\), adding three magnesium ions from the solution would add (+6) charge to balance the negative charge left.
• The chromium therefore has only three valence electrons in the orbital.
• This means that the will be filled with the three electrons and have each one be unpaired, hence, the metal is paramagnetic.

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 \(\frac{mol}{L}\) if the reaction is (a) first order with respect to A or (b) second order with respect to A?

First, for each order, find the rate of the reaction .

A) is independent of the initial concentration of the compound. This means that all first-order reactions will go at the same rate.

Now, by plugging in the give values to the equation we obtain:

B) The rate constant for second order depends on initial concentration.

Plugging into the second order equation we obtain:

Note that the for each order cancels out to yield the units for time. The rule is and .

Q21.2.6

Calculate the density of the nucleus in , assuming that it has the typical nuclear diameter of cm and is spherical in shape.

To find density we use:

The volume of a sphere is given by the equation:

the radius of the sphere is half the diameter, so =.

Therefore,

Note that .

The sum of the masses of the protons and neutrons gives us the Mass:

Mass of proton=

Mass of neutron=

Magnesium has 12 protons and 12 neutrons:

Returning to the original equation:

This is extremely dense! which shows how the majority of the universe is essentially empty space.

Q21.6.1

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

is a dynamic equilibrium?

By introducing a stable solid of into a saturated solution containing either radioactive or . As time passes some of the nonradioactive solid will dissolve and the radioactive ions will precipitate. This would mean that the reaction is going in both directions even though it is at its lowest energy configuration.

Q20.4.7

Identify the oxidants and the reductants in each redox reaction.

1. 
   • The oxidant is the element that gets reduced which is , because it gains two electrons and becomes a solid.
   • The reductant loses electrons so it is , because it becomes an ion in the solution.
2. 
   • The oxidant is because it begins with an oxidation state of (0) as elemental form and becomes(-1).
   • The reductant is because it gives two of electrons and becomes .
3. 
   • is the reductant because it is oxidized into an ion in the solution.
   • is the oxidant because Arsenic starts with a (+5) oxidation state and becomes (-3).
4. 
   • acts as the reductant and the oxidant both gaining an electron and donating one from itself ( gains the electron from , which loses it when it takes an oxygen from )

This reaction is characteristic of a lead storage battery:

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

If we suppose that the battery has a volume of , then:

30.0% of the battery is sulfuric acid by mass, so:

Dividing by the molar mass we obtain moles of sulfuric acid:

Now we can find the molarity to determine how far the conditions are from the standard ones. We use the same volume we initially proposed which is the same as .

Standard conditions are when the molarity of the reactants is equal to one. Therefore, the molarity potential here is greater than the standard one.