Extra Credit 46

Q17.6.5

Why would a sacrificial anode made of lithium metal be a bad choice despite its E∘Li+/Li=−3.04V which appears to be able to protect all the other metals listed in the standard reduction potential table?

Q 17.6.5 Solution

A sacrificial anode is a substance that will be oxidized first, therefore protecting other substances from getting oxidized. This requires a substance to have very small reduction potential, which means it will be easily oxidized and likely to lose electrons. Lithium has a very low reduction potential, so it fits that criteria and should be a good sacrificial anode, but Lithium is an alkali metal. This group is characteristically reactive with water, so it is likely that Lithium would react quickly with water and have to be replaced and not be able to serve its purpose as a sacrificial anode very well.

Q12.3.9

What is the instantaneous rate of production of N atoms Q12.3.8 in a sample with a carbon-14 content of 1.5 × 10⁻⁹ M?

Q12.3.8 for reference

The rate constant for the radioactive decay of ¹⁴C is 1.21 × 10⁻⁴ year⁻¹. The products of the decay are nitrogen atoms and electrons (beta particles):

⁶₁₄C→⁶₁₄N+e⁻

rate=k[⁶₁₄C]

What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of 6.5 × 10⁻⁹ M?

Q 12.3.9 Solution

It is given that the rate law for the radioactive decay of ⁶₁₄C is rate=k [ ⁶₁₄C] and it is given that the rate constant for the radioactive decay of ⁶₁₄C is 1.21 x 10⁻⁴ year⁻¹, which is equal to k

The given equation is:

⁶₁₄C→⁶₁₄N+e⁻

It is also given that carbon-14 content in the sample is 1.5 x 10−9 M

To find the instantaneous rate of production of N atoms in this sample, we need to just plug the given values into the rate law since the rate of consumption of Carbon is equal to the rate of production of Nitrogen.

So, rate=k [⁶₁₄C] becomes rate= (1.21 x10⁻⁴ year⁻¹)[ 1.5 x 10⁻⁹ M]

rate= 1.82 x 10^-13 M/year

Q12.6.1

Why are elementary reactions involving three or more reactants very uncommon?

Q 12.6.1 Solution

Elementary reactions involving three or more reactants are very uncommon because it requires the three reactants to collide with sufficient energy and proper energy for the collision to be successful. The more reactants there are, the more unlikely it is that all of the particles will be at the same place at the same time to collide and create a reaction.

Q21.4.13

Define the term half-life and illustrate it with an example.

Q 21.4.13 Solution

Half-life is defined as the amount of time needed for a reactant concentration to decrease by half compared to its initial concentration. An example is

Given the first-order reaction:

C ⟶ D

The initial concentration of C is 20 M and C has a half life of 10 days. What is the concentration of C after 10 days?

Since the half-life of C is 10 days, we know that the concentration should decrease by half in that time.

The concentration of C after 10 days is 10 M.

Q20.3.1

Is 2NaOH(aq)+H₂SO₄(aq)→Na₂SO₄(aq)+2H₂O(l) an oxidation–reduction reaction? Why or why not?

Q 20.3.1 Solution

2NaOH(aq)+H₂SO₄(aq)→Na₂SO₄(aq)+2H₂O(l) is NOT an oxidation-reduction reaction. An oxidation-reduction reaction is defined as a reaction that involves the transfer of electrons.

To determine whether an equation is an oxidation-reduction reaction, we must assign oxidation states to each of the elements and see if any of the species lose or gain electrons.

On the reactants side, we have 2NaOH (aq) with Na having an oxidation state of +1, O is -2 and H is +1. The other reactant is H₂SO₄ (aq) with H having an oxidation state of +1, S is +6, and O is -2.

On the products side we have Na₂SO₄(aq) with Na having an oxidation state of +1, S is +6 and O is -2. The other product is H₂O(l) and H has an oxidation state of +1 and O is -2.

Now we compare the oxidation states of the elements on the reactants side and the products side.

On the reactants side we have: Na= +1, O=-2, H=+1, S=+6

On the products side we have: Na=+1, O=-2, H=+1, S=+6

Since none of the oxidation states of any of the species change oxidation states, we can conclude there was no transfer of electrons and therefore this is not an oxidation-reduction reaction.

Q20.5.12

How many electrons are transferred during the reaction Pb(s) + Hg₂Cl₂(s) → PbCl₂(aq) + 2Hg(l)? What is the standard cell potential? Is the oxidation of Pb by Hg₂Cl₂ spontaneous? Calculate ΔG° for this reaction.

Q 20.5.12 Solution

Given the equation:

Pb(s) + Hg₂Cl₂(s) → PbCl₂(aq) + 2Hg(l)

We are asked to determine the number of electrons transferred during the reaction. To do this, we must first split up the reaction into oxidation and reduction half reactions.

Oxidation half reaction involves a species being oxidized, or losing an electron and reduction half reaction involves a species being reduced, or gaining an electron.

Oxidation half reaction: Pb(s)→Pb²⁺(aq)+2e^-

Oxidation state of Pb increases from 0 to 2+

Reduction half reaction: Hg²⁺(aq)+2e⁻→Hg(l)

Oxidation state of Hg decreases from 2+ to 0

The number of electrons transferred during these reactions are already the same, therefore number of electrons transferred is 2

Next, we are asked to find the standard cell potential.

To do this, we must use the formula

E⁰_cell=E⁰_cathode−E⁰_anode

The cathode is where reduction occurs so we use Hg²⁺(aq)+2e⁻→Hg(l) which has E⁰ = +0.855V

The anode is where oxidation occurs so we use Pb(s)→Pb²⁺(aq)+2e^- which has E⁰= -0.125V

We then just plug into the above formula and get E⁰_cell=+0.855-(-0.125)= 0.98 V

E⁰_cell=0.98 V

Since E⁰_cell>0, the oxidation of Pb by Hg₂Cl₂ is spontaneous.

Next, we are asked to calculate ΔG°

We must use the formula ΔG°= −nFE⁰_cell and plug in values, with F= Faraday's constant (96.485 kJ/V*mol)

ΔG° = -(2 moles of electrons transferred)*(96.485 kJ/V*mol)*(0.98 V)

ΔG°= -189.1 kJ

Q24.6.8

For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

1. [Cu(NH₃)₄]²⁺
2. [Ni(CN)₄]²⁻

Q 24.6.8 Solution

For each complex, predict its structure, whether it is high spin or low spin, and the number of unpaired electrons present.

a. [Cu(NH₃)₄]²⁺

This complex has a coordination number of 4, so it has a square planar geometry. It has the ligand NH₃ which is a strong ligand. A strong ligand produces a large crystal field splitting energy, so pairing energy will be smaller than the crystal field splitting energy.

The metal Cu²⁺ is d⁹ and therefore will be neither high spin or low spin. When it fills in to the square planar geometry, it has 1 unpaired electron. The first four electrons occupy the lower two orbitals, the next two electrons fill up the next orbitals, then two fill up the next orbital, and finally the last electron is in the last orbital.

b. [Ni(CN)₄]²⁻

This complex has a coordination number of 4, so it has a square planar geometry. It has the ligand CN^- which is a strong ligand. A strong ligand produces a large crystal field splitting energy, so pairing energy will be smaller than the crystal field splitting energy and therefore it will be low spin and completely fill up orbitals of lower energy first.

The metal Ni²⁺ is d⁸ and when it fills in to the square planar geometry, it has no unpaired electrons. The first four electrons fill up the bottom 2 orbitals, the next two electrons fill up the next orbital and last two fill up the next orbital, therefore it has no unpaired electrons.

Q14.7.12

A particular reaction has two accessible pathways (A and B), each of which favors conversion of X to a different product (Y and Z, respectively). Under uncatalyzed conditions pathway A is favored, but in the presence of a catalyst pathway B is favored. Pathway B is reversible, whereas pathway A is not. Which product is favored in the presence of a catalyst? without a catalyst? Draw a diagram illustrating what is occurring with and without the catalyst.

Q 14.7.12 Solution

Path A can be written as X⟶Y and is favored uncatalyzed conditions

Path B can be written as X⇌Z is reversible and is favored with the presence of a catalyst

With or without a catalyst, Y, which is the product of pathway A is favored. When a catalyst is present X produces Z but since Path B is reversible Z will eventually be converted to X. X will then participate in the irreversible reaction in Path A and become Y. When a catalyst is not present, X will simply produce Y. So, in both cases Y, the product of path A, is favored.