Extra Credit 22

Question 17.3.1

For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.

1. \(\ce{Mg}(s)+\ce{Ni^2+}(aq)⟶\ce{Mg^2+}(aq)+\ce{Ni}(s)\)
2. \(\ce{2Ag+}(aq)+\ce{Cu}(s)⟶\ce{Cu^2+}(aq)+\ce{2Ag}(s)\)
3. \(\ce{Mn}(s)+\ce{Sn(NO3)2}(aq)⟶\ce{Mn(NO3)2}(aq)+\ce{Sn}(s)\)
4. \(\ce{3Fe(NO3)2}(aq)+\ce{Au(NO3)3}(aq)⟶\ce{3Fe(NO3)3}(aq)+\ce{Au}(s)\)

Solution 17.3.1

In this question we are asked to determine the standard cell potential at 25 °C for each listed reaction, and whether the reaction is spontaneous at standard conditions.

The first step is to pull up a table of Standard Reduction Potentials (SRPs) and break each reaction into its corresponding half reactions that occur at the cathode and anode. One should know that in these cases, oxidation occurs at the anode, and reduction occurs at the cathode. We should also know that the overall cell potential of a reaction is the difference between the cell potential of the cathode minus the cell potential of the anode:

\[E°_{cell} = E°_{cathode} − E°_{anode}\]

*The equation can also be written as:

*\[E_{cell}^{o}=E_{cathode}^{o}+E_{anode}^{o}\]

*In this case, the sign of the reduction potential value for the oxidation reaction will need to be flipped.

The second step is to determine whether or not the reaction is spontaneous at standard conditions. This can be figured out very easily once you correctly determine the cell potential. If E° is positive, the reaction is spontaneous, if it is negative, it is NOT spontaneous.

Using the table of standard reduction potentials for half reactions at 25 °C one can easily use the reactions to determine for the standard cell potential at 25 °C.

a. \(\ce{Mg}(s)⟶\ce{Mg^2+}(aq)+2e^-\) has an SRP of -2.372 V, and is the oxidation half reaction, meaning it occurs at the anode.

\(\ce{Ni^2+}(aq)+2e^-⟶\ce{Ni}(s)\) has an SRP of -0.257 V, and is the reduction half reaction, meaning it occurs at the cathode.

Using the given E° cell equation above we find that our cell potential is equal to -0.257 minus -2.372 (cathode minus anode). The different results in an E° of +2.115 V. Since it is a positive value, the reaction is spontaneous.

b. \(\ce{2Ag+}(aq)+2e^-⟶\ce{2Ag}(s)\) has an SRP of +0.7996 V, and is the reduction half reaction, meaning it occurs at the cathode.

\(\ce{Cu}(s)⟶\ce{Cu^2+}(aq)+2e^-\) has an SRP of +0.337 V, and is the oxidation half reaction, meaning it occurs at the anode.

Using the given equation again, we find that our cell potential is equal to +0.7996 V minus 0.337 V (cathode minus anode). This results in a E° of +0.4626 V. Which is spontaneous due to its sign.

c. \(\ce{Mn}(s)⟶\ce{Mn(NO3)2}(aq)+2e^-\) has an SRP of -1.185 V, and is the oxidation half reaction, meaning it occurs at the anode.

\(\ce{Sn(NO3)2}(aq)+2e^-⟶\ce{Sn}(s)\) has an SRP of -0.1262 and is the reduction half reaction, meaning it occurs at the cathode.

Using the given equation again, we find that our cell potential is equal to -0.1262 minus -1.185 (cathode minus anode). This results in a E° of +1.0588 V. Which is spontaneous due to its sign.

d. \(\ce{3Fe(NO3)2}(aq)⟶\ce{3Fe(NO3)3}(aq)+3e^-\) has an SRP of +0.771, and is a oxidation half reaction, meaning it occurs at the anode.

\(\ce{Au(NO3)3}(aq)+3e^-⟶\ce{Au}(s)\) has an SRP of +1.498, and is the reduction half reaction, meaning it occurs at the cathode.

Using the given equation again, we find that our cell potential is equal to +1.498 minus +0.771 (cathode minus anode). This results in a E° of +0.727 V. Which is spontaneous due to its sign.

*Edited by Asha Cummings

Question 19.1.20

What is the gas produced when iron(II) sulfide is treated with a non-oxidizing acid?

Solution 19.1.20

To begin, let's define what a non-oxidizing acid is. A non-oxidizing acid is an acid that will not act as an oxidizing agent in a given reaction. A common non-oxidizing acid is hydrochloric acid. Let us write out the balanced reaction of iron(II) sulfide with hydrochloric acid.

\(\ce{FeS}(s)+{2HCl}(aq)⟶\ce{FeCl2}(s)+\ce{H2S}(g)\)

Throughout the reaction, iron kept a 2+ oxidation state, therefore HCl is a non-oxidizing acid. This is a classic and simple double replacement reaction between the two reactants. After balancing the reaction we see that hydrogen sulfide gas is produced.

*Essentially, when iron(II) sulfide reacts with a non-oxidizing gas, a double replacement reaction occurs and a gas with hydrogen and another element is formed.

*Edited by Asha Cummings

Question 19.3.12

Would you expect salts of the gold ion, Au⁺, to be colored? Explain.

Solution 19.3.12

No. Colored ions have unpaired electrons in their outmost orbital. A partially filled d orbital, for example, can yield various colors. After completing the noble gas configuration, we see that Au⁺ has a configuration of [Xe] 4f¹⁴5d¹⁰. Since Au⁺ has a completely filled d sublevel, we are certain that any salts of the gold ion, Au⁺ will be colorless.

*An example of a colored ion would be copper(II). Cu²⁺ has an electron configuration of [Ar]3d⁹. It has one unpaired electron. Copper(II) appears blue.

*Edited by Asha Cummings

Question 12.4.12

Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of 3 × 10⁴ g/mol that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate equation that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing 0.15 µg (0.15 × 10⁻⁶ g) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant.

Solution 12.4.12

This is a pretty classic rate problem and is very straight forward. We are given three molar concentrations of penicillin, and three rates in the units of (mol/L/min).

The first step is to solve for the order or the reaction. This can be done by setting up two expressions which equate the rate to the rate constant times the molar concentration of penicillin raised to the power of it's order. Once we have both expressions set up, we can divide them to cancel out k (rate constant) and use a basic logarithm to solve for the exponent, which is the order. It will look like this.

rate(mol/L/min)=k[M]^x

(1.0 x 10^-10)=k[2.0 x 10^-6]^x

(1.5 x 10^-10)=k[3.0 x 10^-6]^x

Dividing the two equations results in the expression:

(2/3)=(2/3)^x

*A single ratio equation can also be set up to solve for the reaction order:

*\[\frac{rate_{1}}{rate_{2}}=\frac{k[Penicillin]_{1}^{x}}{k[Penicillin]_{2}^{x}}\]

*We then solve for x in a similar fashion.

*\[\frac{1.0x10^{-10}}{1.5x10^{-10}}=\frac{[2.0x10^{-6}]^{x}}{[3.0x10^{-6}]^{x}}\]

We can now use the natural logarithm to solve for x, or simply and intuitively see that in order for the equation to work, x must be equal to one. Thus, the reaction is of the first order.
Now that we have the order of the reaction, we can proceed to solve for the value of the rate constant. Substituting x=1 into our first equation yields the expression:

(1 x 10^-10)=k[2.0 x 10^-6]¹

k=(1 x 10^-10)/(2 x 10^-6)

k= (5 x 10^-5) min^-1

We have a unit of min^-1 because we divided (mol/L/min) by molarity, which is in (mol/L), yielding a unit of min^-1.

We were given two important pieces of information to finish the problem. It is stated that the enzyme has a molecular weight of 3 × 10⁴ g/mol, and that we have a one liter solution that contains (0.15 x 10^-6 g) of penicillinase. Dividing the amount of grams by the molecular weight yields 5 x 10^-12 moles.

(0.15 x 10^-6) g / (3 x 10⁴) g/mol = (5 x 10^-12) mol

Now that we have the amount of moles, we can divide our rate constant by this value.

(5 x 10^-5) min^-1 / (5 x 10^-12) mol = (1 x 10⁷) mol^-1 min ^-1

*Edited by Asha Cummings

Question 21.2.7

What are the two principal differences between nuclear reactions and ordinary chemical changes?

Solution 21.2.7

This is a simple question if one understands the basic principals of nuclear chemistry. Nuclear chemistry is different from normal chemistry in that technically breaks the Law of Conservation of Mass and Energy. *Nuclear chemistry is called "nuclear" because it deals with chemical changes in the nucleus. In a nuclear reaction, mass changes are noticeable, and very large energies are typically involved. Nuclear reactions literally change one nucleus into another. Since the proton number is changed, the atom is now a completely different element. This is referred to as nuclear transmutation, and happens when the proton or neutron number is changed. As stated earlier nuclear reactions also release insanely large quantities of energy for the amount of matter that goes into a nuclear reaction. So, what are the two principal differences between nuclear reactions and ordinary chemical changes?

1. Nuclear reactions usually completely change one type of nucleus into another! New elements and isotopes form! Ordinary chemical changes simply rearrange atoms.

2. Nuclear reactions involve much larger energies than ordinary chemical changes, and have measurable mass changes!

*Edited by Asha Cummings

Question 21.6.2

Technetium-99m has a half-life of 6.01 hours. If a patient injected with technetium-99m is safe to leave the hospital once 75% of the dose has decayed, when is the patient allowed to leave?

Solution 21.6.2

This is a pretty standard half-life decay problem. First, let us write down the information we have. The half life: t_1/2 = 6.01 hrs

We want to know after how long will 75% of the dose the patient received decay?

To solve this problem we will use the percent activity equation: % remaining = (1/2)ⁿ

75% decayed means 25% remains. Which is equal to 0.25. evaluating our expression gets 0.25 = (1/2)ⁿ

ln(0.25) = n(ln(0.5)) Solving for n we get that n is equal to 2.

Now, multiplying n by our half-life yields the number of hours that the patient must wait before leaving the hospital.

2 x 6.01 = 12.02 hours

This answer also makes sense intuitively because after one half life, 50% of the sample remains, after 2 half lives, 25% of the sample remains. Since 75% of it has decayed, 25% remains, therefore, it takes two half lives for 75% to decay. One half life is 6.01 hours, so it will take 12.02 hours for the patient to be able to leave.

Question 20.4.8

Identify the oxidants and the reductants in each redox reaction.

1. Br₂(l) + 2I⁻(aq) → 2Br⁻(aq) + I₂(s)
2. Cu²⁺(aq) + 2Ag(s) → Cu(s) + 2Ag⁺(aq)
3. H⁺(aq) + 2MnO₄⁻(aq) + 5H₂SO₃(aq) → 2Mn²⁺(aq) + 3H₂O(l) + 5HSO₄⁻(aq)
4. IO₃⁻(aq) + 5I⁻(aq) + 6H⁺(aq) → 3I₂(s) + 3H₂O(l)

Solution 20.4.8

To begin, let us define an oxidant and reductant. An oxidant, *or oxidizing agent, gains electrons and is reduced. A reductant, *or reducing agent, loses electrons and is oxidized. *The oxidation states can also be used to determine if a reaction is an oxidant or reductant. *In oxidation reactions, the oxidation state increases. *In reduction reactions, the oxidation state decreases. One can either look at the half reactions, or just look at the species in the overall reaction (it is best to split it into half reactions if you are just learning how to do such a problem).

a. Br₂(l)+e^-→2Br^-(aq)

Here Br goes from an oxidation state of 0 to 1-. It went down in its oxidation state by gaining electrons, thus, was reduced. Therefore, based on the definitions, Br is an oxidant.

2I^-(aq)→ I₂(s) +e^-

Here the oxidation state of I^- went up. It lost electrons and was oxidized. Therefore, it is a reductant.

b. Cu²⁺(aq) + 2Ag(s) → Cu(s) + 2Ag⁺(aq)

Here Cu goes down in its oxidation state by gaining electrons. It is reduced and is therefore an oxidant.

Ag goes up in its oxidation state by losing electrons. It is oxidized and therefore is a reductant.

c. H⁺(aq) + 2MnO₄⁻(aq) + 5H₂SO₃(aq) → 2Mn²⁺(aq) + 3H₂O(l) + 5HSO₄⁻(aq)

Here H⁺ goes to HSO₄. Its oxidation state doesn't change at all and therefore is neither an oxidant or reductant.

Mn goes down in its oxidation state from 7+ to 2+ by gaining electrons, thus, is reduced. Since it is reduced,it is an oxidant.

H₂ goes up in its oxidation state by losing electrons, thus, is oxidized. Since it is oxidized, it is a reductant.

d. IO₃⁻(aq) + 5I⁻(aq) + 6H⁺(aq) → 3I₂(s) + 3H₂O(l)

Here Iodine's oxidation state in IO₃^-(aq) is 5+. It's oxidation state alone, 5I^-(aq), is 1-. I₂ has an oxidation state of zero. Since Iodine's oxidation states go down, they gain electrons, are reduced, and therefore are reductants. Hydrogen's oxidation state doesn't change at all and is neither oxidized nor reduced.

*Edited by Asha Cummings

Question 20.7.5

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?

Solution 20.7.5

So, we know that the potential of a standard cell is based on the molar concentrations of the reaction of the lead storage battery. A standard cell has molar concentrations of one. Therefore, if we solve for a molar concentration of sulfuric acid that is greater than one, we know it will have a potential that is greater than that of the standard cell.

To solve for our molar concentration of sulfuric acid, we will use the given density and percent mass of sulfuric acid in the battery.

1.15g/cm³ x 0.30 = 0.345 g/cm³

Dividing the amount of sulfuric acid by its molar mass yields (3.52 x 10^-3) mol/cm³

Multiplying by 1000 cm³/L yields 3.52 mol/L.

Since sulfuric acid's molar concentration is [3.52] which is greater than [1] (concentration of the standard cell). We know that E > E°. The potential is greater than that of the standard cell.