Extra Credit 22
- Page ID
- 83254
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For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.
- Mg(s)+Ni2+(aq)⟶Mg2+(aq)+Ni(s)
- 2Ag+(aq)+Cu(s)⟶Cu2+(aq)+2Ag(s)
- Mn(s)+Sn(NO3)2(aq)⟶Mn(NO3)2(aq)+Sn(s)
- 3Fe(NO3)2(aq)+Au(NO3)3(aq)⟶3Fe(NO3)3(aq)+Au(s)
S17.3.1
The standard cell potential (E° cell) can be found using the following formula: E° cell= E° cathode - E° anode. The cathode is the location of the reduction reaction and the anode is the location of the oxidation reaction. The values for E° cathode and E° anode can be found in a standard reduction potential (SRP) table. If E° cell is positive, 
G (the metric for Gibbs Free Energy) will be negative and the reaction will be spontaneous. If E° cell is negative, G will be positive and the reaction will be nonspontaneous. This relationship is outlined in the equation G=-nFE° cell.
Here is a sample calculation for part 1.
- First, split the given equation into two half reactions in order to identify what is being reduced and what is being oxidized.
Mg(s)⟶Mg2+(aq) Electrons are lost as the right side is more positive than the left. This is oxidation.
Ni2+(aq)⟶Ni(s) Electrons are gained as the left side is more positive than the right. This is reduction.
- Find the reduction potentials from a table.
Oxidation reaction: -2.37V
Reduction reaction: -0.25V
- Substitute the values into the equation, E° cell= E° cathode - E° anode.
-0.25V-(-2.37V)= 2.12V
Since E° cell is positive, G is negative and the reaction is positive.
2. 2Ag+(aq)+Cu(s)⟶Cu2+(aq)+2Ag(s)
2Ag+(aq)⟶2Ag(s) Reduction. SRP=0.80V
Cu(s)⟶Cu2+(aq) Oxidation. SRP= 0.15V
0.80V-0.15V=.65V
Positive E° cell, so spontaneous reaction.
3. Mn(s)+Sn(NO3)2(aq)⟶Mn(NO3)2(aq)+Sn(s)
Mn(s)⟶Mn(NO3)2(aq) Oxidation. This can be determined by the oxidation states of the element in question. Mn goes from being neutral to being +2, when bonded with 2 NO3, that each have a negative charge. SRP=-1.18V
Sn(NO3)2(aq)⟶Sn(s) Reduction. Sn goes from +2 to neutral, so electrons are gained. SRP= -0.14V
-0.14V-(-1.18V)=1.04V
Positive E° cell, so spontaneous reaction.
4. 3Fe(NO3)2(aq)+Au(NO3)3(aq)⟶3Fe(NO3)3(aq)+Au(s)
3Fe(NO3)2(aq)⟶3Fe(NO3)3(aq). Fe goes from an oxidation state of +2 to +3, without considering the balanced equation coefficients. This is oxidation. SRP= 0.77V
Au(NO3)3(aq)⟶Au(s). Au goes from +3 to neutral. This is reduction. SRP=1.50V
1.50V-0.77V=0.73V
Positive E° cell, so spontaneous reaction.
Q19.1.20
What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid?
S19.1.20
A non-oxidizing acid is also called a strong acid, meaning that it will release hydrogen (H+) atoms in water. The FeSO4 will produce H2SO4 because iron will readily act with strong acids according to the activity series.
Q19.3.12
Would you expect salts of the gold ion, Au+, to be colored? Explain.
S19.3.12
No. Au+ has a complete 5d sublevel. Coordination complexes display colors due to bonding to ligands, which can result in electrons being excited to higher energy levels. Based on whether the bound ligands are strong or weak field, the complex will absorb a certain wavelength of light and reflect the color complementary to that wavelength. Since Au+ has ten d subshell electrons, it cannot bond with any ligands and will be colorless.
Q12.4.12
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.
| [Penicillin] (M) | Rate (mol/L/min) |
|---|---|
| 2.0 × 10−6 | 1.0 × 10−10 |
| 3.0 × 10−6 | 1.5 × 10−10 |
| 4.0 × 10−6 | 2.0 × 10−10 |
S12.4.12
The reaction is first order with k = 2.0 × 104 mol−1 min−1.
The reaction order can be determined analytically with the following formula: rate law=k[M]x. In this problem, k is the reaction constant, M is the concentration of penicillin and x is the reaction order.
To start, figure out the relationship between the reaction order and the rate law. We can do this by dividing the rate for the second trial by that of the first and setting it equal to concentration of penicillin in the second trial divided by that of the first trial. Solve the following equation for x.$$\frac{1.5\times10^{-10}}{1.0\times10^{-10}}=\frac{3.0\times10^{-6}}{2.0\times10^{-6}}^x$$
This calculation renders 1.5 = [1.5]x. x=1. Since x is one this is a first order reaction.
Now that we have the value of x, we can solve for k. Pick any trial and solve for k.
Trial 3 data example:
4.0 × 10−6=k[2.0 × 10−10]
$$\frac{4.0 × 10^{−6}}{2.0 × 10^{−10}}=k$$
k = 2.0 × 104 mol−1 min−1
Q21.2.7
What are the two principal differences between nuclear reactions and ordinary chemical changes?
S21.2.7
Nuclear reactions usually change one type of nucleus into another; chemical changes rearrange atoms. Since nuclear reactions deal with loss and gains of protons, elements can turn into other elements, called transmutation. Isotopes involved in nuclear reactions are very different from each other. Mass is destroyed in nuclear reactions and as a result, nuclear reactions involve much larger energies than chemical reactions and have more measurable mass changes. These large energy generating reactions are unaffected by temperature and the presence of catalysts, unlike chemical reactions.
Q21.6.3
Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine-131 is injected into the body. In larger doses, I-133 is also used as a means of treating cancer of the thyroid. I-131 has a half-life of 8.70 days and decays by β− emission.
- Write an equation for the decay.
- How long will it take for 95.0% of a dose of I-131 to decay?
S21.6.3
$$I_{53}^{133}\Rightarrow Xe_{54}^{133}+e_{-1}^0$$
ß- emission means that the right side of the nuclear equation will have an electron on it. To find the balanced decay equation, you must find an element with an atomic number one greater than Iodine (53). The mass number will be same as an electron has a negligible mass. This can be represented by the following equations:$$131+0=x$$ $$y+(-1)=0$$
x is the mass number and y is the atomic number of the new element. This comes out to be Xenon as it has an atomic number of 54. The mass number will still be 133.
$$Xe_{54}^{133}$$
37.6 days
The half life of an isotope is helpful when determining the time it would take for that particular isotope to decay. It can be plugged into the following equation:$$A=A_0\times\frac{1}{2}^{t/h}$$ $$\frac{A}{A_0}=\frac{1}{2}^{t/h}$$ $$ln\frac{A}{A_0}=ln\frac{1}{2}^{t/h}$$ $$ln\frac{A}{A_0}=\frac{t}{h}\times ln\frac{1}{2}$$
Substitute in 5 for A because if 95% of the sample decays, 5% will be left and 100 for A0 since we are dealing with percentages. h is equal to the half life, which is 8.7 days. Solve the equation for t. $$ln\frac{5}{100}\div ln\frac{1}{2}=\frac{t}{8.7}$$ $$(ln\frac{5}{100}\div ln\frac{1}{2})\times8.7=t$$
Q20.4.8
Identify the oxidants and the reductants in each redox reaction.
a.Br2(l) + 2I−(aq) → → 2Br−(aq) + I2(s)
b.Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(aq)
c.H+(aq) + 2MnO4−(aq) + 5H2SO3(aq) → 2Mn2+(aq) + 3H2O(l) + 5HSO4−(aq)
d.IO3−(aq) + 5I−(aq) + 6H+(aq) → 3I2(s) + 3H2O(l)
S20.4.8
An oxidant is also known as the oxidizing agent in a chemical equation, the substance that gets reduced. A reductant is the reducing agent, the substance that gets oxidized. To solve these problems, split the balanced equation into to half reactions and determine which substances are gaining electrons and which substances are losing electrons.
a. Br2(l) + 2I−(aq) → 2Br−(aq) + I2(s)
- Br2(l) → 2Br−(aq) Gain of electrons from left to right. Br is being reduced and as a result is the oxidant.
- 2I−(aq) → I2(s) Loss of electrons from left to right. I is being oxidized and as a result is the reductant.
b. Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(aq)
- Cu2+(aq) → Cu(s) Cu is being reduced and as a result
- 2Ag(s) → 2Ag+(aq) Ag is being oxidized and as a result is the reductant.
c. H+(aq) + 2MnO4−(aq) + 5H2SO3(aq) → 2Mn2+(aq) + 3H2O(l) + 5HSO4−(aq)
- H+(aq) + 5H2SO3(aq) → 3H2O(l) + 5HSO4−(aq) It's harder to track electrons here, but it can be done via oxidation states. To find the oxidation state of an element, start by using what information you know for sure. H has an oxidation number of +1 and O has an oxidation number of -2. On the right side we have -2(3) O and +1(2)H. In order for the compound to be neutral, S must have an oxidation number of +2, so the negative and positives cancel out. Sulfur is being oxidized. It goes from having an oxidation state of +4 to +6. Sulfur is the reductant.
- 2MnO4−(aq) → 2Mn2+(aq) Mn goes from having an oxidation number of +7 to an oxidation number of +2. This is reduction. Mn is the oxidant.
d. IO3−(aq) + 5I−(aq) + 6H+(aq) → 3I2(s) + 3H2O(l)
- In this reaction, Iodine is acting as both the oxidant AND the reactant. IO3− goes from an oxidation number of +5 to 0. 5I− goes from -1 to 0.
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?
S20.7.5
To solve this problem, use the following version of the Nernst equation:$$Ecell= E° cell-\frac{RT}{nF}\times{lnQ}$$ $$Ecell= E° cell-\frac{0.0257}{n}\times ln{Q}$$
The second equation is simplified to include the values of the gas constant (R) and Faraday's constant (F).
Q is the reaction quotient and to find it we must determined the concentrations of the products and the reactants. Since, Pb, PbO2, 2PbSO4, and 2H2O are all solids or liquids, they will not factor into the calculation of the reaction quotient. All you must do is find the concentration of H2SO4. The molar mass of sulfuric acid is 98.07g. Divide the molar mass by the given density (1.15 g/cm3) and then multiply by (30/100) because the sample is 30% sulfuric acid. Finally multiply this by 1000 to get the cm3 to L.
*A cm3 is equal to a mL and 1000 mL are equal to 1 L*
From the molarity, you can find Q and plug it into the Nernst equation. In this case, Q would just be 1 divided the molarity of sulfuric acid (3.52). Q equals the concentration of the products divided by the concentration of the reactants. Since the products are a solid and a liquid, they will just be represented by a 1. Take the natural log of (1/3.52) and the answer will be a negative number.
You do not have to do the rest of the calculations! Just by looking at the Nernst equation you can tell that you'll be multiplying two negatives since (-0.0257/n) is negative as well. This will result in:$$Ecell= E° cell+\frac{0.0257}{n}\times ln{Q}$$
Since Ecell is a number added to E° cell, you can deduce that it is larger.