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).

S17.2.10

A(increased in mass): active, cathode

B(no change in mass): inert

C(lost mass): active, anode

Active electrodes participate in chemical reactions and can either be oxidized or reduced. Accordingly, these electrodes are found to either lose or gain mass in the process. On the other hand, inert electrodes are those which do not undergo chemical reactions and instead are used to conduct electron flow. These electrodes, such as platinum or graphite, do not lose mass in the reaction. According to these definitions, it is likely that the first metal electrode A and the third metal electrode C are active electrodes and the second electrode B is an inert electrode. The oxidation reaction occurs at the anode, referring to a loss of electrons and likewise loss in mass. The reduction reaction occurs at the cathode, referring to a gain of electrons and likewise an increase in mass. Therefore, it is likely that metal electrode C is the anode and metal electrode A is the cathode.

Q19.1.19

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

a. Zn is added to a solution of Cr₂(SO₄)₃ in acid.

b. FeCl₂ is added to a solution containing an excess of Cr2O2−7Cr2O72− in hydrochloric acid.

c. Cr²⁺ is added to Cr2O2−7Cr2O72− in acid solution.

d. Mn is heated with CrO₃.

e. CrO is added to 2HNO₃ in water.

f. FeCl₃ is added to an aqueous solution of NaOH.

S19.1.19

a. \(\displaystyle Cr_2(SO_4)_3(aq)+2Zn(s)+2H_3O^{+}(aq) \rightarrow 2Zn^{2+}(aq)+H_2(g)+2H_2O(l)+2Cr_2+(aq)+3SO_4^{2-}(aq)\)

b. \(\displaystyle4TiCl_3(s)+CrO_4^{^{2-}}(aq)+8H^{+}(aq)\rightarrow 4Ti^{4+}(aq)+Cr(s)+4H_2O(l)+12Cl^{-}(aq)\)

c. In acid solution between pH 2 and pH 6, \(\displaystyle CrO_4^{2-}\) forms \(\displaystyle HrCO_4^{-}\) , which is in equilibrium with dichromate ion.

The reaction is:\(\displaystyle2HCrO_4^{-}(aq)\rightarrow Cr2O_7^{2-}(aq)+H_2O(l)\)

At other acidic pHs, the reaction is:\(\displaystyle 3Cr^{2+}(aq)+CrO_4^{2-}(aq)+8H_3O^{+}(aq)\rightarrow 4Cr^{3+}(aq)+12H_2O(l)\)

d.\(\displaystyle 8CrO_3(s)+9Mn(s) \overset{\Delta }{\rightarrow}4Cr_2O_3(s)+3Mn_3O_4(s)\)

e. \(\displaystyle CrO(s)+2H_3O^{+}(aq)+2NO_3(aq)\rightarrow Cr^{2+}(aq)+2NO^{-3}(aq)+3H_2O(l)\)

f. \(\displaystyle CrCl_3(s)+3NaOH(aq)\rightarrow Cr(OH)_3(s)+3NaCl(aq)\)

Q19.3.11

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

S19.3.11

You would expect Mg₃[Cr(CN)₆]_{2 to be paramagnetic.}

The first step to determine the magnetism of the complex is to calculate the oxidation state of the transition metal. In this case, the transition metal is Cr.

Before doing so, we need to find charge of the of the complex ion [Cr(CN)₆]₂ given that the oxidation state of Mg_{3 is 2+. Using the subscripts of the \(\displaystyle Mg^{2+}\) ion and the} [Cr(CN)₆]_{2 complex, we find that the oxidation state of} [Cr(CN)₆]_{2 , x, to be:}

_{\(\displaystyle 3(+2)+2(x)=0\)}

_{\(\displaystyle x=3\)}

_{Now that we found the charge of the coordination complex, we are able to find the charge of the transition metal Cr given that the charge of CN is -1. Again, using the subscripts we find the oxidation state of Cr, y, to be:}

_{\(\displaystyle y + 6(-1)=-3\)}

_{\(\displaystyle y=3\)}

_{Therefore, the oxidation state of the transition metal Cr is \(\displaystyle Cr^{3+}\)}

_{Next, using the transition metal \(\displaystyle Cr^{3+}\) and the periodic table as reference, we can determine the electron configuration of \(\displaystyle Cr^{3+}\) to be \(\displaystyle [Ar]d^{3}\). This means that \(\displaystyle Cr^{3+}\) has 3 unpaired electrons in the 3d sublevel. Therefore, we find that since at least one electron is unpaired(in this case all 3 electrons are unpaired),} Mg₃[Cr(CN)₆]_{2 is paramagnetic.}

(Danny: Paramagnetism is when there are unpaired electrons while diamagnetism involves no unpaired 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?

S12.4.11

Reaction: \(\displaystyle A\rightarrow D+E\)
Given: \(\displaystyle t_{1/2}=8.50min\)

initial concentration of A = \(\displaystyle [A]_{_{0}}\) = 0.150 mol/L

final concentration of A = \(\displaystyle [A]\) = 0.0300 mol/L

\(\displaystyle t=?min\)

a) first order with respect to A:

1st order half life: \(\displaystyle t_{1/2}=ln(2/k)=.693/k\)

1st order reaction: \(\displaystyle ln(\frac{[A]}{[A]_{_{0}}})=-kt\), where k is the first-order rate constant

First step is to solve for k using the half life and the 1st order half life equation:

\(\displaystyle 8.50min=\frac{.693}{k}\)

\(\displaystyle k=\frac{.693}{8.50min}\)

\(\displaystyle k\approx .0815 min^{-1}\)

Now that we have a value for k, we can plug it into the 1st order reaction equation to solve for t:

\(\displaystyle ln(\frac{[.0300 mol/L]}{[.150 mol/L]})=-(.0815 min^{-1})t\)

\(\displaystyle -1.6094=-(.0815 min^{-1})t\)

\(\displaystyle t\approx 19.75 min\)

b) Second order with respect to A:

2nd order half life:

2nd order reaction: \(\displaystyle \frac{1}{[A]}=\frac{1}{[A]_{_{0}}}+kt\)

First step is to solve for k using the half life, the initial concentration, and the 2nd order half life equation:

\(\displaystyle 8.50min=\frac{1}{k[.150 mol/L]_{_{0}}}\)

\(\displaystyle k=\frac{1}{(8.50 min)(.150 mol/L)}\)

\(\displaystyle k\approx .784 ((min)(mol/L))^{-1}\)

Now that we have a value for k, we can plug it into the 2nd order reaction equation to solve for t:

\(\displaystyle \frac{1}{[.0300 mol/L]}=\frac{1}{.150 mol/L}+(.784((min)(mol/L))^{-1}))t\)

\(\displaystyle t\approx 34.01 min\)

Q21.2.6

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

S21.2.6

Recall that density is equal to mass divided by volume.

Given the diameter of the nucleus, we can start calculating the density by first determining the volume.

Using the volume of a sphere we find:

\(\displaystyle V=\frac{4}{3}\pi r^{3}\), where the radius is half of the diameter so \(\displaystyle r=5\times 10^{-14}\)

\(\displaystyle V=\frac{4}{3}\pi(5\times 10^{-14})^{3}\)

\(\displaystyle =5.24\times 10^{-40}cm^{3}\)

\(\displaystyle =5.24\times 10^{-40}mL\)

Next, to determine the density we must also calculate the mass of the nucleus:

First, we use the mass number and atomic number to calculate the number of protons and neutrons:

\(\displaystyle 24-12=12\)

There are a total of 12 protons and 12 neutrons in the nucleus.

Next, use the number of protons and neutrons and their individual masses to calculate the mass of the nucleus:

\(\displaystyle mass of nucleus(m)=12(1.6726219\times 10^{-24}g)+12(1.6779286\times 10^{-24}g)=4.0170606\times 10^{-23}g\)

Finally, we plug in the calculated volume and mass to calculate the density:

\(\displaystyle D=\frac{4.0170606\times 10^{-23}g}{5.24\times 10^{-40}cm^{3}mL}\)

\(\displaystyle = 7.67\times 10^{16} g/mL\)

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?

S21.6.1

At a dynamic equilibrium reactants and products are converted at an equal and constant rate. If we were to introduce either radioactive Ag⁺ or radioactive Cl^– into the solution containing the stated reaction, the reaction will shift to the left. According to Le Châtelier's principle, if a dynamic equilibrium is disrupted by a changing condition, such as the introduction of a radioactive element, the position of the equilibrium will shift in order to reestablish its equilibrium. To reestablish equlibrium, the reaction will shift left and produce more of the radioactive precipitate, \(\displaystyle AgCl(s)\).

Q20.4.7

Identify the oxidants and the reductants in each redox reaction.

1. Cr(s) + Ni²⁺(aq) → Cr²⁺(aq) + Ni(s)
2. Cl₂(g) + Sn²⁺(aq) → 2Cl⁻(aq) + Sn⁴⁺(aq)
3. H₃AsO₄(aq) + 8H⁺(aq) + 4Zn(s) → AsH₃(g) + 4H₂O(l) + 4Zn²⁺(aq)
4. 2NO₂(g) + 2OH⁻(aq) → NO₂⁻(aq) + NO₃⁻(aq) + H₂O(l)

S20.4.7

oxidant/oxidizing agent: being reduced/gaining electrons

reductant/reducing agent: being oxidized/losing electrons

1. oxidant: \(\displaystyle Ni^{2+}(aq)\)

\(\displaystyle Ni^{2+}(aq) + 2e^{-} \rightarrow Ni(s)\)

Electrons on the reactant side indicates that \(\displaystyle Ni^{2+}(aq)\) is "gaining electrons".

reductant: \(\displaystyle Cr(s)\)

\(\displaystyle Cr(s)\rightarrow Cr^{2+}+2e^{-}\)

Electrons on the product side indicates that \(\displaystyle Cr(s)\) is "losing electrons".

2. oxidant: \(\displaystyle Cl_2(g)\)

\(\displaystyle Cl_2(g)\rightarrow 2Cl^{-}(aq)+2e^{-}\)

Electrons on the reactant side indicates that \(\displaystyle Cl_2(g)\) is "gaining electrons".

reductant: \(\displaystyle Sn^{2+}(aq)\)

\(\displaystyle Sn^{2+}(aq)\rightarrow Sn^{4+}(aq)+2e^{-}\)

Electrons on the product side indicates that \(\displaystyle Sn^{2+}(aq)\) is "losing electrons".

3. oxidant: \(\displaystyle H_3AsO_4(aq)\)

For this redox reaction, it is best to examine the oxidation states of the elements.

Given that the oxidation state of the hydrogen atom is +1 and the oxidation state of the oxygen atom is -2 in this reaction, we can determine the oxidation of \(\displaystyle As\),x, to be:

reactant((\(\displaystyle H_3AsO_4(aq)\)): \(\displaystyle 3(+1)+(x)+4(-2)=0\)

\(\displaystyle x=5\)

product((\(\displaystyle AsH_3(g)\)): \(\displaystyle (x)+3(+1)=0\)

\(\displaystyle x=-3\)

From reactants to products, the oxidation state of \(\displaystyle As\) changes from 5 to -3, meaning that it has gained electrons. Therefore, \(\displaystyle H_3AsO_4(aq)\) acts as the oxidant.

reductant: \(\displaystyle Zn(s)\)

\(\displaystyle 4Zn(s)\rightarrow 4Zn^{2+}+2e^{-}\)

Electrons on the product side indicates that \(\displaystyle 4Zn(s)\rightarrow 4Zn^{2+}+2e^{-}\) is "losing electrons".

4. oxidant: \(\displaystyle NO_2(g)\)

reductant: \(\displaystyle NO_2(g)\)

For this redox reaction, the only oxidation state that changes from reactants to products is \(\displaystyle N\). Therefore, \(\displaystyle NO_2(g)\) acts as both the oxidant and reductant in this reaction.

(Danny) Please note that the oxidant undergoes reduction, a loss in the charge of the ion. The reductant, on the other hand, experiences oxidation, a increase in the charge of the ion.

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

First, assume that for this reaction we are using 1 liter of solution. Given the density of the electrolyte, we can find the exact mass in grams of the solution:

\(\displaystyle \frac{1 L soln}{1}\cdot \frac{1.15g}{1cm^{3}}\cdot \frac{1cm^{3}}{1mL}\cdot \frac{1000mL}{1L}=1150g\) solution

Next, use the percent by mass(30.0%) of sulfuric acid to calculate its mass:

\(\displaystyle (1150g)(.3)= 345g H_2SO_4\)

Next, calculate the concentration of \(\displaystyle H_2O_4\) using its molar mass and the volume of the solution:

\(\displaystyle 345gH_2SO_4\cdot \frac{1mol}{98.079g}\cdot \frac{1}{1L soln}=3.52M\)

Before plugging in values, use the standard reduction potentials table to calculate \(\displaystyle E^{0}\):

\(\displaystyle E^{0}=E^{0}cathode-E^{0}anode\)

cathode half reaction: \(\displaystyle PbO_2+4H^{+}+SO_4^{2-}+2e^{-}\rightarrow PbSO_4+2H_2O\), \(\displaystyle E^{0}=1.69V\)

anode half reaction: \(\displaystyle PbSO_4+2e^{-}\rightarrow Pb+SO_4^{2-}\), \(\displaystyle E^{0}=-.35V\)

\(\displaystyle E^{0}=1.69V-(-.35V)=2.04V\)

Next, plug in the calculated values above into the Nernst equation \(\displaystyle E=E^{0}-\frac{RT}{nF}lnQ\) to solve for the potential.

Note that the coefficient for \(\displaystyle H_2O_4(aq)\) is 2

\(\displaystyle Q=\frac{prod}{reac}=\frac{1}{3.52M}^{2}\) *Recall that we are only taking into account the molarity of the aqueous solution*

\(\displaystyle E=2.041V-.01284ln((1/3.52)^{2}\)

\(\displaystyle E=2.073V\)

\(\displaystyle E> E^{0}\)

Therefore, the potential \(\displaystyle E\) is greater than that of the standard cell \(\displaystyle E^{0}\)