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

When a galvanic cell circuit is closed, the electrons flow from the anode to the cathode. As a reaction progresses, the anode loses mass since it is oxidized (loses electrons) into metal cations that go into solution, while the cathode gains mass as the cations in solution are reduced (gains electrons) and deposited on the cathode. If the electrode doesn't change in mass, it must be inert since it's only serving as a medium for electron transfer.

Based on that, electrode A is the cathode since it increased in mass; electrode B is the inert cathode since it did not change in mass; and electrode C is the anode since it lost mass.

Q19.1.19

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

a. \(Zn\) is added to a solution of \(Cr_{2}(SO_{4})_{3}\) in acid.
b. \(FeCl_{2}\) is added to a solution containing an excess of \(Cr_{2}O_{7}^{2-}\) in hydrochloric acid.
c. \(Cr^{2+}\) is added to \(Cr_{2}O_{2}^{2-}\) in acid solution.
d. \(Mn\) is heated with \(CrO_{3}\).
e. \(CrO\) is added to \(2HNO_{3}\) in water.
f. \(FeCl_{3}\) is added to an aqueous solution of \(NaOH\).

S19.1.19

a. \(\bf Zn\) is added to a solution of \(\bf Cr_{2}(SO_{4})_{3}\) in acid.
\(Cr_2(SO_4)_3(aq)+2Zn(s)+2H_3O^+(aq)\rightarrow 2Zn^{2+}(aq)+H_2(g)+2H_2O(l)+2Cr^{2+}(aq)+3SO_{4}^{2-}(aq)\)

b. \(\bf FeCl_{2}\) is added to a solution containing an excess of \(\bf Cr_{2}O_{7}^{2-}\) in hydrochloric acid.
\(4TiCl_3(s)+CrO_4^{2-}(aq)+8H^+(aq)\rightarrow4Ti^{4+}(aq)+Cr(s)+4H_2O(l)+12Cl^-(aq)\)

c. \(\bf Cr^{2+}\) is added to \(\bf Cr_{2}O_{2}^{2-}\) in acid solution.
In acid solution between pH 2 and pH 6, \(CrO_{4}^{2-}\) forms \(HrCO_{4}^{-}\), which is in equilibrium with dichromate ion. The reaction is \(2HCrO_{4}^{-}(aq)\rightarrow Cr_2O_{7}^{2-}(aq)+H_2O(l)\). At other acidic pHs, the reaction is \(3Cr^{2+}(aq)+CrO_{4}^{2-}(aq)+8H_3O^+(aq)\rightarrow 4Cr^{3+}(aq)+12H_2O(l)\)

d. \(\bf Mn\) is heated with \(\bf CrO_{3}\).
\(8CrO_3(s)+9Mn(s)\xrightarrow{\triangle}4Cr_2O_3(s)+3Mn_3O_4(s)\)

e. \(\bf CrO\) is added to \(\bf 2HNO_{3}\) in water.
\(CrO(s)+2H_3O^+(aq)+2NO_{3}^{-}(aq)\rightarrow Cr^{2+}(aq)+2NO_3^+(aq)+3H_2O(l)\)

f. \(\bf FeCl_{3}\) is added to an aqueous solution of \(\bf NaOH\).
\(CrCl_3(s)+3NaOH(aq)\rightarrow Cr(OH)_3(s)+3Na^+(aq)+3Cl^-(aq)\)

Q19.3.11

Would you expect the \(Mg_{3}[Cr(CN)_{6}]_{2}\) to be diamagnetic or paramagnetic? Explain your reasoning.

S19.3.11

Figure out the oxidation state of the transition metal, \(Cr\).

First find the charge of the complex ion, \([Cr(CN)_{6}]_{2}\), using the counter ion, \(Mg\). \(Mg\) is an akali earth metal, so it has an oxidation state of \(2+\), and there are three \(Mg^{2+}\) ions, making total charge of the counter ions \(+6\). This means the total charge of the complex ions is \(-6\) since the overall charge of the compound must be \(0\) (\(6 + (-6) = 0\)). However, since there are two complex ions (notice the "2" subscript) each complex ion has a charge of \(-3\) (since the total charge was \(-6\) and there were 2 complex ions, \(-6\div2=-3\)).

Now that we know that each complex ion has a charge of \(-3\), the charges of the ligands and transition metal must add up to \(-3\). \(CN\) (or cyanide) is a monodentate anionic ligand with a \(-1\) charge, and there are six of them. If we temporarily let \(x=\) the oxidation state of \(Cr\), we can set up the following equation and solve for x:

\(x+6(-1)=-3\)
\(x-6=-3\)
\(x=3\)

The oxidation state of \(Cr\) is \(+3\). Using the periodic table, to figure out the electron configuration. "\(+3\)" means \(Cr\)has lost 3 electrons, but it loses its 4s-sublevel electrons before losing one of its 3d-sublevel ones. So its electron configuration is \([Ar]3d^{3}\) (or \(d^3\) in shorthand notation). This means it only has 3 electrons in its 3d sublevel, and if we draw the orbital diagram for the 3d sublevel:

there are 3 unpaired electrons. When we look at CN on the spectrochemical series we see that it is a strong field ligand and therefor it would have low spin(fill uo the bottom row first). In this case although it is an octahedral complex, there are only 3 d electron so they would have t go in the t2g row anyways. Since there are unpaired electrons, \(Mg_{3}[Cr(CN)_{6}]_{2}\) 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 mol/L if the reaction is (a) first order with respect to A or (b) second order with respect to A?

S12.4.11

Organize the given variables:

(half-life of A) \(t_{1/2}=8.50min\)
(initial concentration of A) \([A]_{0}=0.150mol/L\)
(target concentration of A) \([A]=0.0300mol/L\)

Find the the rate constant k, using the half-life formulas for each respective order. After finding k, use the integrated rate law respective to each order and the initial and target concentrations of A to find the time it took for the concentration to drop.

(a) first order with respect to A

(half-life) \(t_{1/2}=\frac{ln(2)}{k}=\frac{0.693}{k}\)
(rearranged for k) \(k=\frac{0.693}{t_{1/2}}\)
(plug in t_1/2 = 8.50 min) \(k=\frac{0.693}{8.50min}=0.0815min^{-1}\)

(integrated rate law) \(ln[A]=-kt+ln[A]_{0}\)
(rearranged for t) \(ln(\frac{[A]}{[A]_{0}})=-kt\)
\(-ln(\frac{[A]}{[A]_{0}})=kt\)
\(ln(\frac{[A]}{[A]_{0}})^{-1}=kt\)
\(ln(\frac{[A]_{0}}{[A]})=kt\)
\(t=\frac{ln(\frac{[A]_{0}}{[A]})}{k}\)
(plug in variables) \(t=\frac{ln(\frac{0.150mol/L}{0.0300mol/L})}{0.0815min^{-1}}=\frac{ln(5.00)}{0.0815min^{-1}}=19.7min\)

(b) second order with respect to A

(half-life) \(t_{1/2}=\frac{1}{k[A]_{0}}\)
(rearranged for k) \(k=\frac{1}{t_{1/2}[A]_{0}}\)
(plug in variables) \(k=\frac{1}{(8.50min)(0.150mol/L)}=\frac{1}{1.275min\cdot mol/L}=0.784L/mol\cdot min\)

(integrated rate law) \(\frac{1}{[A]}=kt+\frac{1}{[A]_{0}}\)
(rearranged for t) \(\frac{1}{[A]}-\frac{1}{[A]_{0}}=kt\)
\(t=\frac{1}{k}(\frac{1}{[A]}-\frac{1}{[A]_{0}})\)
(plug in variables) \(t=\frac{1}{0.784L/mol\cdot min}(\frac{1}{0.0300mol/L}-\frac{1}{0.150mol/L})=\frac{1}{0.784L/mol\cdot min}(\frac{80}{3}L/mol)=34.0min\)

Q21.2.6

Calculate the density of the \(_{12}^{24}Mg\) nucleus in \(g/mL\), assuming that it has the typical nuclear diameter of \(1\times10^{-13}cm\) and is spherical in shape.

S21.2.6

Figure out how many protons and neutrons there are in \(_{12}^{24}Mg\). Since atomic number of \(Mg\) is 12, it has 12 protons. To find the neutrons, subtract the atomic number 12 from the mass number, which is 24.

\(24-12=12\)

So \(_{12}^{24}Mg\) has 12 protons and 12 neutrons. Now find the mass of the \(_{12}^{24}Mg\) nucleus in grams, where \(m_{proton}=1.6726219\times10^{-24}g\) and \(m_{neutron}=1.6749286\times10^{-24}g\).
[Alternatively, you can use atomic mass units, with \(m_{proton}=1.007825u\), \(m_{neutron}=1.008665u\), and the conversion \((\frac{1.6605\times10^{-24}g}{1u})\).]

\(m=12(1.6726219\times10^{-24}g)+12(1.6749286\times10^{-24}g)=4.0170606\times10^{-23}g\)

The volume is assumed to be "spherical in shape" and the nuclear diameter given to be \(1\times10^{-13}cm\). Using the equation for the volume of a sphere, with \(r=\frac{1\times10^{-13}cm}{2}=5\times10^{-14}cm\) (the radius is half the diameter), find the volume.

\(V=\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi(5\times10^{-14}cm)^{3}=\frac{4}{3}\pi(1.25\times10^{-40}cm^{3})=5.24\times10^{-40}cm^{3}\)

Since our answer must be in \(g/mL\), use the conversion \((\frac{1mL}{1cm^{3}})\) to convert cubic centimeters to millimeters.

\(V=(5.24\times10^{-40}cm^{3})(\frac{1mL}{1cm^{3}})=5.24\times10^{-40}mL\)

Density is mass divided by volume.

\(d=\frac{m}{V}=\frac{4.0170606\times10^{-23}g}{5.24\times10^{-40}mL}=7.67\times10^{16}g/mL\)

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

Radiotracers, or radioisotopes (radioactive isotopes) of either \(Ag^+\) or \(Cl^-\) can be introduced into the solution. This will cause the reaction to shift left (according to Le Chatelier's principle), producing more of the \(AgCl(s)\). After giving time for the system to equilibrate, if radioactive precipitates are present, a dynamic equilibrium has been established.

Q20.4.7

Identify the oxidants and the reductants in each redox reaction.

a. \(Cr(s)+Ni^{2+}(aq)→Cr^{2+}(aq)+Ni(s)\)
b. \(Cl_{2}(g)+Sn^{2+}(aq)→2Cl^{-}(aq)+Sn^{4+}(aq)\)
c. \(H_{3}AsO_{4}(aq)+8H^{+}(aq)+4Zn(s)→AsH_{3}(g)+4H_{2}O(l)+4Zn^{2+}(aq)\)
d. \(2NO_{2}(g) + 2OH^{-}(aq) → NO_2^{-}(aq)+NO_2^{-}(aq)+H_{2}O(l)\)

S20.4.7

Find the oxidation states (review oxidation state rules) of each element in each compound on both sides of the equation. Any reactant species that decreases in oxidation state (or becomes more "negative") is reduced, and is an oxidizing agent/oxidant. Any reactant species that increases in oxidation state (or becomes more "positive") is oxidized, and is a reducing agent/reductant (only reactant species can be reductants/oxidants, not products!).

a. \(\bf Cr(s)+Ni^{2+}(aq)→Cr^{2+}(aq)+Ni(s)\)
oxidation states on the reactants side:
\(Cr(s)=0\)
\(Ni^{2+}(aq)=2+\)
oxidation states on the products side:
\(Cr^{2+}(aq)=2+\)
\(Ni(s)=0\)
\(Ni^{2+}(aq)\) was reduced (decreased from 2+ to 0), so it's the oxidant.
\(Cr(s)\) was oxidized (increased from 0 to 2+), so it's the reductant.

b. \(\bf Cl_{2}(g)+Sn^{2+}(aq)→2Cl^{-}(aq)+Sn^{4+}(aq)\)
oxidation states on the reactants side:
\(Cl_{2}(g)=0\)
\(Sn^{2+}(aq)=2+\)
oxidation states on the products side:
\(Cl^{-}(aq)=1-\)
\(Sn^{4+}(aq)=4+\)
\(Cl_{2}(g)\) was reduced (decreased from 0 to 1-), so it's the oxidant.
\(Sn^{4+}(aq)\) was oxidized (increased from 2+ to 4+), so it's the reductant.

c. \(\bf H_{3}AsO_{4}(aq)+8H^{+}(aq)+4Zn(s)→AsH_{3}(g)+4H_{2}O(l)+4Zn^{2+}(aq)\)
oxidation states on the reactants side:
for \(H_{3}AsO_{4}(aq)\):
\(H^+=1+\)
let ox. state of \(As=x\)
\(O=2-\);
\(3(1)+4(-2)+x=0\)
\(3-8+x=0\)
\(x=5\) so \(As=5+\)
\(H=1+\)
\(Zn(s)=0\)
oxidation states on the products side:
for \(AsH_{3}(g)\):
let ox. state of \(As=x\)
\(H=1+\);
\(x+3(1)=0\)
\(x=-3\) so \(As=3-\)
for \(H_{2}O(l)\):
\(H=1+\)
\(O=2-\)
\(Zn^{2+}=2+\)
\(H^{+}(aq)\) did not change oxidation state, so it is neither an oxidant/reductant.
\(As\) was reduced (decreased from 5+ to 3-), so \(H_{3}AsO_{4}(aq)\) is the oxidant.
\(Zn(s)\) was oxidized (increased from 0 to 2+), so it's the reductant.

d. \(\bf 2NO_{2}(g)+2OH^{-}(aq)→NO_{2}^{-}(aq)+NO_{3}^{-}(aq)+H_{2}O(l)\)
oxidation states on the reactants side:
for \(NO_{2}(g)\):
let ox. state of \(N=x\)
\(O=2-\);
\(x+2(-2)=0\)
\(x-4=0\)
\(x=4\) so \(N=4+\)
for \(OH^{-}(aq)\)
\(H=1+\)
\(O=2-\)
oxidation states on the products side:
for \(NO_{2}^{-}(aq)\):
let ox. state of \(N=x\)
\(O=2-\);
\(x+2(-2)=-1\)
\(x-4=-1\)
\(x=3\) so \(N=3+\)
for \(NO_{3}^{-}(aq)\):
let ox. state of \(N=x\)
\(O=2-\);
\(x+3(-2)=-1\)
\(x-6=-1\)
\(x=5\) so \(N=5+\)
for \(H_{2}O(l)\):
\(H=1+\)
\(O=2-\)
\(OH^{-}(aq)\) did not change oxidation state, so it is neither an oxidant/reductant.
\(NO_{2}(g)\) was reduced (decreased from 4+ to 3+) and was oxidized (increased from 4+ to 5+), so it's the oxidant and reductant (This particular reaction is a disproportionation reaction).

Q20.7.5

This reaction is characteristic of a lead storage battery:

\(Pb(s)+PbO_{2}(s)+2H_{2}SO_{4}(aq)→2PbSO_{4}(s)+2H_{2}O(l)\)

If you have a battery with an electrolyte that has a density of \(1.15 g/cm^{3}\) and contains 30.0% sulfuric acid by mass, is the potential greater than or less than that of the standard cell?

S20.7.5

Let's assume that there is one liter of the electrolyte solution. Using the density, convert this volume into mass (use conversion factors \((\frac{1cm^3}{1mL})\) and \((\frac{1000mL}{1L})\)).

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

The solution contains 30.0% sulfuric acid by mass.

\((1150g\ soln.)(0.300)=345g\ H_{2}SO_{4}\)

Find the moles of sulfuric acid in solution (molar mass \(H_{2}SO_{4}=98.079g/mol\)).

\((345g\ H_{2}SO_{4})(\frac{1mol\ H_{2}SO_{4}}{98.079g\ H_{2}SO_{4}})=3.52mol\ H_{2}SO_{4}\)

Since we assumed there was only one liter of electrolyte solution, the molarity is \([H_{2}SO_{4}] = 3.52M\).
Looking at the reaction \(Pb(s)+PbO_{2}(s)+2H_{2}SO_{4}(aq)→2PbSO_{4}(s)+2H_{2}O(l)\), \(H_{2}SO_{4}\) is the only species in solution, so the reaction quotient equation is

\(Q=\frac{[products]}{[reactants]}=\frac{1}{[H_{2}SO_{4}]^2}=\frac{1}{(3.52M)^2}=0.0807\)

Using the Nearst equation, if we plug in \(lnQ=ln(0.0807)=-2.52\),

\(E=E^{\circ}-\frac{RT}{nF}lnQ\)
\(E=E^{\circ}-\frac{RT}{nF}(-2.52)=E^{\circ}+2.52(\frac{RT}{nF})\)

\(\frac{RT}{nF}\) will always be positive since it's a ratio of positive constants, temperature, and moles of electrons. Therefore, \(E>E^{\circ}\).