Extra Credit 48

Q17.7.2

What mass of each product is produced in each of the electrolytic cells of the previous problem if a total charge of 3.33 × 10⁵ C passes through each cell? Assume the voltage is sufficient to perform the reduction.

1. CaCl₂
2. LiH
3. AlCl₃
4. CrBr₃

Solution

The number of moles of electrons is calculated as follows,

Substitute \(Q=3.3\times 10^{5}C\) and \(F=96485C/mol\) into the equation \(Q=n\times F\) to calculate \(n\). \[n=(3.3\times 10^{5}C)\times (\frac{1 mol e^{-}}{96485C})\]

\[n=3.45mol\: e^{-}\]

1. The overall equation can be written as follows,

Anode: \(Ca\rightarrow Ca^{2+}+2e^{-}\)

Cathode: \(Cl_{2}+2e^{-}\rightarrow 2Cl^{-}\)

Overall: \(Ca+Cl_{2}\rightarrow Ca^{2+}+2Cl^{-}\)

One mole of Ca gives 2 moles of electrons to form 1 mole of Ca²⁺ and one mole of Cl₂ gives 2 moles of electrons to form 2 moles of Cl^-. Therefore the total electrons transferred equates to 2.

Therefore,

Mass of Ca= \(3.45\: mol\: e^{-}\times \frac{1\: mol}{2\: mol\: e^{-}}\times \frac{40.078\: g}{mol}=\; 69.1\: g\)

Mass of Cl₂= \(3.45\: mol\: e^{-}\times \frac{1\: mol}{2\: mol\: e^{-}}\times \frac{70.906\: g}{mol}=\; 122\: g\)

2. The overall equation can be written as follows,

Anode: \(2\times (Li\rightarrow Li^{+}+1e^{-})\)

Cathode: \(2H^{+}+2e^{-}\rightarrow H_{2}\)

Overall: \(2Li+2H^{+}\rightarrow 2Li^{+}+H_{2}\)

One mole of Li gives 1 mole of electrons to form 1 mole of Li⁺ and two moles of H⁺ gives 2 moles of electrons to form 1 mole of H₂. Therefore the total electrons transferred equates to 2.

Therefore,

Mass of Li= \(3.45\: mol\: e^{-}\times \frac{2\: mol}{2\: mol\: e^{-}}\times \frac{6.941\: g}{mol}=\; 23.9\: g\)

Mass of H₂= \(3.45\: mol\: e^{-}\times \frac{1\: mol}{2\: mol\: e^{-}}\times \frac{2.0158\: g}{mol}=\; 3.48\: g\)

3. The overall equation can be written as follows,

Anode: \(2\times (Al\rightarrow Al^{3+}+3e^{-})\)

Cathode: \(3\times (Cl_{2}+2e^{-}\rightarrow 2Cl^{-})\)

Overall: \(2Al+3Cl_{2}\rightarrow 2Al^{3+}+6Cl^{-}\)

One mole of Al gives 3 moles of electrons to form 1 mole of Al³⁺ and one mole of Cl₂ gives 2 moles of electrons to form 2 moles of Cl^-. Therefore the total electrons transferred equates to 6.

Therefore,

Mass of Al= \(3.45\: mol\: e^{-}\times \frac{2\: mol}{6\: mol\: e^{-}}\times \frac{26.9815\: g}{mol}=\; 31.0\: g\)

Mass of Cl₂= \(3.45\: mol\: e^{-}\times \frac{3\: mol}{6\: mol\: e^{-}}\times \frac{70.906\: g}{mol}=\;122\: g\)

4. The overall equation can be written as follows,

Anode: \(Cr\rightarrow Cr^{3+}+3e^{-}\)

Cathode: \(3\times (Br^{-}+1e^{-}\rightarrow Br)\)

Overall: \(Cr+3Br^{-}\rightarrow Cr^{3+}+3Br\)

One mole of Cr gives 3 moles of electrons to form 1 mole of Cr³⁺ and one mole of Br₂ gives 1 mole of electrons to form 1 mole of Br. Therefore the total electrons transferred equates to 3.

Therefore,

Mass of Cr= \(3.45\: mol\: e^{-}\times \frac{2\: mol}{6\: mol\: e^{-}}\times \frac{51.9961\: g}{mol}=\;59.8\: g\)

Mass of Br₂= \(3.45\: mol\: e^{-}\times \frac{3\: mol}{6\: mol\: e^{-}}\times \frac{159.808\: g}{mol}=\;276\: g\)

1. mass Ca=69.1g mass Cl₂=122g
2. mass Li=23.9g mass H₂=3.48g
3. mass Al=31.0g mass Cl₂=122g
4. mass Cr=59.8g mass Br₂=276g

Q12.3.11

Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by 25–30%. Women metabolize alcohol a little more slowly than men:

Determine the rate equation, the rate constant, and the overall order for this reaction.

Solution

Start with the general rate law equation: \[rate=k[A]^{m}\]

Where k is the rate constant for the reaction at a certain temperature, and m is the order with respect to A.

Plug \(A=[C_{2}H_{5}OH]\) into the general rate law equation \[rate=k[C_{2}H_{5}OH]^{m}\]

Plug the three sets of values into the rate law equation

First set of values is as follows;

\([C_{2}H_{5}OH]=4.4\times 10^{-2}M\) \(rate=2.0\times 10^{-2}mol/L/h\)

\(2.0\times 10^{-2}=k[4.4\times 10^{-2}]^{m}\)

\(log(2.0\times 10^{-2})=log(k)+mlog(4.4\times 10^{-2})\)

\(-1.699=log(k)-1.357m\)

Second set of values is as follows;

\([C_{2}H_{5}OH]=3.3\times 10^{-2}M\) \(rate=2.0\times 10^{-2}mol/L/h\)

\(2.0\times 10^{-2}=k[3.3\times 10^{-2}]^{m}\)

\(log(2.0\times 10^{-2})=log(k)+mlog(3.3\times 10^{-2})\)

\(-1.699=log(k)-1.481m\)

Subtract \(-1.699=log(k)-1.481m\) from \(-1.699=log(k)-1.357m\) as follows;

\((log(k)-1.357m)-(log(k)-1.481m)=-1.699-(-1.699)\)

\(1.481m-1.357m=0\)

\(0.124m=0\)

\(m=0\)

The order of the reaction is zero. The reaction order being zero means there is no correlation between concentration and reaction rate.

Thus the rate equation for this reaction is \(rate=k[C_{2}H_{5}OH]^{0}\)

\(rate=k\times 1\)

\(rate=k\)

Since the rate of the equation equals the rate constant, the rate constant is \(k=2.0\times 10^{-2}M/L/h\).

rate = k; k = 2.0 × 10⁻² mol/L/h (about 0.9 g/L/h for the average male); The reaction is zero order.

Q12.6.3

Phosgene, COCl₂, one of the poison gases used during World War I, is formed from chlorine and carbon monoxide. The mechanism is thought to proceed by:

1. Write the overall reaction equation.
2. Identify any reaction intermediates.

Solution

1. To write the overall reaction follow this hypothetical stepwise reaction:

The reaction includes these elementary steps,

\[A+B\rightarrow X^{*}\]

\[X^{*}\rightarrow C+D\]

The reaction equation,

\[A+B\rightarrow C+D\]

The overall reaction equation for the mechanism:

\[CO+Cl_{2}\rightarrow COCl_{2}\]

2. An intermediate in a reaction is a species that does not appear in the balanced chemical equation for the overall reaction. A reaction intermediate is transient species within a multi-step reaction mechanism that is produced in the preceding step and consumed in a subsequent step to ultimately generate the final reaction product.

The intermediates are \(Cl\) and \(COCl\).

Q21.4.15

²³⁹Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the ²³⁹Pu present today will be present in 1000 y?

Solution

The half life of \(_{94}^{239}\textrm{Pu}\) is \(t_{\frac{1}{2}}\)=24,000y

The value of \(\lambda\) is calculated as follows,

Substitute \(t_{\frac{1}{2}}\)=24,000y into \(\lambda=\frac{ln(2))}{t\frac{1}{2}}\) to calcuate the value of \(\lambda\)

\[\lambda=\frac{ln(2))}{24000y} =2.888\times 10^{-5}y^{-1}\]

Substitute \(2.888\times 10^{-5}y^{-1}\) for \(\lambda\) and \(t=1000y\) into the expression \(\frac{N}{N_{0}}=e^{-\lambda t}\) to calculate \(\frac{N}{N_{0}}\)

\[\frac{N}{N_{0}}=e^{-(2.888\times 10^{-5}y^{-1})(1000y)}=0.9715\]

The percentage is calculated by \(\frac{N}{N_{0}}\times 100\)%

\(0.9715\times 100=97.15\)%

So, the percentage of \(_{94}^{239}\textrm{Pu}\) left is 97.15%

Q20.3.3

What is the difference between a galvanic cell and an electrolytic cell? Which would you use to generate electricity?

Solution

A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction (ΔG < 0) to generate electricity. This type of electrochemical cell is often called a voltaic cell after its inventor, the Italian physicist Alessandro Volta (1745–1827). A galvanic cell transforms the energy released by a spontaneous redox reaction into electrical energy that can be used to perform work. The oxidative and reductive half-reactions usually occur in separate compartments that are connected by an external electrical circuit; in addition, a second connection that allows ions to flow between the compartments through a porous barrier is necessary to maintain electrical neutrality. The potential difference between the electrodes (voltage) causes electrons to flow from the reductant to the oxidant through the external circuit, generating an electric current.

An electrolytic cell consumes electrical energy from an external source, using it to cause a nonspontaneous redox reaction to occur (ΔG > 0). In an electrolytic cell, an external source of electrical energy is used to generate a potential difference between the electrodes that forces electrons to flow, driving a nonspontaneous redox reaction. A single compartment is employed in most cells.

Both types contain two electrodes, which are solid metals connected to an external circuit that provides an electrical connection between the two parts of the system. The oxidation half-reaction occurs at one electrode (the anode), and the reduction half-reaction occurs at the other (the cathode). When the circuit is closed, electrons flow from the anode to the cathode. The electrodes are also connected by an electrolyte, an ionic substance or solution that allows ions to transfer between the electrode compartments, thereby maintaining the system’s electrical neutrality.

You would use a galvanic (voltaic) cell to generate electricity.

Figure: In both kinds of electrochemical cells, the anode is the electrode at which the oxidation half-reaction occurs, and the cathode is the electrode at which the reduction half-reaction occurs.

Q20.5.14

Explain why the sum of the potentials for the half-reactions Sn²⁺(aq) + 2e⁻ → Sn(s) and Sn⁴⁺(aq) + 2e⁻ → Sn²⁺(aq) does not equal the potential for the reaction Sn⁴⁺(aq) + 4e⁻ → Sn(s). What is the net cell potential? Compare the values of ΔG° for the sum of the potentials and the actual net cell potential.

Solution

The given two half-reactions are:

Sn²⁺(aq) + 2e− → Sn(s) \(E^{o}_{1}\) = -0.14 V

Sn⁴⁺(aq) + 2e− → Sn²⁺(aq) \(E^{o}_{2}\)= 0.13 V

by adding the above two half reactions gives the overall half reaction:

Sn⁴⁺(aq) + 4e− → Sn(s) \(E^{0}\) = -0.01 V

The sum of two half-reactions gives the desired overall half-reaction but we cannot simply add the potentials of two reductive half-reactions to obtain the potential of overall half-reaction because E° is not a state function. While, ΔG° is a state function, therefore by adding the values of ΔG° for the individual reactions provides ΔG° for the overall reaction. To determine the value of E° for the overall half reaction, first add the values of ΔG° (= −nFE°) for each half-reaction to obtain ΔG° for the overall half reaction.

\[\Delta G=-nFE^{0}\]

In both half reactions n = 2, \(E^{o}_{1}\) = -0.14, \(E^{o}_{2}\) = 0.13

\(\Delta G^{0}\) = −2F(-0.14V)

\(\Delta G^{0}\) = −2F(0.13V)

Solving \(\Delta G^{0}\) for overall reaction

Sn⁴⁺(aq) + 4e− → Sn(s)

\(\Delta G^{0}\) = [−2F(-0.14) ]+ [−2F(0.13)]

\(\Delta G^{0}\) = F[(0.28) ]+ [(−0.26)] = F( 0.02V)

Four electrons are transferred in overall half reaction then

\(\Delta G^{0}\) = \(-4FE^{0}\)

F( 0.02V) = \(-4FE^{0}\)

E°l = -0.02V/4 = -0.005 V

The value -0.005 V is different from the value -0.01 V (net cell potential) which is obtained by simply adding \(E^{o}_{1}\) and \(E^{o}_{2}\) for the two half reactions.

Q24.6.9

Q14.4.7

Benzoyl peroxide is a medication used to treat acne. Its rate of thermal decomposition at several concentrations was determined experimentally, and the data were tabulated as follows:

What is the reaction order with respect to benzoyl peroxide? What is the rate law for this reaction?

Solution

Start with the general rate law equation: \[rate=k[A]^{m}\]

Where k is the rate constant for the reaction at a certain temperature, and m is the order with respect to A.

Plug \(A=[C_{14}H_{10}O_{4}]\) into the general rate law equation. \[rate=k[C_{14}H_{10}O_{4}]^{m}\]

Plug the values into the rate law equation

First set of values is as follows;

\(C_{14}H_{10}O_{4}=1M\) \(rate=2.2\times 10^{-4}M/s\)

\(2.2\times 10^{-4}=k[1]^{m}\)

\(log(2.2\times 10^{-4})=log(k)+mlog(1)\)

\(-3.657=log(k)+0m\)

k= \(2.2\times 10^{-4}\)

Plug k into the rate equation \(rate=k[C_{14}H_{10}O_{4}]^{m}\)

\[rate=2.2\times 10^{-4}[C_{14}H_{10}O_{4}]^{m}\]

Find the order by plugging more values into the rate equation

Use the second set of values;

\(C_{14}H_{10}O_{4}=0.70M\) \(rate=1.64\times 10^{-4}M/s\)

\(1.64\times 10^{-4}=2.2\times 10^{-4}[0.70]^{m}\)

\(log(1.64\times 10^{-4})=log(2.2\times 10^{-4})+mlog[0.70]\)

\(-3.78=-3.657-0.154m\)

Use the third set of values;

\(C_{14}H_{10}O_{4}=0.50M\) \(rate=1.12\times 10^{-4}M/s\)

\(1.12\times 10^{-4}=2.2\times 10^{-4}[0.50]^{m}\)

\(log(1.12\times 10^{-4})=log(2.2\times 10^{-4})+mlog[0.50]\)

\(-3.95=-3.657-0.301m\)

Subtract the two equations

\((-3.657-0.301m)-(-3.657-0.154m)=(-3.95)-(-3.78)\)

\(m=1.13\)

Plug the m into the equation \(rate=2.2\times 10^{-4}[C_{14}H_{10}O_{4}]^{m}\)

\[rate=2.2\times 10^{-4}[C_{14}H_{10}O_{4}]^{1.13}\]