Extra Credit 36

Q17.5.4

Consider a battery with the overall reaction:

\[Cu(s)+2Ag^+(aq)⟶2Ag(s)+Cu^{2+}(aq)\]

(a) What is the reaction at the anode and cathode?

(b) A battery is “dead” when it has no cell potential. What is the value of \(Q\) when this battery is dead?

(c) If a particular dead battery was found to have \([Cu^{2+}] = 0.11\, M\), what was the concentration of silver ion?

Solution (a)

Oxidation occurs at the anode in a battery and is when a molecule, atom, or ion loses electrons.

In the overall reaction, \(Cu_{(s)}\) has an oxidation number (charge on ion/atom) of 0 because it is a solid.

\(Cu_{(s)}\) then becomes \(Cu^{2+}(aq)\) ion which has a +2 charge indicating \(Cu_{(s)}\) lost 2 electrons. Therefore \(Cu_{(s)}\) was oxidized. Oxidation occurs at the anode, so the reaction at the anode is... \[Cu(s)⟶Cu^{2+}(aq)+2e^-\]

Reduction occurs at the cathode in a battery and is when a molecule, atom, or ion gains electrons.

In the overall reaction, \(Ag^+(aq)\) has an oxidation number of +1.

\(Ag^+(aq)\) then becomes Ag(s) which has an oxidation number of 0 because it is a solid. \(Ag^+(aq)\) had to gain an electron in order to go from a +1 to 0 charge. Therefore \(Ag^+(aq)\) was reduced. Reduction occurs at the cathode, so the reaction at the cathode is... \(Ag^+(aq) + e^-⟶ Ag(s)\) but in order to balance the overall reaction the number of electrons gained and lost needs to cancel out and therefore be equal.

In order to cancel the electrons, the \(Cu_{(s)}\) lose of 2 electrons means the cathode reaction must also transfer 2 electrons, not 1. To do this, the cathode reaction must be multiplied by 2 (which doesn't effect the standard cell potential of the cathode. So the balanced reaction at the cathode is... \[2 \times (Ag^+(aq) + e^-⟶ Ag(s)) \]

Solution (b)

Using the Nernst Equation: \[E=E^∘-\frac{RT}{nF}lnQ\] with

• \(E\) is defined as the cell potential under general non-standard conditions
• \(E^∘\) is the cell potential under standard conditions calculated by \(E^∘=E^∘(cathode)-E^∘(anode)\)
• \(T\) is the temperature in Kelvin
• \(n\) is the number of electrons transferred in the reaction
• \(R\) is the ideal gas constant 8.314 J/mol•K
• \(F\) is the Faraday constant 96,500 C/mol

The Nernst Equation can be re-written plugging in constants with T=298 K as: \[E=E^∘-\frac{0.0592 V}{n}logQ\]

Using the table of Standard Reduction Potentials the \(E^∘(cathode)\) and \(E^∘(anode)\) are found: \[Ag^{+}{(aq)} + e^− \rightarrow Ag{(s)} \;\;\; E°_{cathode} = 0.7996\; V \] \[Cu{(s)} \rightarrow Cu^{2+}{(aq)} + 2e^- \;\;\; E°_{anode} = 0.34\; V \] so then \[E^∘=E^∘(cathode)-E^∘(anode)\] \[E^∘=0.7996 V-0.34, V\] \[E^∘=0.46\, V\]

In the balanced overall equation 2 electrons are transferred: \(Cu_{(s)}\) loses 2 electrons and \(2Ag^+\) gains the 2 electrons. Therefore \(n\)=2

Because the battery is "dead" there is no cell potential so \(E\)=0 therefore \[0=E^∘-\frac{0.0592 V}{n}logQ\] \[E^∘=\frac{0.0592 V}{n}logQ\] \[logQ=\frac{nE^∘}{0.0592}\] \[Q=10^{\; nE^∘/0.0592}\] plugging in \(n\)=2 and \(E^∘=0.46 V\)... \[Q=10^{\; 2(0.46)/0.0592}\] \[Q= 3.5 \times 10^{15}\]

Solution (c)

Because the battery is dead, it is known that non-standard cell potential of the battery \(E\)=0.

Using the Nernst Equation: \[0=E^∘-\frac{0.0592 V}{n}logQ\] \[logQ=\frac{nE^∘}{0.0592}\] \[Q=10^{\; nE^∘/0.0592}\]

\(Q\) is the reaction quotient and for the general form when \[aA + bB \rightarrow cC + dD\] \[Q=\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\] if A, B, C, and D are aqueous or gases. Solids and liquids =1 in \(Q\).

Therefore for the overall reaction: \[Cu(s)+2Ag^+(aq)⟶2Ag(s)+Cu^{2+}(aq)\] \(Q\) will only include \(Cu^{2+}(aq)\) and \(Ag^+(aq)\) since \(Cu_{(s)}\) and \(Ag_{(s)}\) are solids and =1 in \(Q\). The coefficient before product (in numerator of Q) \(Cu^{2+}(aq)\) is 1 and before reactant (in denominator of Q) \(Ag^+(aq)\) is 2 so... \[Q=\frac{[Cu^{2+}]}{[Ag^+]^2}\] then substituting Q into \(Q=10^{\; nE^∘/0.0592}\): \[\frac{[Cu^{2+}]}{[Ag^+]^2}=10^{\; nE^∘/0.0592}\] To solve for the concentration of \(Ag^+\) the equation can be reworked so that \[[Ag^+]= \sqrt{\frac{[Cu^{2+}]}{10^{\; nE^∘/0.0592}}}\] and n=2 since 2 electrons are transferred and from part (b) it was found that for the overall reaction \(E^∘=E^∘(cathode)-E^∘(anode)=0.46\; V\) and it is given that \([Cu^{2+}]=0.11\; M\), so then \[[Ag^+]= \sqrt{\frac{0.11}{10^{\; 2(0.46)/0.0592}}}\] \[[Ag^+]=5.6 \times 10^{−9}\, M\]

Q12.2.3

In the PhET Reactions & Rates interactive, use the "Many Collisions" tab to observe how multiple atoms and molecules interact under varying conditions. Select a molecule to pump into the chamber and observe the results of this addition. Set the initial temperature and select the current amounts of each reactant. Select "Show bonds" under Options. How is the rate of the reaction affected by concentration and temperature?

Solution

Based on the Collision Theory, a reaction will only occur if the molecules collide with proper orientation and with sufficient energy required for the reaction to occur. The minimum energy the molecules must collide with is called the activation energy (energy of transition state).

Increasing the concentration of reactants increases the probability that reactants will collide since there are more reactants in the same volume of space. Therefore, increasing the concentration of reactants would increase the rate of the reaction. Decreasing the concentration of reactants would decrease the rate of reaction because the overall number of possible collisions would decrease.

Temperature is directly related the the kinetic energy of molecules and activation energy \(E_a\) is the minimum energy required for a reaction to occur and doesn't change for a reaction. Increasing the temperature increases the kinetic energy of the reactants meaning the reactants will move faster and collide with each other more frequently. Therefore, increasing the temperature increase the rate of the reaction. Decreasing the temperature decreases the rate of reaction since the molecules will have less kinetic energy, move slower, and therefore collide with each other less frequently.

Q12.5.8

In an experiment, a sample of \(NaClO_3\) was 90% decomposed in 48 min. Approximately how long would this decomposition have taken if the sample had been heated 20 °C higher?

Solution

The decomposition of \(NaClO_3\) is a first-order reaction: \[2NaCl{O_3} \rightarrow 2NaCl + 3O_2 \] Rate laws are determined experimentally and have the general form: \[aA + bB \rightarrow cC + dD\] \[rate=k[A]^{x}[B]^y\] where

• \(x\) and \(y\) are reaction orders that can only be determined experimentally
• \(k\) is the rate constant for the reaction
• \(x\)+\(y\) is the overall reaction order

Because the decomposition of \(NaClO_3\) is a first-order reaction and the only reactant is \(NaClO_3\) x must equal 1. So the rate law is... \[rate=k[NaClO_3]\] integrating this rate law in terms of time results in... \[rate=- \frac{d[NaClO_3]}{dt} =k[NaClO_3]\] \[ln \frac{[NaClO_3]}{[NaClO_3]_°}=\; -kt\] It is given that in 48 mins 90% of reactant has decomposed so \([NaClO_3]_ °= 1\;M\) and \([NaClO_3]=1((100-10)/100)=\; 0.1\;M\). Therefore... \[k=- \frac{ln \frac{[NaClO_3]}{[NaClO_3]_°}}{t}\] \[k=- \frac{ln \frac{0.1}{1}}{48}\] \[k=\; 0.048\; min^{-1}\]

The Arrhenius equation can be written in two ways (one without the collision factor\(A\), one with this collision factor) is: \[ln(k_1)=\; -\frac{E_a}{RT_1} +ln(A)\;\;\;\;\;\;\; (Eq.\; 1)\] \[ln\; \frac{k_2}{k_1}= \frac{E_a}{R}(\frac{1}{T_1} -\frac{1}{T_2})\;\;\;\;\;\;\; (Eq.\; 2)\] where

• \(E_a\) is the activation energy (must solve for this)
• \(A\) is the collision frequency factor (not given, and can't solve for it)
• \(k_1\) is the rate constant for the reaction at \(T_1\) (found to be \(0.048\; min^{-1}\))
• \(k_2\) is the rate constant for the reaction at \(T_2\) (must solve for this)
• \(R\) is the ideal gas constant 8.314 J/(K•mol)
• \(T_1\) is the temperature in Kelvin of the first reaction (assume it is standard temperature, 298 K)
• \(T_2\) is the temperature in Kelvin of the second reaction (given that it is 20° higher than \(T_1\))

It was found that \(k_1=\; 0.048\; min^{-1}\), and since no temperature was given, it can be assumed that the experiment occurred at \(T_1=298\; K\). Using Eq. 1, the approximate activation energy, \(E_a\) can be found. The exact \(E_a\) can't be found because the collision frequency factor is not given, therefore we will assume \(A\)=1.

Since, \(k_1\) and \(T_1\) are known, and we assume \(A\)=1, the activation energy \(E_a\) can be found... \[ln(k_1)=\; -\frac{E_a}{RT_1} +ln(A)\]Using algebra... \[E_a=\; -RT[ln(k_1)-ln(A)]\] \[E_a=\; -8.314(298)[ln(0.048)-0]\] \[E_a=\; 7500\; J/mol\]Using Eq. 2, \(k_2\) can be found since \(k_1\), \(T_1\), \(T_2\), and \(E_a\) are known... \[ln (\frac{k_2}{k_1})= \frac{E_a}{R}(\frac{1}{T_1} -\frac{1}{T_2})\] \[ln(k_2)-ln(k_1)= \frac{E_a}{R}(\frac{1}{T_1} -\frac{1}{T_2})\] \[ln(k_2)= \frac{E_a}{R}(\frac{1}{T_1} -\frac{1}{T_2})+ln(k_1)\] \[k_2= e^{\frac{E_a}{R}(\frac{1}{T_1} -\frac{1}{T_2})+ln(k_1)}\] \[k_2= e^{\frac{7500}{8.314}(\frac{1}{298} -\frac{1}{(298+20)})+ln0.048}\] \[k_2=\; 0.058\; min^{-1}\]Now using \(k_2\) in \(ln \frac{[NaClO_3]}{[NaClO_3]_°}=\; -kt\), and the same 90% decomposition the time can be found to be...\[t=- \frac{ln \frac{[NaClO_3]}{[NaClO_3]_°}}{k_2}\] \[t=- \frac{ln \frac{0.1}{1}}{0.058}\] \[t= 40\; mins\; approximately \]As temperature increases, it should take less time for a reaction to occur, and since it initially took 48 minutes for the 90% decomposition of \(NaClO_3\) and then only took 40 minutes after the temperature was raised, this answer is plausible.

Q21.4.3

What is the change in the nucleus that results from the following decay scenarios?

a) Emission of a \(\beta\) particle

b) Emission of a \(\beta^{+}\) particle

c) Capture of an electron

Solution (a)

A \(\beta\) particle has a charge of -1 and no mass. Therefore when a \(\beta\) particle is emitted from the nucleus, the daughter nucleus will have no change in mass, but a charge increased by 1. In a nucleus there are neutrons (mass of 1 amu and no charge) and protons (+1 charge and no mass). Therefore when a \(\beta\) particle is emitted from the nucleus a neutron is split into a proton and a \(\beta\) particle... \[\mathrm{\ce{^{1}_{0}n}\rightarrow\ce{^{1}_{1}p}+\ce{^{0}_{-1}ß}}\]

Solution (b)

A \\(\beta^{+}\) particle has a change of +1 and no mass. Therefore, when a \(\beta^{+}\) particle is emitted from the nucleus, the daughter nucleus will have no change in mass, but a charge decreased by 1. In a nucleus where there are protons and neutrons, in order for a \(\beta^{+}\) particle to be emitted a proton must split into a neutron and a \(\beta^{+}\) particle... \[\mathrm{\ce{^{1}_{1}p}\rightarrow\ce{^{1}_{0}n}+\ce{^{0}_{1}ß}}\]

Solution (c)

An electron has a charge of -1 and no mass. When an electron is captured by the nucleus the daughter nucleus will have no change in mass, but a charge decrease of 1. In a nucleus where there are protons and neutrons, in order for an electron to be captured, a proton must capture an emitted electron and become a neutron... \[\mathrm{\ce{^{1}_{1}p}+\ce{^{0}_{-1}e}\rightarrow\ce{^{1}_{0}n}}\]

Q20.2.7

Which of these metals produces H₂ in acidic solution?

1. \(Ag\)
2. \(Cd\)
3. \(Ca\)
4. \(Cu\)

   Solution

   In order to solve this problem we must consult the table of Standard Reduction Potentials. Using the table we cn determine that in order for \(H_2\) to be produced, \(H^+\) must be reduced by a reducing agent. The reduction of \(H^+\) to \(H_2\) is... \[2H^{+}{(aq)} + 2e^− \rightarrow H_{2}{(g)} \;\;\; E° = 0\; V \] A spontaneous reaction (reaction will occur with no external energy source) has an E°>0. \(E°=E°_{cathode}-E°_{anode}\). Since reduction occurs at the cathode and in an acidic solution \(H^+\) gains electrons and is reduced to \(H_2\), the \(E°_{cathode}=\; 0\). \[E°>0 \;\;\; and\; E°_{cathode}=0\; V\] \[0>0-E°_{anode}\] Therefore, any of the 4 metals that has a standard reduction potential E°<0 will reduce \(H^+\) to \(H_2\) in an acidic solution. \[Ag^{+}{(aq)} + e^− \rightarrow Ag{(s)} \;\;\; E° = 0.7996\; V \] \[Cd^{2+}{(aq)} + 2e^− \rightarrow Cd{(s)} \;\;\; E° = -0.403\; V \] \[Ca^{2+}{(aq)} + 2e^− \rightarrow Ca{(s)} \;\;\; E° = -2.84\; V \] \[Cu^{2+}{(aq)} + 2e^− \rightarrow Cu{(s)} \;\;\; E° = 0.3419\; V \]E°<0 for Cd and Ca so Cd and Ca will produce \(H_2\) in an acidic solution.

   Q20.5.2

   (a) What is the relationship between the measured cell potential and the total charge that passes through a cell?

   (b) Which of these is dependent on concentration?

   (c) Which is dependent on the identity of the oxidant or the reductant?

   (d) Which is dependent on the number of electrons transferred?

   Solution (a)

   The measured cell potential of a battery is \(E\), while the total charge that passes through the cell is the number of electrons transferred in the system \(n\) multiplied by Faradays constant \(F\) (96,500 C/mol e-) . The relationship between the two is that the measured cell potential of the cell is: \[ΔG=\; -nFE\]Therefore, the total charge \(nF\) times the measured cell potential \(E\) equals the amount of available energy (work) of the system \(-\Delta G\): \[work=\; (total\; charge\; that\; passes\; through\; cell)(measured\; cell\; potential\; of\; cell)\]

   Solution (b)

   The Quotient factor \(Q\) is dependent on the concentrations of reactants and products in a system: \[aA + bB \rightarrow cC + dD\] \[Q=\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\] and is part of the Nernst Equation \(E=E^∘-\frac{0.0592 V}{n}logQ\) that calculates the measured cell potential of a battery (E ). Therefore, the measured cell potential of a cell is dependent on concentration.

   Solution (c)

   The cell potential is calculated by \(E^∘=E°_{cathode}-E°_{anode}\) where \(E°_{cathode}\) and \(E°_{anode}\) are dependent on the identity of the oxidant \((E°_{anode})\) and the identity of the reductant \((E°_{cathode})\) and are determined using the table of standard reduction potentials. Therefore, the measured cell potential of a cell is dependent on the identify of the oxidant and reductant.

   Solution (d)

   The Nernst Equation \(E=E^∘-\frac{0.0592 V}{n}logQ\) calculates the measured cell potential of a battery \(E\) and is dependent on the number of electrons transferred, n, which is determined from the balanced equation of a oxidation-reduction reaction. Also, the total charge that passes through a cell \(Q\) equals the number of electrons transferred \(n\) multiplied by Faradays constant\(F\): \(Q=\; nF\). Therefore, the measured cell potential of a cell \(E\) and the total charge that passes through the cell (Q) are dependent on the number of electrons transferred.

   Q20.9.11

   Predict the products obtained at each electrode when aqueous solutions of the following are electrolyzed.

   (a) \(AgNO_{3}\)

   (b) \(RbI\)

   Solution (a)

   When \(AgNO_3\) is electrolyzed in an aqueous solution, because the bond between silver and nitrate is ionic, in a polar solution (water), it will dissociate. Therefore \[AgNO_3 \rightarrow Ag^+ + NO_{3}^-\] Electrolysis occurs when a system has an \(E^{o}<0\), since the reaction is non-spontaneous and therefore needs an external energy source to push the reaction in a non-spontaneous direction. Because the solution contains \(Ag^+\) ions and \(Ag^+\) cannot lose another electron due to a high ionization energy, it will gain electrons be reduced: \[Ag^{+}{(aq)} + e^− \rightarrow Ag{(s)} \;\;\; E° = 0.7996\; V \] Using the table of Standard Reduction Potentials, water, the aqueous solution has the reduction half reactions: \[O_2{(g)} + 4H^+{(aq)} +4e^− \rightarrow 2H_2O{(l)} \;\;\; E° = 1.229\; V \] \[H^+{(aq)} +OH^-{(aq)} \rightarrow H_2O{(l)} \;\;\; E° = 0\; V \] \(E°=E°_{cathode}-E°_{anode}\) and since reduction occurs at the cathode, \(E°=\; 0.7996-E°_{anode}\). In order for E°<0, \(E°_{anode}>\; 0.7996\; V\). Only \(O_2{(g)} + 4H^+{(aq)} +4e^− \rightarrow 2H_2O{(l)}\) has \(E°>\; 0.7996\; V\). So oxidation (lose of electrons) occurs at the anode, reduction (gain of electrons) occurs at the cathode: \[Cathode:\;\;\;\; Ag^{+}{(aq)} + e^− \rightarrow Ag{(s)}\] \[Anode:\;\;\;\; 2H_2O{(l)} \rightarrow O_2{(g)} + 4H^+{(aq)} +4e^−\] Therefore, Ag(s) is produced at the cathode, and \(O_2{(g)}\) is produced at the anode.

   Solution (b)

   When \(RbI\) is electrolyzed in an aqueous solution, because the bond between Rubidium and Iodine is ionic, in a polar solution (water), it will dissociate: \[RbI \rightarrow Rb^+ + I^-\]Electrolysis occurs when a system has an \(E^{o}<0\), since the reaction is non-spontaneous and therefore needs an external energy source to push the reaction in a non-spontaneous direction. Because the solution contains \(I^-\) ions and \(I^-\) cannot gain another electron due to a low electronegativity, it will lose electrons and be oxidized: \[2I^-{(aq)} \rightarrow I_2{(s)} + 2e^-\]Using the table of Standard Reduction Potentials, the standard reduction potential for the reduction of \(I_2{(s)}\) is \(E°=\; 0.5355\; V\) and this is the \(E°_{anode}\) because oxidation occurs at the anode. Water, the aqueous solution has a reduction reaction where water \((H_2O{(l)})\) is a reactant that is reduced: \[H_2O{(l)} + 2e^- \rightarrow H_2{(g)} + 2OH^-{(aq)}\;\;\; E°=\; -0.828\; V \]E°<0 for \(E°=E°_{cathode}-E°_{anode}\) for electrolysis and since reduction occurs at the cathode \(E°_{cathode}=\; -0.828\; V\), and since oxidation occurs at the anode \(E°_{anode}=\; 0.5355\; V\): \[Cathode:\;\;\;\; 2H_2O{(l)} + 2e^- \rightarrow H_2{(g)} + 2OH^-{(aq)}\] \[Anode:\;\;\;\; 2I^-{(aq)} \rightarrow I_2{(s)} + 2e^-\] \[E°=E°_{cathode}-E°_{anode}\] \[E°=-0.828-0.5355\; V\] \[E°=-1.364\; V\; so\; E°<0\]Therefore, \(I_2{(g)}\) is produced at the anode, and \(H_2{(g)}\) is produced at the cathode.

   Q14.6.9

   The following reactions are given:

   \[\mathrm{A+B}\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}}\mathrm{C+D}\;\;\;\;\; (Eq.\; 1)\]

   \[\mathrm{D+E}\xrightarrow{k_2}\mathrm F\;\;\;\;\; (Eq.\; 2)\]

   (a) What is the relationship between the relative magnitudes of \(k_{−1}\) and \(k_2\) if these reactions have the following rate law?

   \[\dfrac{Δ[F]}{Δt} = k\dfrac{[A][B][E]}{[C]}\]

   (b) How does the magnitude of \(k_1\) compare to that of \(k_2\)?

   (c) Under what conditions would you expect the rate law to be

   \[\dfrac{Δ[F]}{Δt} =k′[A][B]?\]

   Assume that the rates of the forward and reverse reactions in the first equation are equal.

   Solution (a)

   Since the forward and reverse reactions in the first equation are equal, rate of forward reaction is equal to the rate of the reverse reaction. The overall reaction is broken down into it's mechanics, therefore, the rate law of the reaction can be determined from the mechanism equations, not just experientially. For Eq. 1, the rate laws of the forward and reverse reactions are... \[forward\; reaction:\;\;\;\;\; rate=k_1[A][B]\] \[reverse\; reaction:\;\;\;\;\; rate=k_{-1}[C][D]\]which can be rearranged since the rates are equal to be \[k_1[A][B]=k_{-1}[C][D]\]The rate law for Eq. 2 is... \[rate_1= \frac{Δ[F]}{Δt}=\; k_2[D][E]\] Using algebra we find that... \[k_1[A][B]=k_{-1}[C][D]\] \[[D]=\; \frac{k_1[A][B]}{k_{-1}[C]}\]Substituting [D] into Eq. 2's rate law gives \[\frac{Δ[F]}{Δt}=\; \frac{k_2k_1[A][B][E]}{k_{-1}[C]}\]Given \[\dfrac{Δ[F]}{Δt} = k\dfrac{[A][B][E]}{[C]}\] \[k=\frac{k_1k_2}{k_{-1}}\]Because Eq. 2 is the rate determining step (can tell because overall rate law includes [E] and intermediate D is substituted by reactants and products from faster step) the relationship between the relative magnitude of \(k_{-1}\) and \(k_2\) is... \(k_{-1}\)>>\(k_2\) because the rate determining step is the "slow step" so \(k_2\) is a small constant value so the rate is slower, and \(k_{-1}=k_1\) since the rates are equal, there are equal moles of each product and reactant, and the forward and reverse reaction are at equilibirum.

   Solution (b)

   From part (a)... \[k=\frac{k_1k_2}{k_{-1}}\]And to reiterate, Eq. 2 is the slow step (rate determining reaction), and \(k_{-1}=k_1\), therefore, \(k_1\)>>\(k_2\). For a better explanation refer to part (a, above).

   Solution (c)

   When the overall rate law is... \[\dfrac{Δ[F]}{Δt} =k′[A][B]?\]Which is the same as the rate law for the forward reaction of Eq. 1, \(rate=\;k_1[A][B]\), the forward reaction \(A + B \rightarrow C + D\) with rate constant (\(k_1\)) is the rate determining step (slow step) of the reaction. The slow rate determining step of reaction always has it's rate law equal to the overall reaction's rate law unless there are intermediates, which in this case D is the intermediate and isn't a reactant of forward Eq. 1.