Extra Credit 36

Q17.5.4

Consider a battery with the overall reaction:

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

1. What is the reaction at the anode and cathode?
2. A battery is “dead” when it has no cell potential. What is the value of Q when this battery is dead?
3. If a particular dead battery was found to have [Cu²⁺]=0.11M, what was the concentration of silver ion?

Solution:

1.

The reactions that take place at the anode is oxidation and the reaction that takes place at the cathode is reduction. Recall that oxidation is the loss of electrons and reduction is the gain of electrons.

OIL RIG: Oxidation Is Loss, Reductions Is Gain.

Before writing out the half reactions to determine which one is being reduced or which one is being oxidized, we look at each element's oxidation number. If the oxidation number is more positive on the product side, then the element as been oxidized. If the oxidation number is more negative on the product side, then the element has been reduced.

Easy way to remember what reaction is taking place at the anode/cathode:

AN OX: ANode OXidation

RED CAT: REDuction CAThode

Write out the oxidation number of each element to determine which is being reduced/oxidized.

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

Reactant: Cu(s) has an oxidation number of 0 because it is a solid.

Product: has an oxidation number of .

Conclusion: Copper has been oxidized because its oxidation number is more positive after the reaction has occurred. Its oxidation number went from 0 to +2.

Reactant: has an oxidation number of .

Product: has an oxidation number of 0 because it is a solid.

Conclusion: Silver has been reduced because its oxidation number is more negative after the reaction has occurred. Its oxidation number went from +1 to zero.

After assigning oxidation numbers to find out which reaction is being reduced/oxidized, we can then write out the half reactions keeping in mind that reduction happens in the cathode and oxidation happens in the anode.

Anode/Oxidation:

\[Cu(s)\rightarrow Cu^{2+}(aq)+2e^{-}\]

\[E^{\circ}_{cell}=.3419V\] (Standard Potentials)

Since it is a loss in electrons, the electrons should be added to the product side of the reaction. To determine how many electrons to add to which side of the reaction, we look at the overall charge on each side of the arrow- the reactants and the products.

The reactants have a total charge of 0, whereas the product side has a total charge of +2. In order to balance both sides, we must add two electrons to the product side. Electrons has a -1 charge therefore by adding two electrons to the product side, we will end up with a total charge of 0. Now both sides had been balanced.

Cathode/Reduction:

\[2×(Ag^{+}(aq)+e^{-}\rightarrow Ag(s))\]

\[E^{\circ}_{cell}=.7996V\] (Standard Potentials)

The same process applies to the cathode/reduction. Since it is an reduction reaction, the electrons should appear on the reactant side. It is a gain in electrons. Then we look at the overall charge on each side of the arrow to determine how many electrons need to be added in order to balance out the reaction.

The reactant has a total charge of +1 and the product side has a total charge of 0. To make both sides equal, we must add one electron to the reactant side so that both sides have a charge of 0.

2. If the battery is "dead" then the value of Q can be found by the equation. (n=electrons transferred)

\[E_{cell}= E^{\circ}_{cell}- (\frac{0.0592V}{n}) log(Q)\]

Setting E_cell=0 when the battery is dead because there is no net transfer of electrons when the battery is dead.

\[0= E^{\circ}_{cell}- (\frac{0.0592V}{n}) log(Q)\]

First we have to start by finding E^○_cell, use the standard potential reference tables.

\[E^{\circ}cell= E^{\circ}(Cathode)- E^{\circ}(Anode)\]

\[E^{\circ}cell= E^{\circ}(.7996)- E^{\circ}(.3419)=.4577V\]

Rearranging our equation and inserting the known we get

\[E^{\circ}_{cell}= (\frac{0.0592V}{n}) log(Q)\]

\[log(Q)= (\frac{ E^{\circ}_{cell}(n)}{0.0592V})\]

\[log(Q)=(\frac{.4577V\left ( 2 \right )}{0.0592V})=15.4628\]

\[Q=10^{15.4628}= 2.9*10^{15}\]

3. We can use Q to find the concentration of Ag⁺ by using the given information of the reaction, the Q value calculated above and the definition of Q.

According to this reaction:

\[Q=\frac{[Cu^{2+}]}{[Ag^{+}]^{2}}\]

\[2.9*10^{15}=\frac{[.11M]}{[Ag^{+}]^{2}}\]

\[[Ag^{+}]=(3.79*10^{-17})^{1/2}=6.15*10^{-9}M\] The concentration of [Ag⁺] in the dead battery is 6.15*10^-9M .

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. 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:

The rate of the reaction is directly affected by both the temperature and the concentration, as you increased the temperature the molecules speed up and the reaction occurs quicker. Similarly when you increase the concentration there are more particles and so they are more likely to collide and react making the reaction occur quicker. This is true because in order for a reaction to take place the molecules not only have to collide but they also must be in a specific orientation and they must have enough energy for the reaction to take place. If there are fewer molecules then they are less likely to interact with each other as often. In contrast, when the temperature is increased and the concentration is increased the kinetic energy of molecules increases and they tend to move more and since they are moving faster and more erratically they are more likely to collide with another molecule, both of which will make the rate of the reaction much faster.

Useful link

Q12.5.8

In an experiment, a sample of NaClO₃ 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 NaClO₃ is given by:

\[2NaClO_{3}(s)\rightarrow 2NaCl+3O_{2}\]

Typically, since we are given a percentage of NaClO₃ that decomposes we can assume that only 10% remains after the 48 min. We can use this information to find the rate constant (K) is the rate at which NaClO₃ decomposes. Since it is a first order reaction then we can use the first order reaction equations to find k and then use the Arrhenius equation to relate it to the initial temperature and the temperature increase.

The rate derived from the reaction is

\[Rate=k[NaClO_{3}]\]

The relationship between a first order rate constant and concentration is given by

\[ln[A]_{t}=-kt+ln[A]_{0}\] when rearranged is \[ln(\frac{[A]_{t}}{[A]_{0}})= -kt\]

You can now insert then given information, t=48 min x 60 = 2880 [A]_t=10 [A]₀=100 or and solve for k.

\[k=\frac{ln(\frac{10}{100})}{-2880sec}=7.995*10^{-4}\]

However, since we are not given enough information (an initial temperature or A) we cannot solve for the E_a in order to use the Arrhenius equation. So solving this question as it is stated is not possible. Therefore we need to do so based on intuition. We know that when E_a is positive and temperature is increased the molecules tend to speed up and are much more likely to collide, this is the same thing as saying that the rate constant will increase. This is a rule of thumb for temperature and rate constants, that for every 10 k increase the rate constant (k) doubles. We could compare this scenario to that of the experiment and say that there will be a 4-fold increase in the rate constant since we are increasing our temperature by 20 K. BUT this is not an accurate nor specific answer to the question but when given enough information one is able to use this reasoning to approximate the change of the rate constant when temperature is increased.

source: https://chemistry.stackexchange.com/...onstant-k-when

Q21.4.3

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

1. emission of a β particle
2. emission of a β+ particle
3. capture of an electron

Solution:

Radioactive decay

1. From the emission of a β particle in the nucleus there will only be an increase of +1 because of the addition of a proton in the daughter nucleus. This will increase the atomic number by one and change the element the element.

2. From the emission of a β^- particle in the nucleus will be affected opposite of β decay with only a -1 decrease because of the transformation of a proton to a neutron and emission of a high-energy positron. This will decrease the atomic number by one and change the element the element.

3. Electron capture will only result in a -1 decrease of the nucleus because of the transformation of an electron and a proton into a neutron. This will decrease the atomic number by one and change the element the element.

Q20.2.7

Which of these metals produces H₂ in acidic solution?

1. Ag
2. Cd
3. Ca
4. Cu

Solution:

Looking at the activity series you can identify that Ag and Cu are bellow H₂ and will not be active enough to be the reducing agent. Ca and Cd are both above H₂ In the activity series meaning that they are more likely to be oxidized win comparison with H_2. This is important because if H₂ is oxidized then it will be a reactant rather than a product in the combination with the metal.

Example.

Ca will produce H₂ in an acidic solution because Ca is higher up in the activity series and will be oxidized when combined with H⁺ leaving H₂ as a product rather than a reactant.

\[Ca(s) \rightarrow Ca^{2+}(aq) + 2e^{-} (oxidation)\]

\[2H^{+}(aq)+2e^{-} \rightarrow H_{2}(g) (reduction)\]

Q20.5.2

What is the relationship between the measured cell potential and the total charge that passes through a cell? Which of these is dependent on concentration? Which is dependent on the identity of the oxidant or the reductant? Which is dependent on the number of electrons transferred?

Solution:

The total charge transferred from the reductant to the oxidant is nF where n is the number of moles of electrons and F is Faraday's constant, therefore dependent on the number of electrons transferred and independent of the concentrations. The measured cell potential is E_cell dependent on the concentrations and the identity of the oxidant and the reductant as it is found using the equation\[E_{cell}= E^{\circ}_{cell}- (\frac{0.0592V}{n}) log(Q)\]. They are related in that together they are the maximum amount of work that can be produced(w_max) by the cell and this is equal to the change in free energy(△G) often written as \[w_{max}=\Delta G=-nFE_{cell}\] .

Useful Link

Q20.9.11

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

1. AgNO₃
2. RbI

Solution:

Since Ag is more electronegative than Rb it is more likely that Ag will be reduced. We can confirm this by looking at the reduction potentials list that shows us that Ag is much more likely to be reduced and will there for be oxidizing Rb in the anode. We can use the corresponding half-reactions to predict the products at each electrode

Cathode/Reduction:

\[Ag^{+}(aq) +e^{-}\rightarrow Ag(s)\]

Anode/Oxidation:

\[Rb (s)\rightarrow Rb^{+}(aq) + e^{-}\]

Overall reaction: When writing out the overall reaction, make sure that the electrons are balanced so that when you combine the two half reactions, the electrons on the product and the reactant side will cancel out.

\[Ag^{+}(aq)+ Rb (s)\rightarrow Ag(s) + Rb^{+}(aq)\]

Based on the reaction the products will be Ag(s) And Rb⁺(aq).

Q14.6.9

The following reactions are given:

\[{A+B}\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}}\mathrm{C+D}\]

\[{D+E}\xrightarrow{k_2}\mathrm F\]

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

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

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

First, because we have broken the equations down into elementary steps we can write the rate laws for each step.

Step1:

\[A+B\xrightarrow[]{k_{1}} C+D\]

\[rate=k_{1}[A][B]\]

Step 2:

\[C+D \xrightarrow[]{k_{-1}} A+B\]

\[rate=k_{-1}[C][D]\]

Step 3:

\[D+E \xrightarrow[]{k_{2}} F\]

\[rate=k_{2}[D][E]\]

If we add a these steps together we see that we get overall reaction

\[A+B+E \rightarrow C+F\]

we can see that [D] is an intermediate and \[k_{1}=k_{-1}\]_.

Since we are not told which steps are fast or slow we need to use Steady State Approximation.

If the second step is the slower step (k-₁>>k₂) then our rate determining step would be

\[rate=k_{2}[D][E]\]

Since we can only write rate laws in terms of products and reactants we have to rewrite this so that we are not including an intermediate.

Assume: rate of [D] formation = rate of its disappearance

\[k_{1}[A][B]=k_{-1}[C][D]+k_{2}[D][E]\]

\[k_{1}[A][B]=[D](k_{-1}[C]+k_{2}[E])\]

Solving for [D] we find that

\[[D]= \frac{k_{1}[A][B]}{(k_{-1}[C]+k_{2}[E])}\]

now we can use this to substitute the intermediate [D] in the rate law to get an appropriate rate law.

\[rate=\frac{k_{2} k_{1}[A][B][E]}{(k_{-1}[C]+k_{2}[E])}\]

because we had already established k-₁>>k₂ we can assume that

\[k_{-1}[C]+k_{2}[E]\approx k_{-1}[C]\]

this would give us the observed rate law

\[\frac{\Delta [F]}{\Delta t}=\frac{k_{2}k_{1}[A][B][E]}{k_{-1}[C]}\]

to make this clearer we can set

\[k=\frac{(k_{2})(k_{1})}{(k_{-1})}\]

and we can then simplify it down to

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

we can see that all of these rate constants are related by this ratio k₂k₁/k-_1. Since k₂ is our rate determining step k-₁>>k₂ and since k₁=k-₁ then we can see that k₁>>k_2.

We would expect the rate law to be

\[\dfrac{Δ[F]}{Δt} =k′[A][B]?\]

if the rate determining step aka. Slowest step is that corresponding to K₁ (step 1) since the rate law for this step is k₁[A][B] and this is the exact the same as the rate law that we were given. You can do this only when the slow step is the first step.

Slowest step with the biggest barrier is the always the rate determining step.

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