Extra Credit 35

Q17.5.3

Consider a battery made from one half-cell that consists of a copper electrode in 1 M CuSO₄ solution and another half-cell that consists of a lead electrode in 1 M Pb(NO₃)₂ solution.

a) What are the reactions at the anode, cathode, and the overall reaction?

b) What is the standard cell potential for the battery?

c) Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V. Could this cell be used to make a battery that could replace a dry-cell battery? Why or why not.

d) Suppose sulfuric acid is added to the half-cell with the lead electrode and some PbSO₄(s) forms. Would the cell potential increase, decrease, or remain the same?

S17.5.3

a) First, you have to identify which species is being oxidized and which one is being reduced. (Helpful pneumonic: OIL RIG)

We can obtain the half reactions of each species from the Standard Reduction Potential table, for copper and lead we see that:

\[ \text{ Cu}^\text{ 2+}(\mathit{ aq})+\text{e}^- \rightarrow \text{ Cu}(\mathit{ s})\;\;\; \text{E}^° = 0.34\text{V}\]

\[ \text{ Pb}^\text{ 2+}(\mathit{ aq})+\text{e}^- \rightarrow \text{ Pb}(\mathit{ s})\;\;\; \text{E}^° = -0.13\text{V}\]

Looking at the Standard Reduction Potential Table, we know that the higher the standard reduction potential, the more that element wants to be reduced. In addition, we know that anode is the site where oxidation occurs and cathode is the site where reduction occurs. Based on that knowledge, we can say that since Pb²⁺ has a lower standard reduction potential it will be oxidized. On the other hand, since Cu²⁺ has a higher standard reduction potential it will be reduced. Therefore:

Anode:

\[ \text{ Pb}^\text{ 2+}(\mathit{ aq})+\text{e}^- \rightarrow \text{ Pb}(\mathit{ s})\;\;\; \text{E}^° = -0.13\text{V}\]

Cathode:

\[ \text{ Cu}^\text{ 2+}(\mathit{ aq})+\text{e}^- \rightarrow \text{ Cu}(\mathit{ s})\;\;\; \text{E}^° = 0.34\text{V}\]

Overall Reaction:

\[\text{PbSO}_4(\mathit{s})+\ce{Cu(NO3)2}(\mathit{aq}) \rightarrow \ce{Pb(NO3)2}(\mathit{aq})+\text{CuSO}_4(\mathit{s})\]

b) The equation for standard cell potential for a battery is,

\[ \text{E}^°_\text{cell} = \text{E}^°_\text{cathode} - \text{E}^°_\text{anode} \]

Based on the previous part we found that:

\[ \text{E}^°_\text{cathode} = 0.34\text{ V} \]

\[ \text{E}^°_\text{anode} = -0.13\text{ V} \]

Plugging in the values we get:

\[ \text{E}^°_\text{cell} = 0.34\text{ V} - (-0.13)\text{ V} = 0.47\text{V}\]

c) Yes, this can replace a dry cell battery ONLY if multiple cells are put together (at least 3) because the E°_cell for this reaction is only 0.47V. You would need three of these cells (0.47*3 would fall in the required range for the corresponding system) so it can power a 1.0V-1.5V battery. This cell alone cannot power a device that requires a battery between 1.0-1.5V because the E°_cell is only 0.47V.

d) The only way to determine if the cell potential for a system increases, decreases, or stays the same is to utilize the Nernst Equation and compare the cell potentials under standard and non standard conditions. The Nernst Equation is as follows:

\[E_{cell} = E^{\circ}_{cell} - (\frac{RT}{nF})lnQ\]

The question tells us that sulfuric acid is added to the half-cell with the lead electrode. By adding more sulfuric acid, the only factor that gets affected in the equation is Q (the reaction quotient). The reaction quotient is defined as:

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

At standard conditions the ratio of the reaction quotient, Q, equals 1. If sulfuric acid is added to the lead electrode, we are told that more PbSO₄(s) forms. If more PbSO₄(s) forms then more lead is used which means that Pb(NO₃)₂(aq) is decreased. This scenario will lower the reaction quotient by 1. We know that the ln of any number below 1 is negative, if that number becomes negative the Nernst equation would be affected.

\[E_{cell} = E^{\circ}_{cell} + (\frac{RT}{nF})lnQ\]

(instead of a minus sign, it is now a plus)

Because that sign is now a plus, we can conclude that the cell potential would increase.

Q12.2.2

Go to the PhET Reactions & Rates interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen (O) and carbon monoxide (CO) results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on “Reload Launcher” and change to “Angled shot” to see the difference.

a. What happens when the angle of the collision is changed?

b. Explain how this is relevant to rate of reaction.

S.12.2.2

Proceed to the site that it directs you towards. After going through simulation, you are able to answer the questions.

a. According to the collision theory, there are many factors that cause a reaction to happen. Factors that often affect how the molecules or atoms collide are:

• the orientation of the molecules or atoms
• if there is sufficient energy for the reaction to occur

So, if the angle of the plunger is changed, the atom that is shot (a lone Oxygen atom in this case) will hit the other molecule (CO in this case) at a different spot and at a different angle; therefore, changing the orientation and the number of proper collisions will most likely not cause for a reaction to happen. Depending on the angle selected, the atom may take a long time to collide with the molecule and, when a collision does occur, it may not result in the breaking of the bond and the forming of the other.

b. This is relevant to the rate of reaction because particles of reactant must come into contact with each other before they can react.

Q12.5.7

The rate of a certain reaction doubles for every 10 °C increase in temperature.

a. How much faster does the reaction proceed at 45 °C than at 25 °C?

b. How much faster does the reaction proceed at 95 °C than at 25 °C?

S.12.5.7

A increase in temperature means an increase in the kinetic energy of the molecule and because the kinetic energy of the molecule increases, there are more collisions. These collisions are above the activation energy which cause a reaction to occur. In normal circumstances, the molecules would be moving slower and less reactions would occur; thus, an increase in temperature increases the rate of the reaction. Looking at the question we know that the rate doubles for every 10 °C rise in temperature. Taking that into consideration...

a. A simple mathematical approach is enough to answer this question.

\[45 °C – 25 °C = 20 °C\]

(Notice that 20 is twice the amount of 10)

We are told that the rate DOUBLES for every 10 °C rise in temperature this yields 2² (exponent represents the difference of temperatures) . Therefore, the reaction goes 4 times faster at 45 °C than at 25 °C.

b. We can use the same approach for this question too.

\[95 °C – 25 °C = 70 °C\]

(70 is seven times the amount of 10)

Remember that rate doubles for every 10 °C rise in temperature so that means it will be 2⁷. Therefore, the reaction goes 128 times faster at 95 °C than at 25 °C.

Q21.4.2

What changes occur to the atomic number and mass of a nucleus during each of the following decay scenarios?

a. an α particle is emitted

b. a β particle is emitted

c.γ radiation is emitted

d. a positron is emitted

e. an electron is captured

S21.4.2

a. When an α particle is emitted from a nucleus the nucleus loses two protons and four neutrons.(The original was incorrect) This means the atomic mass decreases by 4 and the atomic number decreases by 2.

b. When a β particle is emitted from the nucleus the nucleus has one more proton. (The original was incorrect) This means the atomic mass number remains unchanged and the atomic number increase by 1.

c. The emission of γ rays does not alter the number of protons or neutrons in the nucleus but instead has the effect of moving the nucleus from a higher to a lower energy state. Therefore, the atomic mass number and atomic number remain unchanged.

d. A positron is the antiparticle of the electron, but the opposite charge. When a positron is emitted, a proton is converted into a neutron so the atomic number decreases by one and the mass number remains the same.

e. When a nucleus assimilates an electron from an inner orbital of its electron cloud it is called electron capture. With that in mind, this affect would decrease the atomic number by 1 because it has one less proton. The atomic mass number will remain the same because one less proton will have no effect on it.

Q20.2.6

Of these elements, which would you expect to be easiest to reduce: Se, Sr, or Ni? Explain your reasoning.

S.20.2.6

One of the rules to remember is that elements on the left hand side (specifically alkaline metals and alkali metals) of the periodic table are more likely to give up their electrons in order for it to "shrink down" and have a full octet. On the opposite end (right side, non-metals and metalloids), these elements are more likely to acquire electrons in order to fill its octet and reach the nearest noble gas. In general, elements towards the right-hand end of the transition metal block have high reduction potentials. Reduction potential meaning that it has a high tendency to acquire electrons and thereby be reduced.

So comparing these three elements: Se, Sr, and Ni...

• Se is on the right hand side of the periodic table and it is easier for that element to GAIN electrons in order to fill its octet rather than give up electrons.
• Sr falls on the left hand side of the periodic table so it is easier for that element to GIVE UP electrons in order to fill its octet rather than accept.
• Ni is towards the right-hand end of the transition metal block so it has a high reduction potential. Another way to determine if this can be easily reduced is to look at the Standard Reduction Potentials tower and find Ni and observe the given volts. The higher the number, the more likely it is to reduce.

Based on the observations made above, we can easily say that Se would be the easiest to reduce because it is more likely to gain electrons in order to fill its orbital and have a full valence shell.

Q20.5.1

State whether you agree or disagree with this reasoning and explain your answer: Standard electrode potentials arise from the number of electrons transferred. The greater the number of electrons transferred, the greater the measured potential difference. If 1 mol of a substance produces 0.76 V when 2 mol of electrons are transferred—as in Zn(s) → Zn²⁺(aq) + 2e⁻—then 0.5 mol of the substance will produce 0.76/2 V because only 1 mol of electrons is transferred.

S.20.5.1

The standard reduction potential is independent of size and amount of matter in a system. THERE IS NO CORRELATION BETWEEN STANDARD ELECTRODE POTENTIALS AND CONCENTRATION. Meaning it remains the same for a metal ion couple. In addition, it does not matter how many electrons were transferred. The only factor that is important to look out for is that the standard reduction potential correlates to the correct metal ion. Based on that intuition, I disagree with this reasoning.

Q20.9.10

Electrolysis of Cr³⁺(aq) produces Cr²⁺(aq). If you had 500 mL of a 0.15 M solution of Cr³⁺(aq), how long would it take to reduce the Cr³⁺ to Cr²⁺ using a 0.158 A current?

S.20.9.10

Use the Standard Reduction Potential table in order to obtain the half reaction for this question, in this case it is:

\[ \text{ Cr}^\text{ 3+}(\mathit{ aq})+\text{e}^- \rightarrow \text{ Cr}^\text{2+}(\mathit{ aq})\]

This helps to see the ratio of moles Cr²⁺ to moles of e^-

Using stoichiometry, we can find the grams of Cr:

\[3.90g=\frac{0.15 mol}{1L}×0.500L×\frac{51.9961g}{1 mol}\]

using the given value of 0.158A, the calculated mass of Cr, and Faraday's constant (96485 C/mol) we can figure out how long it would take to reduce Cr³⁺ to Cr²⁺

\[3.90g\times 96485 C*mol^{-1}\times \frac{1}{0.158A}\times \frac{1 mol Cr^{2+}}{1 mol electrons}\times \frac{1 mol}{51.9961 g*mol^{-1}} = 45803 seconds\]

converting time to seconds yields

\[45803 seconds\times \frac{1min}{60 seconds} = 763.4 min\]

It would take 763.4 minutes to reduce Cr³⁺ to Cr²⁺.

Q14.6.8

Nitramide (O₂NNH₂) decomposes in aqueous solution to N₂O and H₂O. What is the experimental rate law (Δ[N₂O]/Δt) for the decomposition of nitramide if the mechanism for the decomposition is as follows?

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

S14.6.8

The rate determining step is the slowest step and has the highest energy barrier.

\[\text{rate}=\text{rate}_2=\text{k}_2[\mathrm{O_2NNH^-}]\]

In this case, the slowest step is the secondary elementary step. We also know that if the mechanism has a rate determining step as the second elementary step, then the corresponding rate law for that step typically involves an intermediate which is NOT part of the rate law for the mechanism.

In this case, [O₂NNH^-] is the intermediate because this compound appears in the products and reactants in two consecutive steps. Since [O₂NNH^-] is an intermediate, we have to go about this a little differently. Notice that the first step is going forward and backwards so we would have to set them equal to each other:

\[\text{k}_1[\mathrm{O_2NNH_2}]=\text{k}_{-1}[\mathrm{O_2NNH^-}][\mathrm{H^+}]\]

We want to manipulate the equation in a way that [O₂NNH^-] is by itself on one side, so we get:

\[[\mathrm{O_2NNH^-}]=\dfrac{k_1[\mathrm{O_2NNH_2}]}{k_{-1}[\mathrm{H^+}]}\]

Now that we have the intermediate on one side, we substitute the right side for the intermediate (O₂NNH^-) into the slow step rate equation that we found earlier:

\[\textrm{rate}=k_2\dfrac{k_1[\mathrm{O_2NNH_2}]}{k_{-1}[\mathrm{H^+}]}\]

To clean it up a little bit we put the rate in terms of k, which is:

\[\textrm{k}=\dfrac{k_1k_2}{k_{-1}}\]

Thus, the overall rate of the reaction for the corresponding system is

\[\textrm{rate}=k\dfrac{[\mathrm{O_2NNH_2}]}{[\mathrm{H^+}]}\]