Extra Credit 40

SA. At low temperatures the rate of ion exchanged, which is what provided the electrical current, is greatly reduced. This results in more power being dissipated in the cell itself, rather than the desired outlet. This results in less stored chemical energy being available for use.

Batteries work through the movement of electrons from one electrode to another. Decreasing the temperature will inhibit this flow of electrons, thus preventing the battery from working properly.

12.3.3 B. Tripling the concentration of a reactant increases the rate of a reaction nine times. With this knowledge, answer the following questions:

1. What is the order of the reaction with respect to that reactant?
2. Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four times. What is the order of the reaction with respect to that reactant?

BS1. The first reaction is second order. 3^X=9, solve for X.

BS2. The second reaction is first order. 4^X=4, solve for X.

We can solve this by setting up a simple exponential equation.

$$\mathrm{3^x=9}$$

Tripling the concentration of the reactant to some power increases the rate nine times. Solve for x to find the order of the reaction.

$$\mathrm{x=2}$$

With respect to the reactant, the order of the reaction is 2.

The second problem can be solved by applying the same method as before.

$$\mathrm{4^x=4}$$

$$\mathrm{x=1}$$

With respect to that other reactant, the order of the reaction is 1.

12.5.12 C. In terms of collision theory, to which of the following is the rate of a chemical reaction proportional?

1. the change in free energy per second
2. the change in temperature per second
3. the number of collisions per second
4. the number of product molecules

SC. The rate of the chemical reaction is directly proportional to the number of collisions per second. The entire premise of the Collision Theory is defined as the rate of a reaction is based on the frequency of collisions. More collisions equates to a faster reaction.

Yes. Increasing the number of collisions will increase the rate of the reaction, while decreasing it will decrease the rate of the reaction.

Increasing the temperature will also increase the rate of reaction. Higher temperatures increase the movement of the molecules, therefore allowing them to have more energy to collide with other molecules.

21.4.7 D. Which of the following nuclei is most likely to decay by positron emission? Explain your choice.

1. chromium-53
2. manganese-51
3. iron-59

SD. Manganese-51 is most likely to decay by positron emission. The n:p ratio for Cr-53 is 29/24=1.21; for Mn-51, it is 26/25=1.04; for Fe-59, it is 33/26=1.27. Positron decay occurs when the n:p ratio is low. Mn-51 has the lowest n:p ratio and therefore is most likely to decay by positron emission. Besides, Cr 53/24 is a stable isotope, and Fe59/26 decays by beta emission.

Fig 1. Belt of Stability. Uses the neutron:proton ratio to determine the stability of a nucleus. Those below the belt will most likely undergo positron emission or electron capture, while those above the belt will undergo beta decay. The black line is where the stable nuclei lie.

\(\mathrm{^{53}_{24}Cr}\) has 29 neutrons and 24 protons, which as the above explanation states, has a n:p ratio of 1.21. On the belt of stability, this lies on the black line. It is a stable isotope.

\(\mathrm{^{51}_{25}Mn}\) 26 neutrons and 25 protons, with a n:p ratio of 1.04. On the belt, it lies just below the black line. This indicates that it can decay by positron emission or electron capture.

\(\mathrm{^{59}_{26}Fe}\) 33 neutrons and 26 protons, n:p ratio of 1.27. On the belt it is above the black line indicating that it will undergo \(\beta{-}\) decay.

Based on the belt of stability, manganese-51 will decay by positron emission.

20.2.11 E. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction.

1. Platinum wire is dipped in hydrochloric acid.
2. Manganese metal is added to a solution of iron(II) chloride.
3. Tin is heated with steam.
4. Hydrogen gas is bubbled through a solution of lead(II) nitrate.

SE1. Platinum is lower on the Activity series; therefore, it does not displace the Hydrogen. This results in no reaction.

Platinum is mostly unreactive and is considered inert. This is why it is used as an inert electrode.

SE2. Mangenese is located above Iron on the Activity series, resulting in the displacement of the Iron (II).

Mn_(s)+FeCl_2(aq)⇒MnCl_2(aq) +Fe_(s)

$$\mathrm{Mn}(s)+\mathrm{FeCl_2}(aq)\longrightarrow\mathrm{MnCl_2}(aq)+\mathrm{Fe}(s)$$

SE3. Tin is located higher on the Activity series, resulting in the displacement of the Hydrogen from the steam.

2Sn_(s)+2H₂O_(l)⇒2SnOH_(aq)+H_2(g)

$$\mathrm{Sn}(s)+\mathrm{2H_2O}(g)\longrightarrow\mathrm{Sn(OH)_2}(aq)+\mathrm{H_2}(g)$$

SE4. Hydrogen is located lower on the Activity series, resulting in no displacement of the Lead.

20.5.6 F. Although the sum of two half-reactions gives another half-reaction, the sum of the potentials of the two half-reactions cannot be used to obtain the potential of the net half-reaction. Why? When does the sum of two half-reactions correspond to the overall reaction? Why?

lol what. same.

$$\mathrm{Cr^{3+}}(aq)+\mathrm{e^-}\longrightarrow\mathrm{Cr^{2+}}(aq)\qquad\qquad\mathrm{E^o(V)=-0.42}$$

$$\mathrm{Cr^{2+}}(aq)+\mathrm{2e^-}\longrightarrow\mathrm{Cr}(s)\qquad\qquad\mathrm{E^o(V)=-0.90}$$

These half-reactions for chromium shows the how Cr³⁺(aq) losing an electron to become Cr²⁺(aq) and how Cr²⁺(aq) loses two electrons to become elemental chromium. However we do not see the direct change from Cr³⁺(aq) to Cr(s). We can find out by doing a simple cancelation between the two half reactions.

Since the Cr²⁺(aq) in each half-reaction is in the opposite side between the two, we can cancel it and add the half-reactions together to obtain the net half-reaction.

(In the first half-reaction, Cr²⁺(aq) is in the products side, while in the second half-reaction it is in the reactants side.)

We would get:

$$\mathrm{Cr^{3+}}(aq)+\mathrm{3e^-}\longrightarrow\mathrm{Cr}(s)\qquad\qquad\mathrm{E^o(V)=-0.74}$$

Notice how the cell potential for this half-reaction is not the sum of the cell potential for the first two half-reactions.

If we were to add them, we would get \(\mathrm{-0.42 V+(-0.90 V)= -1.32 V}\). This is extremely different from the actual cell potential of the net half-reaction. This discrepancy between the cell potentials is due to how cell potential is a path function. This also explains why the cell potential is not going to be affected even by scaling the reaction (i.e. scaling a reaction by 3 is not going to triple the value of the cell potential.)

In order to find the actual cell value of the net reaction, we would have to indirectly solve it by finding \(\Delta{G^o}\) because Gibbs energy is a state function, unlike cell potential.

$$\mathrm{Cr^{3+}}(aq)+\mathrm{e^-}\longrightarrow\mathrm{Cr^{2+}}(aq)\qquad\qquad\mathrm{E^o(V)=-0.42}$$

$$\Delta{G^o}=-nFE^o_{cell}$$

$$\Delta{G^o}=-(1)(96.485\,KJ \centerdot V^{-1}\centerdot mol^{-1})(-0.42\,V)$$

$$\Delta{G^o}=40.5\,KJ \centerdot mol^{-1} $$

$$\mathrm{Cr^{2+}}(aq)+\mathrm{2e^-}\longrightarrow\mathrm{Cr}(s)\qquad\qquad\mathrm{E^o(V)=-0.90}$$

$$\Delta{G^o}=-(1)(96.485\,KJ \centerdot V^{-1}\centerdot mol^{-1})(-0.90\,V)$$

$$\Delta{G^o}=173.7\,KJ \centerdot mol^{-1} $$

Now we add them together

$$173.7\,KJ \centerdot mol^{-1} + 40.5\,KJ \centerdot mol^{-1} = 214.2\,KJ \centerdot mol^{-1}$$

The last step is to plug this value back into the equation.

$$E^o_{cell}=\dfrac{\Delta{G^o}}{-nF}$$

$$E^o_{cell}=\dfrac{214.2\,KJ \centerdot mol^{-1}}{-3\times96.485\,KJ \centerdot V^{-1}\centerdot mol^{-1}}$$

$$E^o_{cell}=-0.74\,V$$

Finally, we obtained the cell potential value that corresponds to the reaction of Cr³⁺(aq) to Cr(s).

24.6.2 G. In CFT, what causes degenerate sets of d orbitals to split into different energy levels? What is this splitting called? On what does the magnitude of the splitting depend?

Splitting of the d orbitals into different energy levels is caused by the strength of the field. A strong field results in a low spin, low spin meaning the t_2g row is filled first with 3 pairs of electrons, 6 total. This splitting is called Crystal Field Splitting. The magnitude of the spinning depends on the strength of the field.

Fig 2. Energy diagram for an octahedral complex.

Degenerate d orbitals can split into different energy levels which is caused by the ligands around it. Orbitals can be oriented in different ways. There are certain circumstances in which the ligands are oriented in a way that directly faces the orbitals. This increases the repulsion between the orbital and the ligands, which in turn increases the energy of the orbital (this occurs for the d_x²-y² and d_z² orbitals: e_g). On the other hand, when the ligands are not facing the orbitals directly, the energy decreases (d_xy, d_xz, d_yz: t_2g). When the ligands are strong field ligands, they induce a low-spin configuration. This is when the electrons decide fill up the t_2g orbitals first before moving on the e_g orbitals, due to the difference between the spin-pairing energy(P) and \(\Delta{_o}\). If P< \(\Delta{_o}\), then the electrons would be low-spin. Conversely, if P> \(\Delta{_o}\), which is what usually happens with the presence of weak field ligands, then the electrons would rather go up to the higher energy levels first than pair up because doing so costs less energy, resulting in a high-spin configuration.