Extra Credit 40

Q17.5.8

Using the information thus far in this chapter, explain why battery-powered electronics perform poorly in low temperatures.

Q17.5.8Solution:

According to the Nernst equation:

\(E_{cell}=E^{\circ}_{cell}-\left ( \frac{RT}{nF} \right )ln\left ( Q \right )\) or \(E_{cell}=E^{\circ}_{cell}-\left ( \frac{0.0592}{n} \right )log\left ( Q \right )\)

When the temperature (represented as 'T' in the equation) decreases, the cell potential (\(E_{cell}\)) of the battery-powered electronics is also decreased. In other words, temperature is directly proportional to cell potential. If one increase then the other will also increase, and vise-versa for a decreasing trend.

A battery will consist of more than one electrochemical cells. This will convert chemical energy to electrochemical energy which will ultimately create currents. Higher electrical energy is a positive terminal, and vice versa where the negative terminal is a lower electrical energy. Batteries come in two forms of primary and secondary.

Q12.3.3

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

a. What is the order of the reaction with respect to that reactant?

b. 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?

Solution:

a. The order of the reaction is a second-order with respect to the reactant. The order of the reaction is the index raised to its concentration term.

\(Rate\propto [B]^{2}\)

\(\overset{(ex)}{\rightarrow}\) 3^x=9 \(\overset{(Solve for x.)}{\rightarrow}\) 3²=9

b. The order of the reaction is a first-order with respect to the reactant. Rate of a reaction is the speed in which the reaction moves in a specific direction.

\(Rate\propto [B]^{1}\)

\(\overset{(ex)}{\rightarrow}\) 4^x=4 \(\overset{(Solve for x.)}{\rightarrow}\) 4¹=4

Q12.5.12

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

a. the change in free energy per second

b. the change in temperature per second

c. the number of collisions per second

d. the number of product molecules

Solution: In terms of collision theory, the number of collisions per second is proportional to the rate of a chemical reaction. According to the collision theory, in order for a reaction to occur there much be a collision of atoms or molecules. Thus the greater the number of collisions happening per second, the faster the reaction will occur. Rate of reaction is dependent of the concentration in relation to the reactant which will ultimately increase the concentration increasing the number of collisions between the molecules. This will increase the collisions possibility.

\(A+B\rightarrow C+D\)

rate=(fraction of molecules with required orientation)x(fraction of collisions with required energy)x(collision frequency)

Q21.4.7

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

a. chromium-53

b. manganese-51 Atomic number is 25; thus, 51-25=26. Here, there is a low proton ration determining that it is by positron emission.

c. iron-59

Solution:

Proton to Neutron ratio states where the nuclei is located on the Belt of Stability.

(Find the point were the protons and neutrons meet to find the type of decay that is most likely to occur.)

Q20.2.11

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.

a. Platinum wire is dipped in hydrochloric acid.

b. Manganese metal is added to a solution of iron(II) chloride.

c. Tin is heated with steam.

d. Hydrogen gas is bubbled through a solution of lead(II) nitrate.

Solution:

For each scenario write out the possible equation for the overall reaction and then determine whether the first element is higher up or lower down the activity series table.

a.) Balanced: \(Pt_{(s)}+2HCl_{(aq)}{\rightarrow} PtCl_{2(aq)}+H_{2(g)}\)

Compare platinum with hydrogen on the activity series table. As you can see platinum is lower down on the table than hydrogen, thus a reaction WILL NOT occur.

b.) Balanced: \(Mn_{(s)}+FeCl_{2(aq)}{\rightarrow} MnCl_{2(aq)}+Fe_{(s)}\)

Compare manganese with iron on the activity series table. As you can see manganese is higher up on the table than iron, thus a reaction WILL occur.

c.) Balanced: \(Sn_{(s)}+2H_{2}O_{(g)}{\rightarrow} SnO_{2(s)}+2H_{2(g)}\)

Compare tin with hydrogen on the activity series table. As you can see, tin is higher up on the table than hydrogen, thus a reaction WILL occur.

d.) Balanced: \(H_{2(g)}+Pb(NO_{3})_{2(aq)}\) \({\rightarrow}\) \(Pb_{(s)}+2H^{+}_{(aq)}+2(NO_{3})^{-}_{(aq)}\)

Compare hydrogen with lead on the activity series table. As you can see, hydrogen is lower down on the table than lead, this a reaction WILL NOT occur.

Q20.5.6

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?

Solution:

\(\Delta G^{\circ}\) is a state function (a property whose value doesn't depend on how the system got to a specific value), but \(E^{\circ}\) is NOT a state function and thus cannot be added together from two half-reactions to find the potential of the net half-reaction. This can be demonstrated in the equation: \(\Delta G^{\circ}=-nFE^{\circ}\). To obtain the \(E^{\circ}\) of the overall half-reaction, you will need to find what \(\Delta G^{\circ}\) is since \(\Delta G^{\circ}\) is proportional to the potential (\(E^{\circ}\)). Write out the reactions to find \(\Delta G^{\circ}\) and the number of moles in the system. Once those values are found you will be able to rearrange the equation \(\Delta G^{\circ}=-nFE^{\circ}\) to solve for \(E^{\circ}\).

The sum of two half-reactions can correspond to the overall reaction when \(E^{\circ}_{(overall)}=E^{\circ}_{(oxidation)}+E^{\circ}_{(reduction)}\).

When a net reaction occurs in an electrochemical cell, oxidation will occur at one electrode. Then, the reduction will occur in the other electrode called a cathode. Thus, the cell is two half cells joined together by an external circuit.

Q24.6.2

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?

Solution:

The strength of the field determines how the d-orbitals split into their energy levels (i.e. e_g and t_2g). There are two kinds of strengths possible: strong and weak. The way to determine the strength of the field is to see how much spin is in the system. A low spin indicates a strong field while a high spin indicates a weak field. The idea of this split comes from the Crystal Field Theory. When ligand approaches the metal ion, some will experience more opposition from the d-orbital electrons than others. Some ligands will approach from different directions; thus, not all d orbitals will approach directly.

To determine whether the spin is low or high we look at the spectrochemical series: I^- < Br^- < Cl^- < SCN^- < F^- < OH^- < ox^2-< ONO^- < H₂O < SCN^- < EDTA^4- < NH₃ < en < NO₂^- < CN^-. The ion attached to the metal, the charge of the metal, and the orientation of the compound make up the strength of the field.

Low spin indicates that more energy is needed in the system to fill up the orbitals while high spin indicates that less energy is needed in the system to fill up the orbitals.