Extra Credit 33

Q17.5.1

What are the desirable qualities of an electric battery?

S17.5.1

Batteries, also known as galvanic cells, are a series of cells that produce an electrical current. The most desirable qualities of electric batteries are the following: having a constant voltage with high current output, safe to use, light in weight, and affordable.

Q12.1.6

Consider the following reaction in aqueous solution:

\(5Br^-(aq) + BrO^-_3(aq)+6H^+ \to 3 Br_2(aq) + 3H_2O(l)\)

If the rate of disappearance of \(Br^-\)(aq) at a particular moment during the reaction is \(3.5x10^{-4} Ms^{-1}\), what is the rate of appearance of \(Br_2\)(aq) at that moment?

S12.1.6

Step 1. Define the rate of the reaction.

Recall:

For the general reaction: aA + bB → cC+ dD

\(rate =- \frac{\Delta[A]}{a\Delta{t}}=- \frac{\Delta[B]}{b\Delta{t}}= \frac{\Delta[C]}{c\Delta{t}}=\frac{\Delta[D]}{d\Delta{t}}\)

So, for the reaction: \(5Br^−(aq)+BrO^−_3(aq)+6H^+→3Br_2(aq)+3H_2O(l)\)

The rate would be: \(rate =- \frac{\Delta[Br^-]}{5\Delta{t}}=- \frac{\Delta[BrO^-_3]}{\Delta{t}}= -\frac{\Delta[H^+]}{6\Delta{t}}=\frac{\Delta[Br_2]}{3\Delta{t}}=\frac{H_2O}{3\Delta{t}}\)

Step 2. Since we are given the rate for the disappearance of \(Br^-\)(aq) is \(3.5x10^-4 Ms^{-1}\), and we want to find the rate of appearance of \(Br_2\)(aq). Therefore we set the two rates equal to each other.

\(rate =- \frac{\Delta[Br^-]}{5\Delta{t}}= \frac{\Delta[Br_2]}{3\Delta{t}}\)

And,\(-\frac{\Delta[Br^-]}{\Delta{t}}= -3.5x10^{-4} Ms^{-1}\)

So, \(3.5x10^{-4} Ms^{-1}\) = \(\frac{5}{3}\frac{\Delta[Br_2]}{\Delta{t}}\)

Step 3. Now solve the equation.

\(\frac{(3.5x10^{-4})(3)}{5} = \frac{\Delta[Br_2]}{\Delta{t}}\)

\(\frac{\Delta[Br_2]}{\Delta{t}} = 2.1 x 10^{-4} Ms^{-1}\)

Q12.5.5

Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.

S12.5.5

Given the graph:

Based on the Arrhenius equation:

\(ln(\frac{k_1}{k_2}) = -\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})\)

At temperature \(T_1\): \(ln(k_1) = -\frac{E_a}{RT_1} + ln(A)\)

At temperature \(T_2\): \(ln(k_2) = -\frac{E_a}{RT_2} + ln(A)\)

The slope of the line drawn on a ln(k) vs. \(\frac{1}{T}\) is \(-\frac{E_a}{R}\) and so we can determine the activation energy at any desired temperatures in that way.

Q21.3.8

For the reaction \(^{14}_6C \to ^{14}_7N^+\), if 100.0 g of carbon reacts, what volume of nitrogen gas \((N_2)\) is produced at 273 K and 1 atm?

S21.3.8

Recall: PV=nRT

Where:

P=pressure, V=volume, n=number of moles, R= R constant, T= temperature

Given: P = 1 atm

T = 273 K

R = 0.008206 \(\frac{L⋅atm}{mol⋅K}\)

Steps:

Step 1. Convert mass of carbon into moles of \(N_2\) gas using stoichiometry.

\((100g\;C)(\frac{1\;mol\;C}{12g\;C})(\frac{1\;mol\;N_2}{1\;mol\;C}) = 8.33\;moles\;of\;N_2\)

Step 2. Plug all knowns into equation and solve for V.

\((1\;atm)(V) = (8.33\;moles\;of\;N_2)(0.008206\frac{L\cdot atm}{mol \cdot K})(273\;K)\)

V = 18.66 L of \(N_2\) gas

Q20.2.4

Single-displacement reactions are a subset of redox reactions. In this subset, what is oxidized and what is reduced? Give an example of a redox reaction that is not a single-displacement reaction.

S20.2.4

OIL - RIG.

Oxidation is loss of electrons. Reduction is gain of electron.

The element or compound that loses an electron is oxidized, while the element or compound that gains and electron is reduced. We can determine if an element or compound has lost or gained electrons by looking at each element's or compound's oxidation states in the reactant side and then on the product side.

An example of a single-displacement redox reaction is:

\(2Mg(s)+O_2(g) \to 2MgO(s)\)

On the reactant side, Mg has an oxidation state of 0 and \(O_2\) has an oxidation state of 0, too. On the product side, Mg has an oxidation state of +2, while O has an oxidation state of -2. Since Mg lost 2 electrons, it was oxidized. O gained 2 electrons, so it was reduced.

An example of a redox reaction that isn't a single displacement reaction is:

\(2C_4H_{10}(g) + 13O_2(g) \to 8CO_2(g) + 10H_2O(l)\)

This is a combustion reaction. On the reactant side, \(O_2\) has an oxidation state of 0. On the product side, O has an oxidation state of -2. Since O gained electrons, it was reduced.

Q20.4.23

If you place Zn-coated (galvanized) tacks in a glass and add an aqueous solution of iodine, the brown color of the iodine solution fades to a pale yellow. What has happened? Write the two half-reactions and the overall balanced chemical equation for this reaction. What is E°cell?

S20.4.23

Given:

Step 1. Find half reaction of \(Zn^{2+}\;and\;I^-)\ on the standard reduction potential table.

\(Zn^{2+}(aq) + 2e^- \to Zn(s)\quad E^o=-0.76\)

\(I_2(s) + 2e^- \to 2I^-(aq)\quad E^o = 0.54\)

Step 2. Flip one of the equations to cancel out the \(2e^-\). BUT, \(E^o_{cell}\) stays the same, even if we flip the equations.

\(Zn(s) \to Zn^{2+}(aq) + 2e^-\quad E^o = -0.76\)

\(I_2(s) + 2e^- \to 2I^- (aq) \quad E^o=0.54\)

Step 3. Cancel out the \(2e^-\) and add the equations together to get the overall equation.

\(Zn(s) \to Zn^{2+}(aq) + \cancel{2e^-} \quad E^o=-0.76\)

\(I_2(s) + \cancel{2e^-} \to 2I^-(aq) \quad E^o = 0.54\)

\(\implies I_2(s) + Zn(s) \to 2I^-(aq) + Zn^{2+}(aq)\)

Step 4. Use the equation: \(E^o_{cell} = E^o_{cathode} - E^o_{anode}\) to determine \(E^o_{cell}\).

Recall: Cathodes are where reduction happens. Anodes are where oxidation happens. Based on the standard REDUCTION potential, the bigger \(E^o\) is, the more likely it will be reduced. Therefore, based on these equations, Iodine will more likely be reduced than Zinc. So:

\(E^o_{cell} = E^o_{cathode} - E^o_{anode}\)

\(E^o_{cell} = 0.54 - (-0.76) = 1.30\;V\)

\(E^o_{cell} = 1.30\;V\)

Iodine is being reduced while Zinc is being oxidized.

Q20.9.8

What mass of copper metal is deposited if a 5.12 A current is passed through \(Cu(NO_3)_2\) solution for 1.5 hours?

S20.9.8

Use the equation: \(n_e = \frac{It}{F}\)

Where: I = current (amps)

t = time (seconds)

F = Faraday's constant (\(9.65 x 10^4\) C/mole)

\(n_e\) = number of moles of electrons

If the current (I) is 5.12 amps, and time (t) = 1.5 hours...

Step 1. Convert the time in hours to time in seconds using stoichiometry.

\((1.5\;hours)(\frac{60\;min}{1\;hour})(\frac{60\;sec}{1\;min}) = 5400\;sec\)

Step 2. Plug all known variables into the equation and solve for the number of moles of electrons being transferred ().

\(n_e = \frac{(5.12A)(5400\;sec)}{(9.65 x 10^4\;C/mole)}\)

\(n_e\) = 0.2865 moles of electrons

Step 3. Since, in \(Cu(NO_3)_2\), the ration between Cu and \(Cu(NO_3)_2\) is 1, every electron being transferred from \(Cu(NO_3)_2\) is equal to 1 electron from Cu being transferred. Using this conversion factor and the molecular weight of Cu, convert moles of electrons being transferred from \(Cu(NO_3)_2\) to mass of Cu in grams.

(0.2865 moles of electrons \(Cu(NO_3)_2)(\frac{1\;mole\;Cu}{1\;mole\;Cu(NO_3)_2})(\frac{63.55g\;Cu}{1\;mole\;Cu})\) = 18.2 grams of Cu

Q14.6.6

Cyclopropane, a mild anesthetic, rearranges to propylene via a collision that produces and destroys an energized species. The important steps in this rearrangement are as follows:

where M is any molecule, including cyclopropane. Only those cyclopropane molecules with sufficient energy (denoted with an asterisk) can rearrange to propylene. Which step determines the rate constant of the overall reaction?

S14.6.6

Recall: "Rate determining step" = the slowest elementary step, and thus determines the rate constant of the overall reaction.

Since cyclopropane molecules require sufficient energy to be rearranged to propylene, it is suggested that the second step is the slowest step since it requires a lot of energy to react. k₁/k_-1 is an equilibrium equation and an reaction going to equilibrium will most likely to take time. Therefore, is most likely the rate constant for the overall reaction.