Extra Credit 34

STEPS

A battery is an electrochemical cell or series of cells that produces an electric current. In principle, any galvanic cell could be used as a battery. An ideal battery would never run down, produce an unchanging voltage, and be capable of withstanding environmental extremes of heat and humidity. Real batteries strike a balance between ideal characteristics and practical limitations. For example, the mass of a car battery is about 18 kg or about 1% of the mass of an average car or light-duty truck. This type of battery would supply nearly unlimited energy if used in a smartphone, but would be rejected for this application because of its mass. Furthermore, one has to consider the impact the battery has upon the environment. Certain batteries, such as NiCad batteries, have limitations issued by the European Union as a way to encourage usage of more environmental alternatives. NiCad batteries, have limitations issued by the European Union as a way to encourage usage of more environmental alternatives. Thus, no single battery is “best” and batteries are selected for a particular application, keeping things like the mass of the battery, its cost, reliability, and current capacity in mind.

S17.5.2

Considerations include: cost of the materials used in the battery, toxicity of the various components (what constitutes proper disposal), should it be a primary or secondary battery, energy requirements (the “size” of the battery/how long should it last), will a particular battery leak when the new device is used according to directions, and its mass (the total mass of the new device).

STEPS

The rates at which reactants are consumed and products are formed during chemical reactions vary greatly when affected by these factors:

• Chemical reactions typically occur faster at higher temperatures. Food can spoil quickly when left on the kitchen counter. However, the lower temperature inside of a refrigerator slows that process so that the same food remains fresh for days. We use a burner or a hot plate in the laboratory to increase the speed of reactions that proceed slowly at ordinary temperatures.
• An egg cooks more slowly in boiling water in Denver than New York because, at lower pressures, water has a very low boiling point. In order to properly cook an egg, a certain amount of energy must be applied as activation energy for the reaction to work. Until that point, the egg will remain uncooked while the water boils around it. According to collision theory, with gases, pressure has a direct correlation with the reaction rate, as it increases the likelihood of particles colliding with one another.
• The rates of many reactions depend on the concentrations of the reactants. Rates usually increase when the concentration of one or more of the reactants increases.
• A finely divided solid has more surface area available for reaction than does one large piece of the same substance. Objects with large amounts of surface area have more exposed particles that are ready to react with others in the environment. Thus a liquid will react more rapidly with a finely divided solid than with a large piece of the same solid.

S12.2.1

Higher molarity increases the rate of the reaction. Higher temperature increases the rate of the reaction. Smaller pieces of magnesium metal will react more rapidly than larger pieces because more reactive surface exists.

Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the effect of temperature on reaction rate and the effect of pressure on boiling point.)

STEPS

Radioactive Decay happens because certain elements have unstable nuclei - a situation occurring because they have a whole lot of neutrons and so the nuclear forces are not balanced. Atoms relieve the instability by splitting into pieces or emitting radiation and when this occurs, it can form a different isotope of the same element (i.e. it loses neutrons) or it can become a new element (i.e. it loses protons).

What is given off is:

• Alpha particles are two protons bound to two neutrons -- essentially, it is the nucleus of a helium atom. This particle is not very penetrative, and can be blocked by a sheet of paper.
• As with alpha particles, beta radiation comes from the nucleus of an unstable atom. Betas are electrons, and their mass is much smaller than that of alpha particles -- about 1/8,000th as much. Beta particles come in two forms: positrons and electrons. Positrons are beta particles with positive charge, and electrons are beta particles with a negative charge. According to nuclear physics and the study of antimatter, when a positron is released, it is accompanied by a neutrino. When an electron is released, it is accompanied by an anti-neutrino. While doing nuclear accounting, we often write that the mass of a beta particle is 0, as it is so small. This particle is a lot more penetrative than an alpha particle, and can be blocked by a thin sheet of aluminum, or your own hand.
• Gamma rays are a form of electromagnetic radiation, as are visible light, radio waves, infrared and X-rays. Unlike alpha and beta particles, gamma rays have no mass and no electric charge. When an unstable atom gives off gamma radiation, the element remains the same. Gamma rays can get through most substances, and can only be stopped by several feet of concrete, or a few inches of lead.

STEPS

By observing what happens when samples of various metals are placed in contact with solutions of other metals on an activity series, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. The activity series lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table.

On the Activity Series, we see that Li will be the most willing to react. We can also compare these elements and their Standard Reduction Potentials. This comparison shows that Li has the reduction potential of (E^o=-3.05V) which is lower than the reduction potential of Zn (E^o=–1.66) and S (E^o=0.14V).

Because the reducing agent is something that causes reduction, and oxidation is a process in which an element donates electrons, an element very low on the SRP table or very high on the Activity series cause reduction. Li is such an element, and it is the best reducing agent.

STEPS

Use the following equation to find the weight of Pb (s) created. \[W=\frac {I\times t\times A}{F}\] where n is the valence of the dissolved metal in solution in equivalents per mole, F is valence of the dissolved metal in solution in equivalents per mole (F = 96,485.309 coulombs/equivalent), A is the atomic weight of the metal in grams per mole, I is the current in coulombs per second, t is the time in seconds, and w is the weight of plated metal in grams.

Your standard lead-storage battery undergoes the following reaction: \[Pb(s)+PbO_2(aq)+2HSO_4^-(aq)+2H^+(aq)\longrightarrow\ 2PbSO_4(s)+2H_2O(l)\]

As given in Chapter 19.5, we know that the two corresponding half-reactions of a lead-storage battery are

\[PbO_2(s)+HSO^-_4(aq)+3H^+(aq)+2e^-\longrightarrow PbSO_4(s)+2H_2O(l)\;\;\;\;E^\circ_{cathode}=1.685V\]

\[Pb(s)+HSO^-_4(aq)\longrightarrow PbSO_4(s)+H^+(aq)+2e^-\;\;\;\; E^\circ_{anode}=-0.356V\]

The solution above solves for the question how many grams of Pb are oxidized if I were to take 5 amps out of a battery for a time duration of 2 hours. We know by Le Chatelier's principle that if you take some of the reactants out of the reduction equation, then the reaction will favor the production of more reactants. Thus, when we deplete the cell of its electrons, PbO₂ will form at the cathode.

$$\mathrm{Pb^{+2}_{(aq)}+2e^{-}\longrightarrow Pb}_{(s)}$$

Even by this equation -- which matches the answer key -- we find that when electrons are withdrawn, the reverse reaction is favored. The production of Pb (s) is not favored. The oxidation of Pb (s), or the reverse equation is what we solve for below.

This means that 2 moles of e^- transferred is 1 mol of Pb reduced.

$$\\\mathrm{n_{e}=\frac{It}{F}}$$

$$\frac{2hr}{1}\times\frac{60min}{1hr}\times\frac{60sec}{1min} =7200 seconds$$

$$A=\frac{1c}{1sec}$$

$$\mathrm{n_e=\frac{7200Sec\times10\frac{C}{Sec}}{96,485.306\frac{C}{mole}}}=0.373 e^-$$

CORRECTION: $$\mathrm{n_e=\frac{7200Sec\times2\frac{C}{Sec}}{96,485.306\frac{C}{mole}}}=0.373 e^-$$

$$0.373 e^-\times\frac{1molPb}{2mol\;e^-}=0.187\;mol\;Pb$$

\[0.187\;mol\;Pb\times 207.2 \frac{g}{mol} =38.65 \;g\;Pb(s)\; oxidized\; or\; depleted\]

Editor's notes:

PbO₂ will not be reduced if there is a dearth of electrons because of Le Chatelier's principle. If, say, the battery were being charged, then there would definitely be a reduction of PbO₂. By looking at the provided key, it becomes apparent that the question intended to ask about the mass of the Pb (s) oxidized.

STEPS

Because NO₃ is an intermediate which means NO₃ is a product or one of the products of first reaction and a reactant of second reaction and thus does not show up in the overall reaction. From this we know that first reaction has to produce NO₃ using NO₂

$$\mathrm{2NO_{2\;(g)} NO_{3(g)}+\rightleftharpoons NO_{(g)}\;\;\;\;\;\;\;(1) equilibrium}$$

CORRECTION: $$\mathrm{2NO_{2\;(g)}\rightleftharpoons NO_{3(g)}+NO_{(g)}\;\;\;\;\;\;\;(1)\;equilibrium}$$

This step has rate law of

$$\mathrm{Rate\;Law=k[NO_{2}]^{2}=k'[NO]_{3}[NO]}$$

$$\mathrm{CO_{(g)}+NO_{3(g)}\rightarrow CO_{2(g)}+NO_{2(g)}\;\;\;\;\;\;\;(2)}$$

The rate law for this equation is:

$$\mathrm{Rate\;Law=k[NO_{3}][CO]}$$

Overall reaction:

$$\mathrm{CO_{(g)}+NO_{2(g)}\rightarrow CO_{2(g)}+NO_{(g)}\;\;\;\;\;\;\;(overall\;reaction)}$$

We know that at low temperature mechanism's rate is Δ[CO₂]/Δt = k′[NO₂]². Let's assume that first step is slowest step and see if this proposed mechanism is valid. Since first step is in equilibrium we can solve for [NO₃]

First, you want to put your NO₃ in terms of both NO₂ and k's. We know from the equilibrium the two different rate laws.

Reaction Step 1 (forwards):

\[rate_1=k_1{[NO_2]^2}\]

Reaction Step 1 (backwards):

\[rate_1=k_{-1}{[NO_3]}\]

Reaction Step 2:

\[rate_2=k_2{[NO_3]}{[CO]}\]

Using the equations for Reaction Step 1 (equilibrium) above, we can solve for [NO₃].

$$\mathrm{[NO_{3}]=\frac{\frac{k_1}{k_{-1}}[NO_{2}]^{2}}{k_2[CO]}}$$

$$\mathrm{k'\times\frac{\frac{k_1}{k_{-1}}[NO_{2}]^{2}}{k_2[CO]}\times[CO]=K'[NO_{2}]^{2}}$$

From this, we find that our proposed mechanism is possible, and that the first step is the slowest step.

S14.6.7

The first step is the slowest step.