Just How Small Is the Proton?
Galaxy-Scale Bubbles Extending From the Milky Way
CERN Physicists Trap Antimatter Atoms
Sunday, November 21, 2010
Tuesday, November 16, 2010
20.4-20.7: Radioactive Rate of Decay-Nuclear Fission and Nuclear Fusion
Radioactive Rate of Decay:
half-life: the time it takes from one-half of the nuclei in a sample to decay. Independent of the amount of sample, denoted t½.
First Order Half Life:
t½ = 0.693 / k
Example:
k = 1.45 x 10-12 s-1 for Ta-183, find t½ in years.
t½ = 0.693 / 1.45 x 10-12 s-1
t½ = 4.78 x 1011 s
Now convert to years:
(4.78 x 1011 s) x (1h / 3600s) x (1 day / 24h) x (1 year / 365 days) = 1.52 x 104 years
Once we know the decay constant (k) for a radioactive isotope, we can calculate the fraction of the radioactive nuclei left after a given time by the following equation:
ln (Nt / No) = –kt
No = the number of nuclei in the original sample
Nt = the number of nuclei left after time t
Never really need to find Nt or No, it is the fraction that is important.
Example:
When t½ = 10.76 years, what is the fraction remaining after 25 years?
Remember k = 0.693 / t½:
k = 0.693 / (10.76 years)
ln (Nt / No) = –(0.693t / t½)
ln (Nt / No) = –(0.693x 25years / 10.76 years)
ln (Nt / No) = –1.61
(Nt / No) = e–1.61
(Nt / No) = 0.200
This means 20% is left.
Example:
How long until there is only 5% left?
(Nt / No) = 5% = 0.05
ln(0.05) = –(0.693t / 10.76 years)
t = 46.6 years
Radioactive Dating:
Carbon–14, t½ = 5730 years
The upper atmosphere is constantly bombarded by cosmic ray radiations from space, which creates high energy neutrons when it reacts with Nitrogen–14 (the most abundant nitrogen nuclide).
147N + 10n → 146C + 11H
Then goes into carbon dioxide:
146CO2 → plants → animals
There is a set amount of Carbon–14 per kg of living matter. When something dies, it no longer takes in Carbon–14. The Carbon–14 then decays:
146C → 147N + 0-1e
In living matter, we measure 15.3 disintegrations per kg of C per minute. Anything lower → must be dead. In this way, this ratio of carbon isotopes becomes a clock measuring the time since death of an organism.
Nt / No = (less than 15.3 dis/kg • min) / (15.3 dis/kg • min)
Example:
You want to carbon date a caveman's foot, where you observe (4.50 dis/kg • min). When did the caveman lose his foot?
ln(Nt / No) = –(0.693t / t½)
ln(4.50 / 15.3) = –(0.693t / 5730 years)
t = 10118 years
Applications of Radioactive Isotopes:
Chemical Analysis:
Radioactive Tracer: a very small amount of radioactive isotope added to a chemical, biological, or physical system to study the system.
Evidence of dynamic equilibrium can be seen using radioactive tracers. In a beaker, imagine a solution of PbI2 in contact with the solid. This beaker contains only iodine atoms with nonradioactive isotopes. Now add a radioactive iodide ion to the solution. The solution is still saturated, and the amount of solid remains constant. But, after time, the solid lead iodide, which was initially nonradioactive, is radioactive. Nonradioactive iodide ions in the solid have swapped with radioactive ions in solution.
Tracers are also used to determine chemical mechanisms, like photosynthesis.
Isotope Dilution: a technique to determine the quantity of a substance in a mixture or of the total volume of solution by adding a known amount of an isotope to it. After time, removal of a sample of the mixture and measurement of the fraction by which the isotope has diluted provides a way to determine the quantity of substance or volume of solution.
Neutron Activation Analysis: an analysis of elements in a sample based on the conversion of stable isotopes to radioactive isotopes by bombarding a sample with neutrons. Example: measuring trace amounts of arsenic in human hair. The arsenic is converted into a metastable nucleus via bombardment with a neutron. As the arsenic returns to its ground state, it emits gamma rays. These measurements can be interpreted to find who the hair belongs to, or how much arsenic is in it.
7533As + 10n → 76m33As
76m33As → 7533As + 00γ
Mass-Energy Equivalence:
When nuclei decay, they form products of lower energy. The change of energy is related to changes in mass, according to the mass-energy equivalence derived by Albert Einstein.
E = mc2
Where c = the speed of light = 3.00 x 108 m/s
E = energy
m = mass
For any mass, there is an associated energy. For any energy, there is an associated mass.
When the energy changes by an amount ΔE, the mass changes by an amount Δm.
ΔE = Δmc2
When energy is given off, mass decreases.
Example:
Chemical reaction:
When carbon burns in oxygen, it releases heat energy:
C (gr) + O2 (g) → CO2 (g); ΔH = –393.5 kJ
ΔE = –3.935 x 105 J (converted from kJ to J)
Recall that J = (kg•m2 / s2)
Δm = ΔE / c2
Δm = –(3.935 x 105 kg•m2 / s2) / (3.00 x 108 m/s)2
Δm = –4.37 x 10-12 kg
Mass changes in nuclear reactions are approximately a milion times larger per mole of reactant thatn those in chemical reaction.
Example:
Nuclear Reaction:
Consider the alpha decay of Uranium–238 to Thorium–234
23892U → 23490Th + 42He
Use the nuclear mass in amu to find the change in mass for this reaction. You would think the mass of the reactants is the same as the mass of the products, but not so. There is a mass loss.
23892U = 238.0003 amu
23490Th = 233.9942 amu
42He = 4.00150 amu
Δm = (233.9942 amu + 4.00150 amu) – (238.00003 amu)
Δm = – 0.0046 amu
(Assume 1 mole of isotope.)
Δm = – 0.0046 amu = – 0.0046g = – 4.6 x 10-6 kg
ΔE = Δmc2 = (– 4.6 x 10-6 kg)(3.00 x 108 m/s)2
ΔE = – 4.14 x 1011 J = – 4.14 x 108 kJ
Nuclear Fission: spontaneous break up of isotopes, naturally occurring, does not require energy input. Powers nuclear reactors and bombs.
Nuclear Fusion: Fusing of two nuclei into a larger nucleus. Takes GIGANTIC amounts of energy.
21H 31Th→ 42He + 10n
half-life: the time it takes from one-half of the nuclei in a sample to decay. Independent of the amount of sample, denoted t½.
First Order Half Life:
t½ = 0.693 / k
Example:
k = 1.45 x 10-12 s-1 for Ta-183, find t½ in years.
t½ = 0.693 / 1.45 x 10-12 s-1
t½ = 4.78 x 1011 s
Now convert to years:
(4.78 x 1011 s) x (1h / 3600s) x (1 day / 24h) x (1 year / 365 days) = 1.52 x 104 years
Once we know the decay constant (k) for a radioactive isotope, we can calculate the fraction of the radioactive nuclei left after a given time by the following equation:
ln (Nt / No) = –kt
No = the number of nuclei in the original sample
Nt = the number of nuclei left after time t
Never really need to find Nt or No, it is the fraction that is important.
Example:
When t½ = 10.76 years, what is the fraction remaining after 25 years?
Remember k = 0.693 / t½:
k = 0.693 / (10.76 years)
ln (Nt / No) = –(0.693t / t½)
ln (Nt / No) = –(0.693x 25years / 10.76 years)
ln (Nt / No) = –1.61
(Nt / No) = e–1.61
(Nt / No) = 0.200
This means 20% is left.
Example:
How long until there is only 5% left?
(Nt / No) = 5% = 0.05
ln(0.05) = –(0.693t / 10.76 years)
t = 46.6 years
Radioactive Dating:
Carbon–14, t½ = 5730 years
The upper atmosphere is constantly bombarded by cosmic ray radiations from space, which creates high energy neutrons when it reacts with Nitrogen–14 (the most abundant nitrogen nuclide).
147N + 10n → 146C + 11H
Then goes into carbon dioxide:
146CO2 → plants → animals
There is a set amount of Carbon–14 per kg of living matter. When something dies, it no longer takes in Carbon–14. The Carbon–14 then decays:
146C → 147N + 0-1e
In living matter, we measure 15.3 disintegrations per kg of C per minute. Anything lower → must be dead. In this way, this ratio of carbon isotopes becomes a clock measuring the time since death of an organism.
Nt / No = (less than 15.3 dis/kg • min) / (15.3 dis/kg • min)
Example:
You want to carbon date a caveman's foot, where you observe (4.50 dis/kg • min). When did the caveman lose his foot?
ln(Nt / No) = –(0.693t / t½)
ln(4.50 / 15.3) = –(0.693t / 5730 years)
t = 10118 years
Applications of Radioactive Isotopes:
Chemical Analysis:
Radioactive Tracer: a very small amount of radioactive isotope added to a chemical, biological, or physical system to study the system.
Evidence of dynamic equilibrium can be seen using radioactive tracers. In a beaker, imagine a solution of PbI2 in contact with the solid. This beaker contains only iodine atoms with nonradioactive isotopes. Now add a radioactive iodide ion to the solution. The solution is still saturated, and the amount of solid remains constant. But, after time, the solid lead iodide, which was initially nonradioactive, is radioactive. Nonradioactive iodide ions in the solid have swapped with radioactive ions in solution.
Tracers are also used to determine chemical mechanisms, like photosynthesis.
Isotope Dilution: a technique to determine the quantity of a substance in a mixture or of the total volume of solution by adding a known amount of an isotope to it. After time, removal of a sample of the mixture and measurement of the fraction by which the isotope has diluted provides a way to determine the quantity of substance or volume of solution.
Neutron Activation Analysis: an analysis of elements in a sample based on the conversion of stable isotopes to radioactive isotopes by bombarding a sample with neutrons. Example: measuring trace amounts of arsenic in human hair. The arsenic is converted into a metastable nucleus via bombardment with a neutron. As the arsenic returns to its ground state, it emits gamma rays. These measurements can be interpreted to find who the hair belongs to, or how much arsenic is in it.
7533As + 10n → 76m33As
76m33As → 7533As + 00γ
Mass-Energy Equivalence:
When nuclei decay, they form products of lower energy. The change of energy is related to changes in mass, according to the mass-energy equivalence derived by Albert Einstein.
E = mc2
Where c = the speed of light = 3.00 x 108 m/s
E = energy
m = mass
For any mass, there is an associated energy. For any energy, there is an associated mass.
When the energy changes by an amount ΔE, the mass changes by an amount Δm.
ΔE = Δmc2
When energy is given off, mass decreases.
Example:
Chemical reaction:
When carbon burns in oxygen, it releases heat energy:
C (gr) + O2 (g) → CO2 (g); ΔH = –393.5 kJ
ΔE = –3.935 x 105 J (converted from kJ to J)
Recall that J = (kg•m2 / s2)
Δm = ΔE / c2
Δm = –(3.935 x 105 kg•m2 / s2) / (3.00 x 108 m/s)2
Δm = –4.37 x 10-12 kg
Mass changes in nuclear reactions are approximately a milion times larger per mole of reactant thatn those in chemical reaction.
Example:
Nuclear Reaction:
Consider the alpha decay of Uranium–238 to Thorium–234
23892U → 23490Th + 42He
Use the nuclear mass in amu to find the change in mass for this reaction. You would think the mass of the reactants is the same as the mass of the products, but not so. There is a mass loss.
23892U = 238.0003 amu
23490Th = 233.9942 amu
42He = 4.00150 amu
Δm = (233.9942 amu + 4.00150 amu) – (238.00003 amu)
Δm = – 0.0046 amu
(Assume 1 mole of isotope.)
Δm = – 0.0046 amu = – 0.0046g = – 4.6 x 10-6 kg
ΔE = Δmc2 = (– 4.6 x 10-6 kg)(3.00 x 108 m/s)2
ΔE = – 4.14 x 1011 J = – 4.14 x 108 kJ
Nuclear Fission: spontaneous break up of isotopes, naturally occurring, does not require energy input. Powers nuclear reactors and bombs.
Nuclear Fusion: Fusing of two nuclei into a larger nucleus. Takes GIGANTIC amounts of energy.
21H 31Th→ 42He + 10n
20.2-20.3: Nuclear Bombardment Reactions-Radiations and Matter
Transuranium Elements:
DOES NOT INCLUDE URANIUM!!!
Includes elements with an atomic number greater than that of uranium (Z = 92), the naturally occurring element of greatest Z.
Radiation Counters:
Because nuclear radiations can ionize molecules and break chemical bonds, they adversely affect biological organisms. So it's good to know when they're there.
Geiger Counter: a kind of ionization counter used to count particles emitted by radioactive nuclei, consists of a metal tube filled with gas, such as argon. Can detect both alpha and beta particles, and under special circumstances, neutrons.
10n + 105B → 73Li + 42He
Scintillation Counter: a device that detects nuclear radiation from flashes of light generated in a material by the radiation. A phosphor is a substance that emits flashes of light when struck by radiation. Can detect both beta and gamma particles using a photomultiplier tube.
One e– can produce 106 e–.
This vintage reel on Scintillation Counters is way better than a diagram:
Radiation counters measure the number of nuclear disintegrations in a radioactive sample.
Activity: (activity of a radioactive source) is the number of nuclear disintegrations per unit time occurring in a radioactive material.
Curie: Denoted Ci, is a unit of activity equal to 3.700 x 1010 disintegrations per second.
Nuclear disintegrations / second = nuc / s = Ci
Biological Effects of Radiation Dosage:
Radiation dosage effects can be quite damaging. DNA is especially affected, which interferes with cell division.
rad → (radiation absorbed dose) the dosage of radiation that deposits 1 x 10-2 J of energy per kg of tissue.
Depends not only on the amount of energy deposited in the tissue, but also on the particle. Neutrons are more destructive than gamma rays of the same radiation dosage measured in rads.
rem → a unit of radiation dosage used to relate various kinds of radiation in terms of biological destruction. It equals the rad times a factor for the type of radiation, called the relative biological effectiveness, or RBE (basically, how efficient it is at killing you).
rems = rads x RBE → total damage
RBE = 1 for γ (gamma) and β (beta).
RBE = 5 for no
RBE = 10 for α (alpha)
Here's that movie mentioned in class:
It really is an awesome movie, go rent it!
Rate of Radioactive Decay:
The rate equation for radioactive decay has the same form as the rate law for a first order chemical reaction.
R = kNt
R = Rate = Activity
Nt = the number of radioactive nuclei at time t
k = radioactive decay constant, the rate constant for radioactive decay
Example:
A 2.50 mg sample of Tc-99 has an activity of 2.70 x 10-5 Ci and is decaying by beta emission. What is the decay constant?
R = kNt
First convert the activity to nuc/s:
R = 2.70 x 10-5 Ci x (3.700 x 1010 nuc/s / 1s) = 9.99 x 105 nuc/s
Now convert the given sample to Nt:
2.50 mg Tc-99 = 2.50 x 10-3g Tc-99
(2.50 x 10-3g Tc-99) x (1 mol Tc-99 / 99g Tc-99) x (6.023 x 1023 nuclei / 1 mol Tc-99) = 1.52 x 1019 nuclei = Nt
k = Rate / Nt
k = (9.99 x 105 nuc/s) / ( 1.52 x 1019 nuclei)
k = 6.57 x 10-14 s-1
Units for k will always be inverted seconds because it is first order.
DOES NOT INCLUDE URANIUM!!!
Includes elements with an atomic number greater than that of uranium (Z = 92), the naturally occurring element of greatest Z.
Radiation Counters:
Because nuclear radiations can ionize molecules and break chemical bonds, they adversely affect biological organisms. So it's good to know when they're there.
Geiger Counter: a kind of ionization counter used to count particles emitted by radioactive nuclei, consists of a metal tube filled with gas, such as argon. Can detect both alpha and beta particles, and under special circumstances, neutrons.
10n + 105B → 73Li + 42He
Scintillation Counter: a device that detects nuclear radiation from flashes of light generated in a material by the radiation. A phosphor is a substance that emits flashes of light when struck by radiation. Can detect both beta and gamma particles using a photomultiplier tube.
One e– can produce 106 e–.
This vintage reel on Scintillation Counters is way better than a diagram:
Radiation counters measure the number of nuclear disintegrations in a radioactive sample.
Activity: (activity of a radioactive source) is the number of nuclear disintegrations per unit time occurring in a radioactive material.
Curie: Denoted Ci, is a unit of activity equal to 3.700 x 1010 disintegrations per second.
Nuclear disintegrations / second = nuc / s = Ci
Biological Effects of Radiation Dosage:
Radiation dosage effects can be quite damaging. DNA is especially affected, which interferes with cell division.
rad → (radiation absorbed dose) the dosage of radiation that deposits 1 x 10-2 J of energy per kg of tissue.
Depends not only on the amount of energy deposited in the tissue, but also on the particle. Neutrons are more destructive than gamma rays of the same radiation dosage measured in rads.
rem → a unit of radiation dosage used to relate various kinds of radiation in terms of biological destruction. It equals the rad times a factor for the type of radiation, called the relative biological effectiveness, or RBE (basically, how efficient it is at killing you).
rems = rads x RBE → total damage
RBE = 1 for γ (gamma) and β (beta).
RBE = 5 for no
RBE = 10 for α (alpha)
Here's that movie mentioned in class:
It really is an awesome movie, go rent it!
Rate of Radioactive Decay:
The rate equation for radioactive decay has the same form as the rate law for a first order chemical reaction.
R = kNt
R = Rate = Activity
Nt = the number of radioactive nuclei at time t
k = radioactive decay constant, the rate constant for radioactive decay
Example:
A 2.50 mg sample of Tc-99 has an activity of 2.70 x 10-5 Ci and is decaying by beta emission. What is the decay constant?
R = kNt
First convert the activity to nuc/s:
R = 2.70 x 10-5 Ci x (3.700 x 1010 nuc/s / 1s) = 9.99 x 105 nuc/s
Now convert the given sample to Nt:
2.50 mg Tc-99 = 2.50 x 10-3g Tc-99
(2.50 x 10-3g Tc-99) x (1 mol Tc-99 / 99g Tc-99) x (6.023 x 1023 nuclei / 1 mol Tc-99) = 1.52 x 1019 nuclei = Nt
k = Rate / Nt
k = (9.99 x 105 nuc/s) / ( 1.52 x 1019 nuclei)
k = 6.57 x 10-14 s-1
Units for k will always be inverted seconds because it is first order.
20.1: Radioactivity-Nuclear Bombardment Reactions
Nuclear Stability:
Why don't all the protons in the nucleus repel each other and cause the nucleus to blow apart?
Nuclear Force: a strong force of attraction between nucleons that acts only at very short distances, overwhelms the electrostatic force (the positive charges repelling each other.)
Shell Model of the Nucleus: a nuclear model in which protons and neutrons exist in levels, or shells, analogous to the shell structure that exists for electrons in an atom.
First Test For Stability:
Magic Numbers: the number of nuclear particles in a completed shell of protons or neutrons. Think of the way full electron shells make noble gases stable, this is similar. If a nucleus has two magic numbers, it is very stable.
Magic Numbers for p+ → 2, 8, 20, 28, 50, 82, (114)
Magic Numbers for no → 2, 8, 20, 28, 50, 82, 126
Example:
42He
p+ → 2
no → 2
Both are magic numbers, very stable.
Example:
Which is more stable?
10251Sn or 10151Sn?
10251Sn → p+ = 51, no = 51
10151Sn → p+ = 51, no = 50
10151Sn has a magic number and 10251Sn doesn't.
Second Test for Stability:
Even/Odd Test:
Nuclei with an even number of protons and an even number of neutrons are very likely to be stable.
Nuclei with an odd number of protons and an odd number of neutrons are not at all likely to be stable.
Nuclei with variably even and odd numbers of protons or neutrons may or may not be stable.
Example:
Which is more likely to be stable, 6231Ga or 6432Ge?
6432Ge because it has both an even number of protons and an even number of neutrons.
Third Test for Stability:
Band of Stability: the region in which stable nuclides lie in a plot of number of protons against number of neutrons.
When it turns out to be unstable, what does the nucleus do about it?
The Six Types of Radioactive Decay:
1. Alpha Emission: emission of a 42He nucleus, or alpha particle from an unstable nucleus. Happens for very large nuclei.
Example:
22688Ra → 22286Rn + 42He
2. Beta Emission: emission of a high-speed electron (β–) from an unstable nucleus.
Example:
146C → 147N + 0-1β
Equivalent to the conversion of a neutron to a proton.
10n → 11p + 0-1e
3. Positron Emission: emission of a positron (β+, or 01e) from an unstable nucleus.
Example:
9543Tc → 9542Mo + 01e
Equivalent to the conversion of a proton to a neutron.
11p → 10n + 01e
Professor said a positron is an electron traveling backward in time, but I have no idea what that means.
4. Electron Capture: the decay of an unstable nucleus by capturing, or picking up, an electron from an inner orbital of an atom.
Example:
4019K + 0-1e → 4018Ar
In effect, a proton is changed to a neutron, as in positron emission:
11p → 01e- + 10n
5. Gamma Emission: emission from an excited nucleus of a gamma photon (denoted γ). Often, gamma emission occurs very quickly after radioactive decay. In some cases, however, an excited state has a significant lifetime before it emits a gamma photon.
Metastable Nucleus: a nucleus in an excited state with a lifetime of at least one nanosecond (10-9s). In time, the metastable nucleus decays by gamma emission.
Example:
99m43Tc → 9943Tc + 00γ
99m43Tc → the m denotes metastable, excited state for nucleus
9943Tc → ground state
6. Spontaneous Fission: the spontaneous decay of an unstable nucleus in which a heavy nucleus of mass number greater than 89 splits into lighter nuclei and energy is released.
Example:
23692U → 9639Y + 13633I + 410n
When A is greater than 89 → spontaneous fission
Recall the Band of Stability:
*To the left of the band, nuclides have a neutron to proton ratio larger than that needed for stability, so they tend to decay by beta emission.
*To the right of the band, nuclides have a smaller neutron to proton ratio that that needed for stability, so they tend to decay by either positron emission or electron capture.
*As the curve follows Z as it becomes larger than 83, decay is often by alpha emission.
Radioactive Decay Series: a sequence in which one radioactive nucleus decays to a second, which then decays to a third, and so forth. Eventually, a stable nucleus, which is an isotope of lead, is reached.
Each step will involve an alpha or a beta decay.
Example:
What is the nucleus formed from Uranium–238 after six alpha (42He) and two beta (0-1β) emissions?
23892U → □□□ + 642He + 20-1e
□□□ = 21482Pb
Nuclear Bombardment Reactions:
Alchemists → wanted to turn lead to gold and live forever
Transmutation: the change of one element to another by bombarding the nucleus of the element with nuclear particles or nuclei.
Notation:
Target Nucleus + Subatomic Particle → Product Nucleus + Subatomic Particle
Example:
147N +42He → 178O + 11H
147N(42He, 11H)178O
Example:
94Be +42He → 126C + 10n
94Be(42He, 10n)126C
Elements with large atomic numbers deflect alpha particles (large positive nucleus deflect positive alpha particles.) To shoot large particles into heavy nuclei, charged particles must be accelerated.
Particle Accelerator: a device used to accelerate electrons, protons, and alpha particles and other ions to very high speeds. The kinetic energies of these particles is units of electron volts.
Electron Volt: Denoted eV, the quantity of energy that would have to be imparted to an electron to accelerate it by one volt potential difference.
1 eV = 1.602 x 10-19 J
Cyclotron: a type of particle accelerator consisting of two hollow, semicircular metal electrodes called dees (because the shape resembles the letter D), in which charged particles are accelerated by stages to higher and higher kinetic energies.
Ions introduced at the center of the cyclotron are accelerated in the space between the two dees. A magnetic field keeps the ions moving in a spiral path.
The dees are connected to a high-frequency electric current that changes their polarity so that each time the ion moves in the space between the dees, it is accelerated. Then it leaves the cyclotron at high speed and hits its target.
This guy really loves cyclotrons:
Why don't all the protons in the nucleus repel each other and cause the nucleus to blow apart?
Nuclear Force: a strong force of attraction between nucleons that acts only at very short distances, overwhelms the electrostatic force (the positive charges repelling each other.)
Shell Model of the Nucleus: a nuclear model in which protons and neutrons exist in levels, or shells, analogous to the shell structure that exists for electrons in an atom.
First Test For Stability:
Magic Numbers: the number of nuclear particles in a completed shell of protons or neutrons. Think of the way full electron shells make noble gases stable, this is similar. If a nucleus has two magic numbers, it is very stable.
Magic Numbers for p+ → 2, 8, 20, 28, 50, 82, (114)
Magic Numbers for no → 2, 8, 20, 28, 50, 82, 126
Example:
42He
p+ → 2
no → 2
Both are magic numbers, very stable.
Example:
Which is more stable?
10251Sn or 10151Sn?
10251Sn → p+ = 51, no = 51
10151Sn → p+ = 51, no = 50
10151Sn has a magic number and 10251Sn doesn't.
Second Test for Stability:
Even/Odd Test:
Nuclei with an even number of protons and an even number of neutrons are very likely to be stable.
Nuclei with an odd number of protons and an odd number of neutrons are not at all likely to be stable.
Nuclei with variably even and odd numbers of protons or neutrons may or may not be stable.
Example:
Which is more likely to be stable, 6231Ga or 6432Ge?
6432Ge because it has both an even number of protons and an even number of neutrons.
Third Test for Stability:
Band of Stability: the region in which stable nuclides lie in a plot of number of protons against number of neutrons.
When it turns out to be unstable, what does the nucleus do about it?
The Six Types of Radioactive Decay:
1. Alpha Emission: emission of a 42He nucleus, or alpha particle from an unstable nucleus. Happens for very large nuclei.
Example:
22688Ra → 22286Rn + 42He
2. Beta Emission: emission of a high-speed electron (β–) from an unstable nucleus.
Example:
146C → 147N + 0-1β
Equivalent to the conversion of a neutron to a proton.
10n → 11p + 0-1e
3. Positron Emission: emission of a positron (β+, or 01e) from an unstable nucleus.
Example:
9543Tc → 9542Mo + 01e
Equivalent to the conversion of a proton to a neutron.
11p → 10n + 01e
Professor said a positron is an electron traveling backward in time, but I have no idea what that means.
4. Electron Capture: the decay of an unstable nucleus by capturing, or picking up, an electron from an inner orbital of an atom.
Example:
4019K + 0-1e → 4018Ar
In effect, a proton is changed to a neutron, as in positron emission:
11p → 01e- + 10n
5. Gamma Emission: emission from an excited nucleus of a gamma photon (denoted γ). Often, gamma emission occurs very quickly after radioactive decay. In some cases, however, an excited state has a significant lifetime before it emits a gamma photon.
Metastable Nucleus: a nucleus in an excited state with a lifetime of at least one nanosecond (10-9s). In time, the metastable nucleus decays by gamma emission.
Example:
99m43Tc → 9943Tc + 00γ
99m43Tc → the m denotes metastable, excited state for nucleus
9943Tc → ground state
6. Spontaneous Fission: the spontaneous decay of an unstable nucleus in which a heavy nucleus of mass number greater than 89 splits into lighter nuclei and energy is released.
Example:
23692U → 9639Y + 13633I + 410n
When A is greater than 89 → spontaneous fission
Recall the Band of Stability:
*To the left of the band, nuclides have a neutron to proton ratio larger than that needed for stability, so they tend to decay by beta emission.
*To the right of the band, nuclides have a smaller neutron to proton ratio that that needed for stability, so they tend to decay by either positron emission or electron capture.
*As the curve follows Z as it becomes larger than 83, decay is often by alpha emission.
Radioactive Decay Series: a sequence in which one radioactive nucleus decays to a second, which then decays to a third, and so forth. Eventually, a stable nucleus, which is an isotope of lead, is reached.
Each step will involve an alpha or a beta decay.
Example:
What is the nucleus formed from Uranium–238 after six alpha (42He) and two beta (0-1β) emissions?
23892U → □□□ + 642He + 20-1e
□□□ = 21482Pb
Nuclear Bombardment Reactions:
Alchemists → wanted to turn lead to gold and live forever
Transmutation: the change of one element to another by bombarding the nucleus of the element with nuclear particles or nuclei.
Notation:
Target Nucleus + Subatomic Particle → Product Nucleus + Subatomic Particle
Example:
147N +42He → 178O + 11H
147N(42He, 11H)178O
Example:
94Be +42He → 126C + 10n
94Be(42He, 10n)126C
Elements with large atomic numbers deflect alpha particles (large positive nucleus deflect positive alpha particles.) To shoot large particles into heavy nuclei, charged particles must be accelerated.
Particle Accelerator: a device used to accelerate electrons, protons, and alpha particles and other ions to very high speeds. The kinetic energies of these particles is units of electron volts.
Electron Volt: Denoted eV, the quantity of energy that would have to be imparted to an electron to accelerate it by one volt potential difference.
1 eV = 1.602 x 10-19 J
Cyclotron: a type of particle accelerator consisting of two hollow, semicircular metal electrodes called dees (because the shape resembles the letter D), in which charged particles are accelerated by stages to higher and higher kinetic energies.
Ions introduced at the center of the cyclotron are accelerated in the space between the two dees. A magnetic field keeps the ions moving in a spiral path.
The dees are connected to a high-frequency electric current that changes their polarity so that each time the ion moves in the space between the dees, it is accelerated. Then it leaves the cyclotron at high speed and hits its target.
This guy really loves cyclotrons:
20.1-Radioactivity
Nuclear Chemistry:
Radioactive Decay and Nuclear Bombardment Reactions:
Antoine Henri Becquerel discovered in 1896 that photographic plates developed bright spots when exposed to uranium minerals and concluded that they must give off radiation.
Terms:
α = alpha particle = positively charged, consist of helium-4 nuclei (nuclei with two protons and two neutrons)
β = beta particle = negatively charged, consist of high speed electrons
γ = gamma particle = no charge = a form of electromagnetic radiation similar to x-rays except higher in energy.
Nuclide Symbol: AZX
Example: 23892U → Uranium–238
X = element
Z = atomic number → number of protons → p+
A = mass number → number of protons and neutrons → no
number of neutrons = N = A – Z
A gives the mass of the isotope
Z gives the charge
isotopes → same Z, same element. Ex: 126C, 136C, 146C
nuclides → technically any combination 126C, 147N, 188O
*But now these terms are used interchangeably
Notation:
protons → 11H, 11p, p+
neutrons → 10n, no, n
electrons → 0– 1e, 0– 1β (high energy beta particle), β–
positron → 01e, 0+ 1β (like a positive electron), β+
gamma photon → 00γ, γ
alpha particle → 42He, α
deuterium → 21H, D, d (H plus one neutron)
tritium → 31H, t (H plus two neutrons)
Radioactive Decay: has one reactant, occurs naturally, no energy input needed. Usually, high energy particles come out.
Example:
Nuclear Equation:
23892U → 23490Th + 42He
These must be mass balanced and charge balanced (superscripts and subscripts on the left side of the equation equal the sum of the superscripts and subscripts on the right side of the equation.)
Nuclear Bombardment Reactions: has two reactants. Usually synthesized and energy input is required.
Example:
□□□ + 21H → 2411Na + 42He
□□□ = 2612Mg
Radioactive Decay and Nuclear Bombardment Reactions:
Antoine Henri Becquerel discovered in 1896 that photographic plates developed bright spots when exposed to uranium minerals and concluded that they must give off radiation.
Terms:
α = alpha particle = positively charged, consist of helium-4 nuclei (nuclei with two protons and two neutrons)
β = beta particle = negatively charged, consist of high speed electrons
γ = gamma particle = no charge = a form of electromagnetic radiation similar to x-rays except higher in energy.
Nuclide Symbol: AZX
Example: 23892U → Uranium–238
X = element
Z = atomic number → number of protons → p+
A = mass number → number of protons and neutrons → no
number of neutrons = N = A – Z
A gives the mass of the isotope
Z gives the charge
isotopes → same Z, same element. Ex: 126C, 136C, 146C
nuclides → technically any combination 126C, 147N, 188O
*But now these terms are used interchangeably
Notation:
protons → 11H, 11p, p+
neutrons → 10n, no, n
electrons → 0– 1e, 0– 1β (high energy beta particle), β–
positron → 01e, 0+ 1β (like a positive electron), β+
gamma photon → 00γ, γ
alpha particle → 42He, α
deuterium → 21H, D, d (H plus one neutron)
tritium → 31H, t (H plus two neutrons)
Radioactive Decay: has one reactant, occurs naturally, no energy input needed. Usually, high energy particles come out.
Example:
Nuclear Equation:
23892U → 23490Th + 42He
These must be mass balanced and charge balanced (superscripts and subscripts on the left side of the equation equal the sum of the superscripts and subscripts on the right side of the equation.)
Nuclear Bombardment Reactions: has two reactants. Usually synthesized and energy input is required.
Example:
□□□ + 21H → 2411Na + 42He
□□□ = 2612Mg
19.9-19.11: Electrolysis of Molten Salts-Stoichiometry of Elecrolysis
Electrolysis: the process of producing a chemical change in an electrolytic cell, which is an electrochemical cell in which an electric current drives an otherwise non-spontaneous reaction.
Downs Cell: a commercial electrochemical cell used to obtain sodium metal by the electrolysis of molten sodium chloride.
The half-reactions are:
Na+ (l) + e– → Na (l)
Cl– (l) → (½)Cl2 (g) + e–
Where the reduction of Na+ to Na occurs at the cathode and the oxidation of Cl– to Cl2 occurs at the anode. No water is used.
CaCl2 is added to reduce the melting point of NaCl from 801oC to 580oC. And the electrodes do not need to be separated because the non-spontaneous reaction does not want to proceed anyway. This is the only way to commercially produce Na(s).
The cell reaction is:
Na+ (l) + Cl– (l) → Na (l) + (½)Cl2 (g)
Overvoltage:
“The presence of water in a solution of sodium chloride must be examined in respect to its reduction and oxidation in both electrodes. Usually, water is electrolysed as mentioned in electrolysis of water yielding gaseous oxygen in the anode and gaseous hydrogen in the cathode. On the other hand, sodium chloride in water dissociates in Na+ and Cl– ions, cation, which is the positive ion, will be attracted to the cathode (+), thus reducing the sodium ion. The anion will then be attracted to the anode (–) oxidizing chloride ion.
The following half reactions describes the process mentioned:
1. Cathode: Na+(aq) + e– → Na(s) (E° = –2.71 V)
2. Anode: 2 Cl–(aq) → Cl2(g) + 2 e– (E° = +1.36 V)
3. Cathode: 2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq) (E° = –0.83 V)
4. Anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e– → (E° = +1.23 V)
Reaction 1 is discarded as it has the most negative value on standard reduction potential thus making it less thermodynamically favorable in the process.
When comparing the reduction potentials in reactions 2 and 4, the reduction of chloride ion is favored. Thus, if the Cl– ion is favored for reduction, then the water reaction is favored for oxidation producing gaseous oxygen, however experiments show gaseous chlorine is produced and not oxygen.
Although the initial analysis is correct, there is another effect that can happen, known as the overvoltage effect. Additional voltage is sometimes required, beyond the voltage predicted by the E°cell. This may be due to kinetic rather than thermodynamic considerations. In fact, it has been proven that the activation energy for the chloride ion is very low, hence favorable in kinetic terms. In other words, although the voltage applied is thermodynamically sufficient to drive electrolysis, the rate is so slow that to make the process proceed in a reasonable time frame, the voltage of the external source has to be increased (hence, overvoltage.)
Finally, reaction 3 is favorable because it describes the proliferation of OH– ions thus letting a probable reduction of H+ ions less favorable an option.
The overall reaction for the process according to the analysis would be the following:
Anode (oxidation): 2 Cl–(aq) → Cl2(g) + 2 e–
Cathode (reduction): 2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
Overall reaction: 2 H2O + 2 Cl–(aq) → H2(g) + Cl2(g) + 2 OH–(aq)
As the overall reaction indicates, the concentration of chloride ions is reduced in comparison to OH– ions (whose concentration increases). The reaction also shows the production of gaseous hydrogen, chlorine and aqueous sodium hydroxide.”
Source
Electroplating of Metals:
Copper (II) ion leaves the anode and plates out on the cathode. Unreactive substances (such as gold) collect as anode mud under the anode (very valuable).
Stoichiometry of Electrolysis:
Ampere: Denoted A, is the SI base unit for current. The coulomb ( C ) is the SI unit of electric charge and is equivalent to an ampere-second.
Coulombs = Amperes x Seconds
(Electric Charge = Electric Current x Time Lapse)
and
Amperes = Coulombs / Seconds
One faraday (96,485 Coulombs) is equivalent to the charge on one mole of electrons.
Example:
System: Cu2+ (aq) + 2e– → Cu (s)
A constant current passes through this system for 6.50 hours and 6.75g Cu (s) was deposits. What was the current?
Amperes = Coulombs / Seconds
To get the electric current, convert Cu from grams to moles, use stoichiometry to determine moles of e– to moles of Cu, then use moles of e– to convert to Coulombs:
6.75g Cu x (1 mol Cu / 63.5g Cu) x (2 mol e– / 1 mol Cu) x (96485 C / 1 mol e–) = 20512.6 C
To get the time, convert 6.50h to seconds:
6.50h x (3600 s / 1 h) = 23400s
Divide Coulombs by Seconds to get Amperes:
Amperes = 20512.6 C / 23400s = 0.876 Amps
Downs Cell: a commercial electrochemical cell used to obtain sodium metal by the electrolysis of molten sodium chloride.
The half-reactions are:
Na+ (l) + e– → Na (l)
Cl– (l) → (½)Cl2 (g) + e–
Where the reduction of Na+ to Na occurs at the cathode and the oxidation of Cl– to Cl2 occurs at the anode. No water is used.
CaCl2 is added to reduce the melting point of NaCl from 801oC to 580oC. And the electrodes do not need to be separated because the non-spontaneous reaction does not want to proceed anyway. This is the only way to commercially produce Na(s).
The cell reaction is:
Na+ (l) + Cl– (l) → Na (l) + (½)Cl2 (g)
Overvoltage:
“The presence of water in a solution of sodium chloride must be examined in respect to its reduction and oxidation in both electrodes. Usually, water is electrolysed as mentioned in electrolysis of water yielding gaseous oxygen in the anode and gaseous hydrogen in the cathode. On the other hand, sodium chloride in water dissociates in Na+ and Cl– ions, cation, which is the positive ion, will be attracted to the cathode (+), thus reducing the sodium ion. The anion will then be attracted to the anode (–) oxidizing chloride ion.
The following half reactions describes the process mentioned:
1. Cathode: Na+(aq) + e– → Na(s) (E° = –2.71 V)
2. Anode: 2 Cl–(aq) → Cl2(g) + 2 e– (E° = +1.36 V)
3. Cathode: 2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq) (E° = –0.83 V)
4. Anode: 2 H2O(l) → O2(g) + 4 H+(aq) + 4 e– → (E° = +1.23 V)
Reaction 1 is discarded as it has the most negative value on standard reduction potential thus making it less thermodynamically favorable in the process.
When comparing the reduction potentials in reactions 2 and 4, the reduction of chloride ion is favored. Thus, if the Cl– ion is favored for reduction, then the water reaction is favored for oxidation producing gaseous oxygen, however experiments show gaseous chlorine is produced and not oxygen.
Although the initial analysis is correct, there is another effect that can happen, known as the overvoltage effect. Additional voltage is sometimes required, beyond the voltage predicted by the E°cell. This may be due to kinetic rather than thermodynamic considerations. In fact, it has been proven that the activation energy for the chloride ion is very low, hence favorable in kinetic terms. In other words, although the voltage applied is thermodynamically sufficient to drive electrolysis, the rate is so slow that to make the process proceed in a reasonable time frame, the voltage of the external source has to be increased (hence, overvoltage.)
Finally, reaction 3 is favorable because it describes the proliferation of OH– ions thus letting a probable reduction of H+ ions less favorable an option.
The overall reaction for the process according to the analysis would be the following:
Anode (oxidation): 2 Cl–(aq) → Cl2(g) + 2 e–
Cathode (reduction): 2 H2O(l) + 2 e– → H2(g) + 2 OH–(aq)
Overall reaction: 2 H2O + 2 Cl–(aq) → H2(g) + Cl2(g) + 2 OH–(aq)
As the overall reaction indicates, the concentration of chloride ions is reduced in comparison to OH– ions (whose concentration increases). The reaction also shows the production of gaseous hydrogen, chlorine and aqueous sodium hydroxide.”
Source
Electroplating of Metals:
Copper (II) ion leaves the anode and plates out on the cathode. Unreactive substances (such as gold) collect as anode mud under the anode (very valuable).
Stoichiometry of Electrolysis:
Ampere: Denoted A, is the SI base unit for current. The coulomb ( C ) is the SI unit of electric charge and is equivalent to an ampere-second.
Coulombs = Amperes x Seconds
(Electric Charge = Electric Current x Time Lapse)
and
Amperes = Coulombs / Seconds
One faraday (96,485 Coulombs) is equivalent to the charge on one mole of electrons.
Example:
System: Cu2+ (aq) + 2e– → Cu (s)
A constant current passes through this system for 6.50 hours and 6.75g Cu (s) was deposits. What was the current?
Amperes = Coulombs / Seconds
To get the electric current, convert Cu from grams to moles, use stoichiometry to determine moles of e– to moles of Cu, then use moles of e– to convert to Coulombs:
6.75g Cu x (1 mol Cu / 63.5g Cu) x (2 mol e– / 1 mol Cu) x (96485 C / 1 mol e–) = 20512.6 C
To get the time, convert 6.50h to seconds:
6.50h x (3600 s / 1 h) = 23400s
Divide Coulombs by Seconds to get Amperes:
Amperes = 20512.6 C / 23400s = 0.876 Amps
Sunday, November 14, 2010
19.4-19.7: Cell Potential-Dependence of Cell Potential on Concentration
Wmax = ∆Go = –nFEcell
Example:
What is the Wmax that can be obtained from 351g of zinc metal in the galvanic cell with the overall cell reaction of
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
Wmax = ∆Go = –nFEcell
n = moles of e– transferred = 2
F = 96485 C / mol e–
to find the Wmax we need Ecell.
To find Ecell, reference Table 19.1: Standard Electrode (Reduction) Potentials(Eo), page 786.
First, write the two half-reactions from the overall cell reaction:
Zn (s) → Zn2+ (aq) + 2e– (oxidation half-reaction)
Cu2+ (aq) + 2e– → Cu (s) (reduction half-reaction)
This table gives Eo values only for the reduction half-reactions. In order to get the Eo value for an oxidation half-reaction, reverse the order to find it as it is written in the table, and take the negative of the given Eo number given for that reaction.
Zn (s) → Zn2+ (aq) + 2e– is not a reaction that is in Table 19.1 because it is an oxidation reaction. Reverse the reaction to find it as it is written in the table:
Zn2+ (aq) + 2e– → Zn (s)
The given Eo value is – 0.76 V. Since the reaction we are working with is the reverse of this one, take the negative of this value. So:
Zn (s) → Zn2+ (aq) + 2e– (Eo = + 0.76 V)
Cu2+ (aq) + 2e– → Cu (s) (Eo = + 0.34 V)
To get the value for Ecell for the overall reaction, add these two Eo values together.
0.76 V + 0.34 V = 1.10 V = Ecell
Now back to the original question:
Wmax = ∆Go = –nFEcell
Wmax = –(2)(96485)(1.10) = –212 kJ per mole of Zn
Remember that the problem stated that we had 351g of Zn, not one mole of Zn, so:
(351g Zn) x (1 mol Zn / 65.4g Zn) x (–212 kJ / 1 mol Zn) = 1.14 x 103kJ
Standard Cell Potential: Denoted Eocell, it is the cell potential of a voltaic cell operating under standard-state conditions (solute concentrations are each 1 M, gas pressures are each 1 atm, and the temperature has a specified value, usually 25oC.) The degree sign signifies standard-state conditions.
Standard Electrode Potential: Denoted Eo, it is the electrode potential under standard-state conditions. The degree sign signifies standard-state conditions.
Standard electrode potentials help determine strengths of oxidizing and reducing agents.
-The strength of reducing agents increases going up the table.
-Example: Lithium is the strongest reducing agent.
-The strength of oxidizing agents increases going down the table.
-Example: F2 is the strongest oxidizing agent.
Example:
Which is the stronger reducing agent? Li (s) or Li+?
Li, it has more e– to give.
To determine the direction of spontaneity with this information as well:
-The stronger oxidizing agent will be on the reactant side when the equation is written as a spontaneous reaction.
-The stronger reducing agent will be on the reactant side of the spontaneous reaction.
Equilibrium Constants from Cell Potentials:
If Wmax = ∆Go
and
∆Go = –nFEocell
and
∆Go = –RtlnK
then
–nFEocell = –RtlnK
nFEocell = RtlnK
Since we're at standard conditions, we can rearrange the equation and plug in known values to get:
Eocell = (0.0592 / n)logK
Example:
Find K for the overall reaction:
2H3+ (aq) + 3Fe (s) → 2Al (s) + 3 Fe2+ (aq)
Eocell = (0.0592 / n)logK
To find K, need n and Eocell. So start, like always, with writing the half-reactions:
2Al3+ (aq) + 6e– → 2Al (s) (Eo = –1.66 V)
3Fe (s) → 3Fe2+ (aq) + 6e– (Eo = +0.41 V)
Eocell = –1.66 V + 0.41 V = –1.25 V
n = 6
(–1.25 V) = (0.0592 / 6)(logK)
logK = –127
K = 10–127
Nernst Equation:
What if we're at non-standard conditions (assuming temperature remains at 298K)? Luckily, the monopoly guy derived an equation that relates cell potential to standard cell potential.
Walther Hermann Nernst
Ecell = Eocell – (0.0592 / n)(logQ)
where Q is the reaction quotient (remember this is the same as K except not at equilibrium).
Example:
Zn (s) │Zn2+ (0.155M) ║ Ag+ (0.0200M) │ Ag (s)
Find Ecell. (Must use Nernst equation because it is at non-standard conditions.)
Ecell = Eocell – (0.0592 / n)(logQ)
First convert notation into half-reactions and find Eocell:
Zn (s) → Zn2+ (aq) + 2e– (Eo = +0.76 V)
2Ag+ (aq) + 2e– → 2Ag (s) (Eo = +0.80 V)
Eocell = +1.56 V
n = 2
Add the half-reactions to get the overall reaction:
Zn (s) + 2Ag+ (aq) → Zn2+ (aq) + 2Ag (s)
Write Q using the same format used to write K:
Q = [Zn2+] / [Ag+]2
Q = (0.155) / (0.0200)2
Q = 387.5
Ecell = Eocell – (0.0592 / n)(logQ)
Ecell = (1.56) – (0.0592 / 2)(log 387.5)
Ecell = 1.48 V
Determination of pH:
A pH meter actually measures Ecell and then backtracks to pH.
pH = (0.76 – Ecell) / (0.0592)
Example:
If the pH = 4.3, what is the Ecell of the test system?
4.3 = (0.76 – Ecell) / (0.0592)
Ecell = 0.505 V
Example:
What is the Wmax that can be obtained from 351g of zinc metal in the galvanic cell with the overall cell reaction of
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s)
Wmax = ∆Go = –nFEcell
n = moles of e– transferred = 2
F = 96485 C / mol e–
to find the Wmax we need Ecell.
To find Ecell, reference Table 19.1: Standard Electrode (Reduction) Potentials(Eo), page 786.
First, write the two half-reactions from the overall cell reaction:
Zn (s) → Zn2+ (aq) + 2e– (oxidation half-reaction)
Cu2+ (aq) + 2e– → Cu (s) (reduction half-reaction)
This table gives Eo values only for the reduction half-reactions. In order to get the Eo value for an oxidation half-reaction, reverse the order to find it as it is written in the table, and take the negative of the given Eo number given for that reaction.
Zn (s) → Zn2+ (aq) + 2e– is not a reaction that is in Table 19.1 because it is an oxidation reaction. Reverse the reaction to find it as it is written in the table:
Zn2+ (aq) + 2e– → Zn (s)
The given Eo value is – 0.76 V. Since the reaction we are working with is the reverse of this one, take the negative of this value. So:
Zn (s) → Zn2+ (aq) + 2e– (Eo = + 0.76 V)
Cu2+ (aq) + 2e– → Cu (s) (Eo = + 0.34 V)
To get the value for Ecell for the overall reaction, add these two Eo values together.
0.76 V + 0.34 V = 1.10 V = Ecell
Now back to the original question:
Wmax = ∆Go = –nFEcell
Wmax = –(2)(96485)(1.10) = –212 kJ per mole of Zn
Remember that the problem stated that we had 351g of Zn, not one mole of Zn, so:
(351g Zn) x (1 mol Zn / 65.4g Zn) x (–212 kJ / 1 mol Zn) = 1.14 x 103kJ
Standard Cell Potential: Denoted Eocell, it is the cell potential of a voltaic cell operating under standard-state conditions (solute concentrations are each 1 M, gas pressures are each 1 atm, and the temperature has a specified value, usually 25oC.) The degree sign signifies standard-state conditions.
Standard Electrode Potential: Denoted Eo, it is the electrode potential under standard-state conditions. The degree sign signifies standard-state conditions.
Standard electrode potentials help determine strengths of oxidizing and reducing agents.
-The strength of reducing agents increases going up the table.
-Example: Lithium is the strongest reducing agent.
-The strength of oxidizing agents increases going down the table.
-Example: F2 is the strongest oxidizing agent.
Example:
Which is the stronger reducing agent? Li (s) or Li+?
Li, it has more e– to give.
To determine the direction of spontaneity with this information as well:
-The stronger oxidizing agent will be on the reactant side when the equation is written as a spontaneous reaction.
-The stronger reducing agent will be on the reactant side of the spontaneous reaction.
Equilibrium Constants from Cell Potentials:
If Wmax = ∆Go
and
∆Go = –nFEocell
and
∆Go = –RtlnK
then
–nFEocell = –RtlnK
nFEocell = RtlnK
Since we're at standard conditions, we can rearrange the equation and plug in known values to get:
Eocell = (0.0592 / n)logK
Example:
Find K for the overall reaction:
2H3+ (aq) + 3Fe (s) → 2Al (s) + 3 Fe2+ (aq)
Eocell = (0.0592 / n)logK
To find K, need n and Eocell. So start, like always, with writing the half-reactions:
2Al3+ (aq) + 6e– → 2Al (s) (Eo = –1.66 V)
3Fe (s) → 3Fe2+ (aq) + 6e– (Eo = +0.41 V)
Eocell = –1.66 V + 0.41 V = –1.25 V
n = 6
(–1.25 V) = (0.0592 / 6)(logK)
logK = –127
K = 10–127
Nernst Equation:
What if we're at non-standard conditions (assuming temperature remains at 298K)? Luckily, the monopoly guy derived an equation that relates cell potential to standard cell potential.
Walther Hermann Nernst
Ecell = Eocell – (0.0592 / n)(logQ)
where Q is the reaction quotient (remember this is the same as K except not at equilibrium).
Example:
Zn (s) │Zn2+ (0.155M) ║ Ag+ (0.0200M) │ Ag (s)
Find Ecell. (Must use Nernst equation because it is at non-standard conditions.)
Ecell = Eocell – (0.0592 / n)(logQ)
First convert notation into half-reactions and find Eocell:
Zn (s) → Zn2+ (aq) + 2e– (Eo = +0.76 V)
2Ag+ (aq) + 2e– → 2Ag (s) (Eo = +0.80 V)
Eocell = +1.56 V
n = 2
Add the half-reactions to get the overall reaction:
Zn (s) + 2Ag+ (aq) → Zn2+ (aq) + 2Ag (s)
Write Q using the same format used to write K:
Q = [Zn2+] / [Ag+]2
Q = (0.155) / (0.0200)2
Q = 387.5
Ecell = Eocell – (0.0592 / n)(logQ)
Ecell = (1.56) – (0.0592 / 2)(log 387.5)
Ecell = 1.48 V
Determination of pH:
A pH meter actually measures Ecell and then backtracks to pH.
pH = (0.76 – Ecell) / (0.0592)
Example:
If the pH = 4.3, what is the Ecell of the test system?
4.3 = (0.76 – Ecell) / (0.0592)
Ecell = 0.505 V
Subscribe to:
Posts (Atom)