Extra Credit 23
- Page ID
- 82993
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For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.
- Mn(s)+Ni2+(aq)⟶Mn2+(aq)+Ni(s)Mn(s)+Ni2+(aq)⟶Mn2+(aq)+Ni(s)
- 3Cu2+(aq)+2Al(s)⟶2Al3+(aq)+2Cu(s)3Cu2+(aq)+2Al(s)⟶2Al3+(aq)+2Cu(s)
- Na(s)+LiNO3(aq)⟶NaNO3(aq)+Li(s)Na(s)+LiNO3(aq)⟶NaNO3(aq)+Li(s)
- Ca(NO3)2(aq)+Ba(s)⟶Ba(NO3)2(aq)+Ca(s)Ca(NO3)2(aq)+Ba(s)⟶Ba(NO3)2(aq)+Ca(s)
Q17.3.2
For each reaction listed, determine its standard cell potential at 25 °C and whether the reaction is spontaneous at standard conditions.
Step 1: Write the half reactions for each process
Step 2: Look up the standard potentials for the reduction half-reaction. Determine which half reaction is being oxidized and which is being reduced. (oxidation is loss of electrons and reduction is gain of electrons)
Step 3: Add the cell potentials together to get the overall standard cell potential.
Eocell = Eoreduction - Eooxidation
E cell = E cathode - E anode
Step 4: The reaction is spontaneous if the standard cell potential is positive.
1. Mn(s)+Ni2+(aq)⟶Mn2+(aq)+Ni(s)Mn(s)+Ni2+(aq)⟶Mn2+(aq)+Ni(s)
Step 1: First equation= Mn(s)--> Mn2+ (aq) +2e-
Second equation= Ni2+(aq) + 2e- --> Ni(s)
Step 2: Standard reduction potential for first half reaction= -1.18 (oxidation)
Standard reduction potential for second half reaction=-0.25 (reduction)
Step 3: Eocell=(-0.25) - (-1.18) = 0.93
Step 4: Reaction is spontaneous in standard conditions because when you apply the formula G=-nFE Gibbs energy is negative
2. 3Cu2+(aq)+2Al(s)⟶2Al3+(aq)+2Cu(s)3Cu2+(aq)+2Al(s)⟶2Al3+(aq)+2Cu(s)
Step 1: First equation= Cu2+(aq) + 2e--->Cu(s)
Second equation= Al(s)-->Al3+(aq) +3e-
Step 2: Standard reduction potential for first half reaction= 0.34 (reduction/cathode)
Standard reduction potential for second half reaction= -1.66 (oxidation/anode)
Step 3: Eocell=(0.34) - (-1.66) = 2
Step 4: Reaction is spontaneous in standard conditions because when you apply the formula G=-nFE Gibbs energy is negative
3. Na(s)+LiNO3(aq)⟶NaNO3(aq)+Li(s)Na(s)+LiNO3(aq)⟶NaNO3(aq)+Li(s)
Step 1: First equation= Na(s) --> Na+(aq) +e-
Second equation= Li+(aq) +e- --> Li(s)
Step 2: Standard reduction potential for first half reaction= -2.71 (oxidation)
Standard reduction potential for second half reaction=-3.05 (reduction)
Step 3: Eocell=(-3.05) - (-2.71) = -0.34
Step 4: Reaction is not spontaneous in standard conditions because when you apply the formula G=-nFE Gibbs energy is positive
4. Ca(NO3)2(aq)+Ba(s)⟶Ba(NO3)2(aq)+Ca(s)Ca(NO3)2(aq)+Ba(s)⟶Ba(NO3)2(aq)+Ca(s)
Step 1: First equation= Ca2+ (aq) +2e- --> Ca(s)
Second equation= Ba(s) --> Ba2+ (aq) +2e-
Step 2: Standard reduction potential for first half reaction= -2.87 (reduction)
Standard reduction potential for second half reaction= -2.9 (oxidation)
Step 3: Eocell=(-2.87) - (-2.9) = 0.03
Step 4: Reaction is spontaneous in standard conditions.
Q19.1.21
Predict the products of each of the following reactions and then balance the chemical equations.
- Fe is heated in an atmosphere of steam.
- NaOH is added to a solution of Fe(NO3)3.
- FeSO4 is added to an acidic solution of KMnO4.
- Fe is added to a dilute solution of H2SO4.
- A solution of Fe(NO3)2 and HNO3 is allowed to stand in air.
- FeCO3 is added to a solution of HClO4.
- Fe is heated in air.
Q19.1.21
Predict the products of each of the following reactions and then balance the chemical equations.
Transition metals are elements that belong to the D block of the periodic table. They are also known to have multiple oxidation states.
Step 1: Put the elements in the statement as a reactant.
Step 2: Predict the product.
Step 3: Balance the equation by making sure the elements are equal on both sides of the equation.
1.Fe is heated in an atmosphere of steam.
3Fe(s) + 4H2O(g) --> Fe3O4(s) +4H2(g)
2. NaOH is added to a solution of Fe(NO3)3.
Fe(NO3)3 +3NaOH --> FeOH3 +NaNO3
3. FeSO4 is added to an acidic solution of KMnO4.
10FeSO4 +2KMnO4 + 8H2SO4 -->5Fe2(SO4) +2MnSO4 +K2SO4 +8H2O3
4.Fe is added to a dilute solution of H2SO4.
Fe + dil. H2SO4 -->FeSO4 +H2 + H2O
5. A solution of Fe(NO3)2 and HNO3 is allowed to stand in air.
4Fe(NO3)2 + 4HNO3 +O2 --> 4Fe3+ + 4NO3- +H2O
6. FeCO3 is added to a solution of HClO4.
FeCO3 +HClO4 --> Fe(ClO4)2 +H2O +CO2
7. Fe is heated in air.
3Fe +4O2 -->Fe3O4
Q19.3.13
[CuCl4]2− is green. [Cu(H2O)6]2+is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting?
Q19.3.13
[CuCl4]2− is green. [Cu(H2O)6]2+is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting?
If [CuCl4]2- is green, this means that it absorbs the color opposite of green on the color wheel. After referring to the color wheel we see that the complex absorbs red.
If [Cu(H2O)6]2+ is blue, this means that it absorbs the color opposite of blue on the color wheel. After referring to the color wheel we see that the complex absorbs orange.
A complex that absorbs red absorbs wavelength 680, and a complex that absorbs orange absorbs wavelength 610.
Wavelength and energy have an inverse relationship so the higher the wavelength, the lower the energy.
Red has a higher wavelength so it has a lower energy. Therefore, [Cu(H2O)6]2+ absorbs higher energy photons.
[CuCl4]2- has a larger crystal field splitting because their photons have less energy. It requires more energy to pair electrons in a ligand field.
Q12.4.13
Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?
Q12.4.13
Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?
The number of half lives in a species of interest can be calculated as follows: n=t/t1/2
The amount of radioactivity that would remain after n half lives can be calculated as follows: (0.5)n[A]0%
- The half life of technetium-99 is 6 hours, so the number of half lives in 48 hours is: 48/6=8
- Considering that the initial concentration of technetium-99 is 100%, the amount remaining after 48 hours is: (0.5)8[100]%= 0.391%
- The half life of thallium-201 is 73 hours, so the number of half lives in 48 hours is: 48/73=0.6575
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- Considering that the initial concentration of thallium-201 is 100%, the amount remaining after 48 hours is: (0.5)0.6575[100]%= 63.39%
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Q21.2.8
The mass of the atom 2311Na is 22.9898 amu.
1. Calculate its binding energy per atom in millions of electron volts.
2. Calculate its binding energy per nucleon.
Q21.2.8
The mass of the atom 2311Na is 22.9898 amu.
1. Calculate its binding energy per atom in millions of electron volts.
proton mass ~ 1.00728 amu
neutron mass ~ 1.00866 amu
electron mass ~ electron mass = 0.000549 amu
To calculate the binding energy per atom we first take the mass of the constituents and subtract it from the the mass of the atom for Na (this will give us the mass defect). Na has 11 neutrons and 12 protons.
22.9898-(12*1.00728)-(11*1.00866)= 22.9898-12.08736- 11.09526=-0.19282 is the mass defect.
Now we will use the equation E=mc2 (m= .19282, c=300000000)
Because the question wants the answer in MeV we will need to the convert the units. mass of 1 u is equivalent to an energy of 931.5 MeV.
=0.19282 * 931.5= 179.61183 MeV
2. Calculate its binding energy per nucleon.
Binding energy is the total energy (calculated above) divided by the number of nucleons: 179.61183/23=7.80921 MeV/nucleon
Q21.6.3
Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine-131 is injected into the body. In larger doses, I-133 is also used as a means of treating cancer of the thyroid. I-131 has a half-life of 8.70 days and decays by β− emission.
1. Write an equation for the decay.
2. How long will it take for 95.0% of a dose of I-131 to decay?
Q21.6.3
Iodine that enters the body is stored in the thyroid gland from which it is released to control growth and metabolism. The thyroid can be imaged if iodine-131 is injected into the body. In larger doses, I-133 is also used as a means of treating cancer of the thyroid. I-131 has a half-life of 8.70 days and decays by β− emission.
1. Write an equation for the decay.
A general equation for β− decay is 10n --> 11p + 0-1β
(the top numbers (mass) have to equal to each other on both sides of the the equation and the bottom numbers (atomic number) have to equal each other as well)
The equation for the decay of I-131 is: 13153I --> 13154Xe + 0-1β
2. How long will it take for 95.0% of a dose of I-131 to decay?
Consider N to be the initial amount of the sample. Therefore, after the decay of 95% the sample will be 0.05 N.
The decay constant for I- 131 is as follows: wavelength= ln2/t1/2=ln2/8.7=0.07967 per day
The time required for the decay is calculated as follows: 1/wavelength(ln(no/nt))= (1/.07967)ln(N/0.05N)= 37.6 days
Q20.4.9
All reference electrodes must conform to certain requirements. List the requirements and explain their significance.
Q20.4.9
All reference electrodes must conform to certain requirements. List the requirements and explain their significance.
A reference electrode is an electrode that is known to have a stable and well known electrode potential. The standard hydrogen electrode is usually what is used as the reference electrode. The reference electrode helps determine the value for Eo in electrochemical cells. It is important for a reference electrode to be stable and have a well known electrode potential because if the value of the potential changes, then this could skew the value of the Eo we determine for the electrochemical cell we are measuring. The observed changes in the reference electrode's measured potential should be due solely to changes of the analyte concentration.
Q20.8.1
Do you expect a bent nail to corrode more or less rapidly than a straight nail? Why?
Q20.8.1
Do you expect a bent nail to corrode more or less rapidly than a straight nail? Why?
When a nail is bent, atoms dislocate and layers slip. This causes there to be stress at the bending point of the nail, allowing for the exposed, ungalvanized metals to be easily accessible to the oxygen in the environment.Therefore, at points of stress in a metal, oxidation occurs more rapidly. This is because the point of stress becomes an anode for the reaction of iron with oxygen, which ultimately results in the corrosion of the nail.