Extra Credit 8

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Q19.2A

A Lithium-ion battery has a theoretical cell of 1.5 V. Find the for the reduction half-reaction:

Consult the activities series under the Standard Reduction Potential page. In the activities series, the equations with a higher Standard Reduction Potential are better reducers (oxidizing agents), while the equations with lower Standard Reduction Potentials are better oxidizers (reducing agents). On the chart, this equation has a Standard Reduction Potential of -3.04V, the lowest in the series.

Q19.35B

Find

in cell if the voltaic cell's

Hint: use Nernst's Equation.

In order to solve this problem, we need to denote the cell diagram as two chemical equations: that of reduction (cathode) and oxidation (anode). Solid zinc loses electrons to become a zinc ion, so this is the area of oxidation and is the anode. The equation is:

Standard Reduction Potential page

Since hydrogen peroxide involves the reduction of hydrogen, the conversion of hydrogen peroxide to water is the cathode. The equation is:

Standard Reduction Potential page

Now that we know the and , we can now determine the overall cell potential through the equation

- = E

Plug the overall cell potential into the Nernst equation

\begin{matrix} (Nernst Equation @ 298 K) & E = E^{o} - \frac{0.0592 V}{n} \log Q \end{matrix}

We are looking for the concentration of hydrogen peroxide, which we will call x. We are given that E=1.5V, and we just found that

Q20.21B

Write the possible half equations to represent each of the following

Cu²⁺ as an oxidizing agent
Mn²⁺ as a reducing agent

a. If Cu²⁺ is to act as an oxidizing agent, it must undergo reduction, or the addition of electrons (since the charge is lowered). Copper will most likely gain two electrons, since it is a +2 ion.

Cu²⁺ + 2e^- -> Cu

b. If Mn²⁺ is to act as a reducing agent, it must undergo oxidation, or the losing of electrons (since the charge is raised). Magnesium will most likely lose two electrons.

Mn²⁺ --> Mn⁴⁺ + 2e^-

Refer to the Half-Reactions portion of the Libretext for more information.

Q21.27A

Which of the following complex ions would you expect to have the largest overall Kf and why? $[Cr {(CO)}_{2}^{} {({NH}_{3}^{})}_{2}^{} (en)]$ $[Cr {(CO)}_{2}^{} {({NH}_{3}^{})}_{2}^{} (en)]$

$[Cr {(CO)}_{2}^{} {({NH}_{3}^{})}_{2}^{} (en)]$

Ethylenediamine attaches onto central metals in two places, and are therefore polydentate ligands. Therefore, [Cr(en)3]2+ has the highest Kf out of these ions because of the higher number of polydentate ligands. $[Cr {(CO)}_{2}^{} {({NH}_{3}^{})}_{2}^{} (en)]$ $[Cr {(CO)}_{2}^{} {({NH}_{3}^{})}_{2}^{} (en)]$

k

Q24.37A

For a certain decomposition reaction, the following observations have been made: at , ; at , ; at , ; and at , . Determine the order and half-life of this reaction.

When you graph the concentration of the reactants versus time, you get a linear line. Therefore, according to the The Rate Law the reaction has an order of zero.

In order to find the half-life of the reaction, we can look at the equation for half-life for zero order reactions, which is

t_1/2=[A]₀/2k

k is the slope of the linear line, therefore we can take two points on the graph and find the slope to find k.

k=(.11-.69)/(264-148)=-.005 M/s

[A]₀ is the initial amount of the reactant, which is 1.43M. When we put these into the equation, we get:

t_1/2=(1.43)/(2*.005) = 143s

$\begin{matrix} (17) & t_{1 / 2} = \frac{[A]_{0}}{2 k} \end{matrix}$

Q25.13D

Write out a nuclear equation that show the formation of an isotope element 97 with mass number 255 by the bombardment of bismuth-219 by nickel-64 nuclei followed by a succession of $α$ a-particle emission, hence find number of $α$ a-particle emission.

Isotopes are denoted as:

^A_ZX

Where A=mass number (the number of protons and neutrons), Z=number of protons, and X=element.

Berkelium (Bk) has an atomic number of 97, denoting the number of protons in the element. Combined with the mass number of 255, we know that Berkelium will be denoted as ²⁵⁵₉₇Bk.

The 219 in bismuth-219 denotes the mass number of the element. The periodic table indicates bismuth has 83 protons, therefore the bismuth isotope will be denoted as ²¹⁹₈₃Bi.

Using this same logic, nickel-64 is denoted as ⁶⁴₂₈Ni

Alpha particles are helium particles, denoted as ⁴₂He. In order to find the number of alpha particles, we must make sure that the A's and Z's in the final chemical equation balance out. So the equation right now looks like:

²¹⁹₈₃Bi + ⁶⁴₂₈Ni ->²⁵⁵₉₇Bk + X⁴₂He

where X is the number of alpha particles. When you add the atomic masses of the isotopes on the reactants side of the equation, we get 283. Berkelium has a mass of 255, so:

283-255=28

Since alpha particles have an atomic mass of 4, we do 28/4=7, so 7 alpha particles are necessary for this reaction.Therefore, the final reaction is:

²¹⁹₈₃Bi + ⁶⁴₂₈Ni ->²⁵⁵₉₇Bk + 7⁴₂He

Q19.2

The isotope ¹³⁷₅₅Cs ${}_{55}^{137}{Cs}$ undergoes beta emission with a half-life of 30 years.

Write a balanced nuclear equation for this reaction.
What fraction of $Cs-137$

$[Cr {(CO)}_{2}^{} {({NH}_{3}^{})}_{2}^{} (en)]$

c. What mass of Cs will be left in a 24g sample of ¹³⁷₅₅Cs after 90 years?

d. What fraction of Cs-137 has decayed after 120 years?

a. Beta emission occurs when an isotope loses electrons. The beta particle is an electron, denoted as ⁰_-1e. Since the isotope on the products side gains a proton, the proton number becomes 56, which corresponds to the element barium (Ba). The mass number between the two isotopes must stay the same. Therefore the balanced nuclear equation is:

¹³⁷₅₅Cs -> ⁰_-1e + ¹³⁷₅₆Ba

b. Since we know that the half-life of the isotope is 30 years, we can figure out how many half-lives are in 60 years by dividing it by the half-life. Let's call this value n. Therefore:

60/30 = 2 = n

After every half-life, half of the original material will be left (50%). So, to find the percent mass remaining, we do:

(1/2)ⁿ=(1/2)²=1/4=25%

c. We use the same process in part (b) to first find the percent of the original amount that will remain after this reaction.

90/30=3=n

(1/2)ⁿ=(1/2)³=1/8

Then, you multiply this amount by the original to see how much will remain.

(1/8)*(24.0) = 3.00g remaining

d. Using the same process as part (b):

120/30 = 4

(1/2)⁴=1/16 = 6.25%

Q21.3.3

Propose an explanation for the observation that elements with even atomic numbers are more abundant than elements with odd atomic numbers.

Many of the elements we know of today formed from nuclear fusion reactions within stars, which use helium (alpha particles), because it is the element with the smallest amount of even protons (Z=2). Fusion with helium produces even numbered atomic numbers. This is why elements with even atomic numbers are more abundant than those with odd atomic numbers.

$3 {ArSO}_{2}^{} H \to Ar {SO}_{2}^{} SAr + {ArSO}_{3}^{} H + H_{2}^{} O$ $[Cr {(CO)}_{2}^{} {({NH}_{3}^{})}_{2}^{} (en)]$ $[Cr {(CO)}_{2}^{} {({NH}_{3}^{})}_{2}^{} (en)]$

$[Cr {(CO)}_{2}^{} {({NH}_{3}^{})}_{2}^{} (en)]$

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Q20.21B

Q21.27A

Q24.37A

Q25.13D

Q19.2

Q21.3.3