Extra Credit 25

Q17.3.4

Determine the overall reaction and its standard cell potential at 25 °C for the reaction involving the galvanic cell made from a half-cell consisting of a silver electrode in 1 M silver nitrate solution and a half-cell consisting of a zinc electrode in 1 M zinc nitrate. Is the reaction spontaneous at standard conditions?

A: The fact that the reaction occurs at 25 °C and both of the electrodes have a value of 1M, this reaction is occurring under standard conditions.

Step 1: split the reaction into the silver nitrate solution and zinc nitrate. Due to the fact that Nitrate is in both metals, it is a spectator ion and due to this will not appear in the overall reaction.

(In Ag half reaction I just put e^- because she put e-)

\[Ag^{+} + e^{-}\rightarrow Ag(s)\] \[Zn^{2+} (aq) + 2e^{-}\rightarrow Zn(s)\]

Step 2: to determine if the reaction is spontaneous you must determine whether or not Ecell is negative or positive. A negative Ecell signifies a nonspontaneous reaction whereas a positive Ecell means that the reaction is spontaneous.

Determine the standard reduction potentials of each reaction by referring to an SRP table.

The value for Ag⁺(aq) + e- →Ag(s) is 0.7996V and Zn²⁺(aq) + 2e^-→Zn(s) has a value of -0.7618 V

Step 3: The equation for Ecell is Ecell=cathode-anode which could also be written as Ecell=reduction-oxidation

In this case The electrons for the reactions are both on the reactant side of the equation. A trick to discern which one is the anode and the cathode is the anode is always the smaller value, in this case that is Zn²⁺(aq) + 2e^-→Zn(s) -0.7618 V

Step 4: Substitute in the values and solve

Ecell= 0.7996-(-0.7618)

Ecell= 1.5614 V Since it is positive the reaction is spontaneous.

Agree with solution

Q19.1.23

Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility.

A: Step 1: look at the question and begin to write out a general product to reactant formula for this reaction.

Step 2: try to reason out why a precipitate will form but only for a finite period of time before reforming in an aqueous substance.

Step 3: With step 2 you should have noticed that the reaction is a multiple step reaction and using the rough formula that you derived in step 1, you should try and see what the series of steps are that lead to the overall product of liquid AgCN₂

In this reaction we see how NaCN is added to AgNO₃ .A precipitate forms but then disappears with the addition of even more NaCN, this must mean that its an intermediate reaction which will not appear as the final product. The silver and the cyanide temporarily bond, but the bond is too weak to hold them together so they are pulled apart again when NaCN is added because a new, more stronger and stable compound is formed: [Ag(CN)₂] (aq).

The actual reaction equation when it is first taking place is \[AgCl(aq)+NaCN(aq)\rightarrow AgCN(s)+NaCl(aq)\] She wrote a simplified form

This can be written out in the following way: as CN⁻ is added, the silver and the cyanide combine : Ag⁺(aq)+CN⁻(aq)→AgCN (s)

As more CN^- is added the silver and two cyanide combine to create a more stable compound: Ag⁺(aq)+2CN⁻(aq)→[Ag(CN)₂]^- (aq)

AgCN(s) + CN^- (aq) → [Ag(CN)₂] (aq)

Q12.4.16

1. What is the rate constant for the decomposition of fluorine-18?
2. If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h?
3. How long does it take for 99.99% of the ¹⁸F to decay?

A: a. The formula for the rate constant decomposition is characteristic of a nuclear reaction and is solved by using the first order equation.

k=0.693/t_1/2

_Edit: \[k=\frac{0.693}{t\frac{1}{2}}\]

In this case t_1/2 is 109.7 minutes. Keep the value of minutes in mind when solving the formula

Here all of the information is given and we just compute.

k=0.693/109.7 min

edit: \[k=\frac{0.693}{109.7 min}\]

solving this yields k=0.00632 min^-1

b. The formula for solving this equation is [A]=[A₀]e^-kt By taking the Ln of both sides this equation can be rewritten into a more useful form for our purposes

ln(A)=ln(A₀)e^-kt dividing out the ln(A₀) we get our final equation: e^-kt =([A]/[A₀]) this formula is especially useful because the ([A]/[A₀]) is the percent of reactivity and is the percent that remains. It will tell us how much will remain after 5.59 hours.

Now it is just a matter of converting into the appropriate units and substituting into the equation. Looking back at our rate constant we see that it in terms of min^-1 so first convert 5.59 hours into minutes by multiplying by 60. This yields a value of 335.4 minutes

Next substitute the values into the equation and you will get e-^{0.00632/min*335.4min} =([A]/[A₀])

Solving we get ([A]/[A₀])= 0.12 or 12%

c. In order to determine how long it will take 99.99% of F to decay recognize that you can use the previously formulated formula e^-kt =([A]/[A₀]) to solve for the value of t.

Before you can do that you much subtract 1-0.9999= 0.0001 this is the percent that remains.

Now you simply substitute the known values into the formula and solve

e^-kt =([A]/[A₀]) turns into e^-0.00632*t=0.0001

This yields an answer of 1457 minutes

Q 21.2.10

A: To determine the stability of an isotope you can use the ratio of neutrons to protons N:Z with N being the number of Neutrons and Z symbolizing the atomic number. Therefore to determine the N:Z Ratio you must look at the Neutron to Proton ratio. The golden rule is that even number of protons and neutrons means that the isotope is stable whereas an odd number of protons and neutrons signifies an unstable isotope. Furthermore, If the previously stated ratio of neutrons to protons is 1:1, then it is stable.

Follow these series of steps:

Step 1: identify the elements atomic number by using the periodic table.

Step 2: find the number of neutrons by subtracting the atomic mass number from the number of protons (the atomic number.) Remember, the atomic number is always the number of protons and elements are defined by their number of protons.

Step 3: Look at the description above and discern whether the number of neutrons are positive, negative, or have a 1:1 ratio. From this information you should be able to tell whether or not the isotope is stable.

a. argon-40 has an atomic number of 18 and the number of protons can be calculated by subtracting the number of nucleons from the atomic number: 40-18=22 neutrons. In this case both the protons and neutrons are positive, this means that the isotope is stable.

1. oxygen-16. Oxygen has the atomic number of 8. Subtract mass from atomic number: 16-8=8 neutrons. Here the proton to neutron ratio is 8:8 or 1:1. It is stable.
2. ¹²²Ba. The atomic number is 56. 122-56= 66. Therefore there are 66 neutrons, 56 protons, therefore it is stable because they are even.
3. ⁵⁸Ni. The atomic number is 28. 58-28= 30. Once again the number of protons and neutrons is even so the the isotope is stable.
4. ²⁰⁵Tl. Atomic number is 81. 205-81 protons= 124 neutrons. In this case the number of neutrons is even and the number of protons is odd. Taking the ratio we see that the isotope has a ratio of 81:124. This is not 1:1 so the isotope is unstable.
5. ²¹⁰Tl atomic number is 81. 210-81= 129. Here the protons and neutrons have an odd amount so the isotope is unstable.
6. ²²⁶Ra. The atomic number is 88. 226-88 protons= 138 neutrons. Both protons and neutrons are even so it is stable.
7. magnesium-24. Atomic number is 12. 24-12= 12. Here the ratio of protons to neutrons is 12:12 or 1:1 so the isotope is stable.

No edits, nice work

Q21.7.2

Based on what is known about Radon-222’s primary decay method, why is inhalation so dangerous?

An unstable radioactive nucleus may emit particles like alpha, beta, and gamma to get to a stable configuration.

A, Alpha particles can be stopped by very thin shielding but have much stronger ionizing potential than beta particles, X-rays, and γ-rays. When inhaled, there is no protective skin covering the cells of the lungs, making it possible to damage the DNA in those cells and cause cancer.

Q20.4.12

Draw the cell diagram for a galvanic cell with an SHE and a zinc electrode that carries out this overall reaction: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g).

A: A cell diagram is a convenient method of writing out the reaction in its separate anode and cathode forms. Anode refers to oxidation (the loss of electrons) and cathode refers to reduction(the gain of electrons).

Step 1: Distinguish between the anode and the cathode of this reaction.

Here we see that Zn(s) goes from a charge of 0 to a charge of +2, this means that it has been oxidized.

In the same sense we see H(aq) go from the charge of + to 0 which means that it has been reduced.

Step 2: The general outline of a cell diagram is the anode on the left, a double bar of separation in the middle, cathode on the right, electrodes on the ends, and within the anode and cathode reactions there may be single bars if a change in state occurs.

Here the anode is Zn(s) → Zn²⁺(aq) Notice how there is a change in the state-from (s) to (aq) and the cathode is 2H⁺(aq) → H₂(g) Once again, a change in state.

We can rewrite this into the cell diagram form: Zn(s) l Zn²⁺(aq) ll H⁺(aq) l H₂(g)

Here is the drawn cell, KCl is just used for the salt bride as positive and negative ions to maintain the charge balance

Q 20.4.14

a. Cu⁺(aq) + Ag⁺(aq) → Cu²⁺(aq) + Ag(s)

Step 1: separate the reaction.

Cu⁺(aq)→ Cu²⁺(aq) and Ag⁺(aq) →Ag(s)

Step 2: Check to make sure that the chemical elements are balanced. In this case Cu and Ag both have a 1:1 ratio on the reactant and product sides so it is balanced.

Step 3: Balance the charges by adding electrons. The reactant side for Cu has a +1 charge whereas the product side has a +2 charge so one e- must be added. The reactant side for Ag has a +1 charge and the product side has no change so one e- must be added to the reactant side.

Cu⁺(aq)→ Cu²⁺(aq) +e- and e- + Ag⁺(aq) →Ag(s)

Step 4: add the two equation together, Since the same value of e- exist both both equations they should cancel each other out. In this case the equation was already balanced.

Cu⁺(aq) + Ag⁺(aq) → Cu²⁺(aq) + Ag(s)

Step 5: to find the Standard reduction potential identify the oxidation and reduction reactions. Oxidation is a loss of electrons and reduction is a gain of electrons.

Cu⁺(aq)→ Cu²⁺(aq) Here the charge of Cu goes from +1 to +2. Therefore it is oxidation

Ag⁺(aq) →Ag(s) here the charge of Ag goes from +1 to 0. Therefore it is reduction.

Step 6: Go to the table of standard reduction potentials and find the values for Cu⁺(aq)→ Cu²⁺(aq) +e- and e- + Ag⁺(aq) →Ag(s)

Step 7: these values are 0.7996 for Ag and 0.159 for Cu

The formula for SRP is Ecell=cathode-anode

Official equation is \[E^o_{Cell}= E^o_{Red,Cathode} - E^o_{Red,Anode}\]

this means Ecell=recuction-oxidation. In this case Ecell= 0.7996-0.159

The answer for Ecell=0.6406 V

b. Sn(s) + Fe³⁺(aq) → Sn²⁺(aq) + Fe²⁺(aq)

Step 1: separate the reaction.

Sn(s)→ Sn²⁺(aq) and Fe³⁺(aq) →Fe²⁺(aq)

Step 2: Check to make sure that the chemical elements are balanced. In this case Sn and Fe both have a 1:1 ratio on the reactant and product sides so it is balanced.

Step 3: Balance the charges by adding electrons. The reactant side for Sn has a 0 charge whereas the product side has a +2 charge so 2e- must be added. The reactant side for Fe has a +3 charge and the product side has a +2 change so one e- must be added to the reactant side.

Sn(s)→ Sn²⁺(aq)+2e- and e- + Fe³⁺(aq) →Fe²⁺(aq)

Step 4: Since the number of e- for the Sn reaction and Fe reaction are different you must multiply them by a common coefficient to get the same number of electrons for both reactions. In this case multiply the Fe reaction by 2.

Sn(s)→ Sn²⁺(aq)+2e- and 2e- + 2Fe³⁺(aq) →2Fe²⁺(aq)

Add the two equation together, Since the same value of e- exist in both equations they should cancel each other out.

Sn(s) + 2Fe³⁺(aq) → Sn²⁺(aq) + 2Fe²⁺(aq)

Step 5: to find the Standard reduction potential identify the oxidation and reduction reactions. Oxidation is a loss of electrons and reduction is a gain of electrons.

Sn(s)→ Sn²⁺(aq) Here the charge of Sn goes from 0 to +2. Therefore it is oxidation

2Fe³⁺(aq) →2Fe²⁺(aq) here the charge of Fe goes from +6 to +4 (You must multiply the charge by the coefficients) . Therefore it is reduction.

Step 6: Go to the table of standard reduction potentials and find the values for Sn(s)→ Sn²⁺(aq)+2e- and 2e- + 2Fe³⁺(aq) →2Fe²⁺(aq)

Step 7: these values are -0.14 for Sn and 0.771 for Fe

The formula for SRP is Ecell=cathode-anode this means Ecell=recuction-oxidation. In this case Ecell= 0.771-(-0.14)

The answer for Ecell=0.0.911 V

c. Mg(s) + Br₂(l) → 2Br⁻(aq) + Mg²⁺(aq)

Step 1: separate the reaction.

Mg(s)→ Mg²⁺(aq) and Br₂(l) → 2Br⁻(aq)

Step 2: Check to make sure that the chemical elements are balanced. In this case Br has a 2:2 ratio and Mg has a 1:1 ratio on the reactant and product sides so they are balanced.

Step 3: Balance the charges by adding electrons. The reactant side for Mg has a 0 charge whereas the product side has a +2 charge so 2 e- must be added. The reactant side for Br has a 0 charge and the product side has -2 change so 2 e- must be added to the reactant side.

Mg(s)→ Mg²⁺(aq)+2e- and 2e-+ Br₂(l) → 2Br⁻(aq)

Step 4: add the two equation together, Since the same value of e- exist both both equations they should cancel each other out. In this case the equation was already balanced.

Mg(s) + Br₂(l) → 2Br⁻(aq) + Mg²⁺(aq)

Step 5: to find the Standard reduction potential identify the oxidation and reduction reactions. Oxidation is a loss of electrons and reduction is a gain of electrons.

Mg(s)→ Mg²⁺(aq) Here the charge of Mg goes from 0 to +2. Therefore it is oxidation

Br₂(l) → 2Br⁻(aq) here the charge of Br goes from 0 to -2. Therefore it is reduction.

Step 6: Go to the table of standard reduction potentials and find the values for Mg(s)→ Mg²⁺(aq)+2e- and 2e-+ Br₂(l) → 2Br⁻(aq)

Step 7: these values are -2.356 for Mg and 1.087 for Br

The formula for SRP is Ecell=cathode-anode this means Ecell=recuction-oxidation. In this case Ecell= 1.087-(-2.356)

The answer for Ecell=3.443 V

Q14.1.3

What is the relationship between each of the following factors and the reaction rate: reactant concentration, temperature of the reaction, physical properties of the reactants, physical and chemical properties of the solvent, and the presence of a catalyst?

A: Reaction rates generally increase with increasing reactant concentration, increasing temperature, and the addition of a catalyst. Physical properties such as high solubility also increase reaction rates. Solvent polarity can either increase or decrease the reaction rate of a reaction, but increasing solvent viscosity generally decreases reaction rates.

Correct! I'm just giving reasons

Reactions cannot occur if their particles don't "collide" with one another. An increase is reaction concentration would increase the chance for a collision to occur while increasing the amount of collisions happening simultaneously.

Increase in temperature increases the average kinetic energy of its particles, causing particles to move faster and creating more collisions.

The addition of a catalyst lowers the activation energy of a reaction so it can get to the activation energy level quicker

A highly viscous solvent causes particles to diffuse more slowly causing a slower reaction rate, while a less viscous solvent has fast moving particles so it can react quicker