Extra Credit 22

Q. 19.7B : Determine the value of the ΔG∘ for the following reactions carried out in voltaic cells. [Use the equations of 22]

1. Al(s) + Cu²⁺→ Cu(s) + Al³⁺
2. Fe³⁺ + Ag(s) → Ag⁺+ Fe²⁺
3. Cu(s) + 2Fe³⁺→ Cu²⁺+ 2Fe²⁺(aq)

S. 19.7B : To solve the problem, you have to first write the balanced equation.

1.

Oxidation: [ Al(s) → Al³⁺ + 3e^- ] x 2 E= −1.676 V (anode)

Reduction: [ Cu²⁺ + 2e^- → Cu(s) ] x 3 E = 0.340 V (cathode)

E∘_cell = 0.340 - (-1.676) = 2.106 V

ΔG∘=−nFE∘_cell

= −6mol∗96485C/mol∗2.106V
= −1.2 x 10⁶ J

When you split the reaction, you will then determine which is the cathode and which is the anode. Look up the values given in the table of standard reduction potential values and do: Cathode - Anode, which will give you the standard cell potential in volts. Once you have this value, you can plug it into the equation given above. When you balance the half reactions, you will do so by balancing the amount of electrons present. If you notice, there are 3e^- in the anode and 2e^- in the cathode. To balance these, multiple to get 6e- on both sides, making your n=6. F is Faradays' constant which is 96485 C/mol.

2.

Reduction: Fe³⁺ + e⁻ → Fe²⁺ E = 0.771 V (cathode)

Oxidation: Ag(s) → Ag⁺(aq) + e⁻ E = 0.800 V (anode)

E∘cell = 0.800 V - 0.771 V = .029 V

ΔG∘=−nFE∘_cell

=−1mol∗96485 (KJ/mol*V) ∗ 0.029V
=−2605.095V

When you split the reaction, you will then determine which is the cathode and which is the anode. Look up the values given in the standard reduction potential values and do: Cathode – Anode, which will give you the standard cell potential in volts. Once you have this value, you can plug it into the equation given above. When you balance the half reactions, you will do so by balancing the amount of electrons present. If you notice, there is 1e^- in the anode and 1e^- in the cathode. Since they are already balanced, you will get 1e^- on both sides, making n=1. F is Faraday’s constant which is 96485 C/mol.

3.

Oxidation: Cu(s) → Cu²⁺ + 2e⁻ E=0.340 V

Reduction: [ Fe³⁺ + e⁻ → Fe²⁺] x 2 E=0.771 V

E∘_cell = 0.771 V - 0.340 V = 0.029 V

ΔG∘=−nFE∘_cell

=−2mol∗96485 (KJ/mol*V) ∗ 0.029V
=−83170.07J

When you split the reaction, you will then determine which is the cathode and which is the anode. Look up the values given in the standard reduction potential values and do: Cathode – Anode, which will give you the standard cell potential in volts. Once you have this value, you can plug it into the equation given above. When you balance the half reactions, you will do so by balancing the amount of electrons present. If you notice, there are 2e- in the anode and 1e- in the cathode. To balance, you will need to multiply the cathode reaction by 2, making n=2. F is Faraday’s constant which is 96485 C/mol.

Q. 19.63A : Calculate the amount in grams of metal that is deposited at the cathode by running a current of 3.15 A for 78 min in an electrolysis reaction for an aqueous solution containing a) Zn²⁺ b) Sn²⁺ c) Fe³⁺ d) Ni²⁺

S. 19.63A : You are given the current (I) which is 3.15 A for 78 min which is time. For each part, the number of (mol e^-) is equal to the charge of the ion. Use the equation n=IT/F. To convert from mol to grams use the molar mass of each element. You want your units to be in seconds so you convert minutes to seconds which will be the same for every equation, 4680 seconds.

# of mols of electrons=(3.15A)(4680sec) / (96485C) = 0.153 mol e⁻ (since 78min×60sec/1min)

1. 0.153 mol electrons × (1 mol Zn / 2 mol e⁻ ) = 0.0765 mol Zn × (65.41 g / 1mol ) = 5.00g Zn
2. 0.153 mol electrons × (1 mol Sn / 2 mol e⁻ ) = 0.0765 mol Sn × (118.7 g / 1mol ) = 9.07g Sn
3. 0.153 mol electrons × 1 mol Fe / 3 mol e⁻ ) = 0.051 mol Fe × (2.85 g / 1mol ) = 8.55g Fe

4. 0.153 mol electrons × (1 mol Ni / 2 mol e⁻ ) = 0.0765 mol Ni × (58.69g / 1mol ) = 4.49g Ni

Q. 21.4 B

1. How many unpaired electrons would you expect to find in the octahedral complex [CoF₆]³⁻

S.21.4B In order to determine how many total electrons the octahedral complex has you must determine the charge of the transition metal, in this case it is Co. If you notice, there are six Fluorine ligands. The charge of the Fluorine ligand is -1 making the total charge that the Fluorine ligands contribute, -6. Also, if you notice, the charge of the overall complex is -3. So if you have a -6 charge on the Fluorine ligand and want it to equal the overall charge of the complex, you have to do some math to determine the charge of the transition metal. In this case:

x + (-6) = -3

To solve for x, add six to the other side giving you a +3. You have found the charge of the transition metal.

Using this information, use it to calculate the total number of d electrons the transition metal has. Once you see this, you will notice that because Co has a +3 charge, it will cause both the electrons in the s orbital and one electron in its d orbital to go away leaving you with six total d electrons. Since the complex features Fluorine, a weak field ligand, it will yield a high spin complex. A high spin complex follows Hund's rule meaning that the orbitals can't be paired up until every orbital is occupied. So move back to the bottom orbital and fill the first orbital with a second arrow. If you notice, there are four unpaired electrons.

2. How many unpaired electrons would you expect to find in the tetrahedral complex [FeBr₄]²⁻? Would you expect more, fewer, or the same number of unpaired electrons as in the octahedral complex [Co(NH₃)₆]³⁺

S.21.4B : Tetrahedral complexes have a tendency for the ligands to interact with the dxz, dxy, and dyz orbitals, producing an electron-electron repulsion which yields a higher energy. In this complex, the ∆t is smaller which means it rarely surpasses the value of the pairing energy. As a result, the electrons will move up higher energy orbitals causing them to be high spin complexes which have more unpaired electrons.

S24.11D

1. m and n are the respective orders according to A and B:
   R1=1.261×10⁻⁴ = [0.241]^m[0.153]ⁿ
   R2=5.044×10⁻⁴=[0.241]^m[0.306]ⁿ
   R3=2.522×10⁻⁴=[0.482]^m[0.153]ⁿ
   R4=6.741×10⁻⁴=[0.482]^m[0.306]ⁿ

-Now utilize the rate laws / solve:

-Divide reaction 2 by reaction 1 (A is constant B changes concentration)

R2/R1=(5.044×10⁻⁴) / (1.261×10⁻⁴⁾ = [0.241]^m[0.306]ⁿ / ( [0.241]^m[0.151]ⁿ )
R2/R1= 4=2ⁿ (so n=2)

a) Reaction is second order with respect to B because the value of the exponent is equal to 2. Do the same for A. Use reactions 3 and 1 ([B] is constant, [A] changes)

R3/R1=2.522×10⁻⁴/ (1.26×10⁻⁴) = [0.482]^m[0.153]ⁿ / ( [0.241]^m[0.153]ⁿ )
R3/R1=2=2m (so m=1)

a) Reaction is first order with respect to A because the exponent value is equal to 1.

2. Overall reaction order = (reaction order of A + reaction order of B) = 1 + 2 = 3rd order overall

Since the reaction order of A is 1, and the reaction order of B is 2, all you would need to do is to add the two which look like exponents to get a +3. This means that the overall reaction order is 3 making it a third order reaction.

3. Substitute in for one of the reactions
Rate law=k[A][B]²

For example use reaction 1 to get your rate. Remember your A is raised to the first power and B is raised to the second power. To find k, you would divide the rate by the product of A and B. Make sure you take into consideration your units.

Rate: 1.261×10⁻⁴=k[A][B]²

1.261×10⁻⁴ = k[0.241][0.153]²

k=0.0224 M⁻²s⁻¹

Q24.59B

1. Do catalysts take part in the reaction they catalyze? Do catalysts always speed up a reaction? Explain.

a) A catalyst is a substance that speeds up a chemical reaction and participates in the reaction but doesn't get completely used up. It is possible that it will appear in the reaction mechanism but not in the overall reaction mechanism.

1. What is the function of a catalyst?

a) A catalyst works by lowering the activation energy which enhances the rate of the forward and backward reaction.

Q.25.27D

1. Calculate the energy of the alpha decay of ²¹²Po given the following atomic mass and conversion factors:

²¹²₈₄Po → ²⁰⁸₈₂Pb + ⁴₂He

209.0231u 207.2211u 4.0026u

1u=1.4924×10⁻¹⁰ J and 1u=931.5MeV

S.25.27D

1. Δm=(mass of ²⁰⁸Pb + ⁴He) − (mass of ²¹²Po)

Δm=207.2211u+4.0026u−209.0231u=2.2006u

E=2.2006u(1.4924×10⁻¹⁰ J/u)

E=3.284×10⁻¹⁰ J

Explanation: The first thing to do is to calculate the mass deficit (Δm). This is equal to the total atomic mass of ²⁰²Pb and ⁴He minus the total atomic mass of ²¹²Po given. The difference will give you your Δm. Because the mass deficit is in "u", you must use the first conversion in the upper right corner to get the units to J (energy).

2. How much energy, in megaelectronvolts, would be released if 3 alpha particles were destroyed?

You can use the conversion 1u = 931.5 MeV. However, since three alpha particles are being destroyed, you must multiply this amount to the atomic mass of the alpha particle, which will increase the atomic mass:

atmoic mass of alpha particle: 4.0026 u

⁴₂He=alpha particle
=4.0026u
=3×4.0026u=12.0078u

12.0078u×(931.50 MeV/u)=11185 MeV

*note that conversation factors are derived from equation E=mc².

Q21.1.1

Why are many radioactive substances warm to the touch? Why do many radioactive substances glow?

S21.1.1 : Many substances are warm to the touch because they are constantly releasing nuclear radiation that is unstable in the nucleus. They glow because the kinetic energy of radioactive decay gets converted to visible light which can actually be dangerous and cause health problems.

Q21.4.6 : The first-order decomposition of hydrogen peroxide has a half-life of 10.7 h at 20°C. What is the rate constant (expressed in s⁻¹) for this reaction? If you started with a solution that was 7.5 × 10⁻³ M H₂O₂, what would be the initial rate of decomposition (M/s)? What would be the concentration of H₂O₂ after 3.3 h?

A) Since this is a first order reaction involving the half life and temperature, you can use t_1/2 = ln(2)/k , making sure to convert hours to s^-1.

10.7 h * 60 min/h * 60 s/min = 38520 s

38520 s = ln(2)/k

k = 1.80 x 10^-5 s^-1

B) Now that you have found your rate constant, and been given the molarity of the starting solution along with the temperature in the previous example, you can use rate = k[A]¹ since it is a first order reaction.

rate = (1.8 x 10^-5 s^-1) [ 7.5 x 10^-3 M ]¹ = 1.35 x 10^-7 M/s

C) Since you are given the starting concentration (initial), you will need to find the final concentration after 3.3 hours. Knowing that it is a first order reaction allows you to use the following equation: [A]_t = [A]_oe^-kt and use the information that you found in the previous examples. Use the conversion to change 3.3 hours to seconds.

3.3 hours * 60 min/hour * 60 s/min = 11880 s

[A]_t = [ 7.5 x 10^-3 M] * e ^{-(1.80 x 10^(-5) s^(-1)(11880s)} = 6.05 x 10^-3 M