1) Calculate the cell potential, at 25 C, based upon the overall reaction

Zn2+(aq) + 2 Fe2+(aq) -> Zn(s) + 2 Fe3+(aq)
if [Zn2+] = 1.50 x 10^-4 M, [Fe3+] = 0.0200 M, and [Fe2+] = 0.0100 M. The standard reduction potentials are as follows:
Zn2+(aq) + 2 e– -> Zn(s) E = –0.763 V
Fe3+(aq) + e– -> Fe2+(aq) E = +0.771 V
a. –1.665 V
b. –1.534 V
c. –1.439 V
d. –0.008 V
e. +0.008 V

Answer is B

2) Al3+ is reduced to Al(s) at an electrode. If a current of 2.75 ampere is passed for 36 hours, what mass of aluminum is deposited at the electrode? Assume 100% current efficiency.
a. 9.2 x 10^–3 g
b. 3.3 x 10^1 g
c. 9.9 x 10^1 g
d. 1.0 x 10^2 g
e. 3.0 x 10^2 g

Answer is E

3) One kind of battery used in watches contains mercury(II) oxide. As current flows, the mercury(II) oxide is reduced to mercury.
HgO(s) + H2O() + 2 e– ->Hg() + 2 OH–(aq)
If 2.3 x 10^–5 amperes flows continuously for 1200 days, what mass of Hg(l) is produced?
a. 2.5 g
b. 5.0 g
c. 9.9 g
d. 13 g
e. 15 g

Answer is A

4) If a nucleus undergoes beta particle emission
a. its atomic number decreases by four and its mass number decreases by two.
b. its atomic number decreases by two and its mass number decreases by four.
c. its atomic number increases by one and its mass number is unchanged.
d. its atomic number decreases by one and its mass number increases by one.
e. its atomic number is unchanged and its mass number decreases by one.

Answer is B

5)The molar nuclear mass of fluorine-19 is 18.99840 g/mol. The molar mass of a proton is 1.007825 g/mol. The molar mass of a neutron is 1.008665 g/mol. Calculate the binding energy (in J/mol) of F-19. (c = 2.998 x 10^8 m/s)
a. 6.753 x 1012 J/mol
b. 7.131 x 1012 J/mol
c. 1.426 x 1013 J/mol
d. 8.609 x 1014 J/mol
e. 8.538 x 1014 J/mol

Answer is C

6) Lead-200 has a half-life of 21.5 hours. Starting with 3.77 mg of this isotope, how much would remain after 72 hours?
a. 0.0 mg
b. 0.098 mg
c. 0.37 mg
d. 0.47 mg
e. 1.1 mg

Answer is C

I don't agree with #1.

I don't agree with 2.

I agree with 3.

I don't agree with 4.

I agree with 5 and 6.

To solve these questions, you need to apply the concepts of electrochemistry, atomic structure, radioactive decay, and binding energy.

1) To calculate the cell potential, you need to use the Nernst equation, which relates the cell potential to the concentrations of the reactants and the standard reduction potentials. The equation is given as follows:

Ecell = Eºcell - (0.0592/n) * log(Q)

where Ecell is the cell potential, Eºcell is the standard cell potential (sum of the standard reduction potentials), n is the number of electrons transferred in the balanced equation, and Q is the reaction quotient.

In this case, the balanced equation is Zn2+(aq) + 2 Fe2+(aq) -> Zn(s) + 2 Fe3+(aq), and the standard reduction potentials are given as Eº(Zn2+) = -0.763 V and Eº(Fe3+)/2(Fe2+) = 0.771 V.

First, calculate the reaction quotient Q using the concentrations given:

Q = [Zn(s)][Fe3+]^2 /[Zn2+][Fe2+]^2

Then substitute the values into the Nernst equation to calculate Ecell:

Ecell = -0.763 V - (0.0592/2) * log(Q)

After calculating Ecell at 25°C, you will find that the answer is -1.534 V, which corresponds to choice B.

2) To calculate the mass of aluminum deposited at the electrode, you need to use Faraday's laws of electrolysis. The first law states that the mass of a substance produced at an electrode is directly proportional to the amount of electricity passed through the cell.

The equation to calculate the mass of aluminum is given as:

Mass = (Current * Time * Molar Mass) / (n * Faraday's Constant)

In this case, the current is given as 2.75 A, time is given as 36 hours, molar mass of aluminum is 26.98 g/mol, n is the number of electrons transferred (which is 3 for the reduction of Al3+ to Al), and Faraday's Constant is 96,485 C/mol.

Plug these values into the equation to calculate the mass of aluminum, and you will find that the answer is 3.0 x 10^2 g, which corresponds to choice E.

3) To calculate the mass of Hg(l) produced, you need to use Faraday's laws of electrolysis again. The equation and values are similar to the previous question.

Mass = (Current * Time * Molar Mass) / (n * Faraday's Constant)

In this case, the current is given as 2.3 x 10^-5 A, time is given as 1200 days (convert to seconds), molar mass of Hg is 200.59 g/mol, n is 2 (as stated in the balanced equation), and Faraday's Constant is 96,485 C/mol.

Plug these values into the equation to calculate the mass of Hg(l), and you will find that the answer is 2.5 g, which corresponds to choice A.

4) When a nucleus undergoes beta particle emission, a neutron inside the nucleus is converted into a proton. This results in a decrease in atomic number by one. However, the mass number remains the same because the mass of both the neutron and the proton are nearly the same.

Therefore, the answer is choice B, where the atomic number decreases by two and the mass number decreases by four.

5) To calculate the binding energy of F-19, you need to use Einstein's equation, E = mc^2, where E is the binding energy, m is the mass defect, and c is the speed of light.

The mass defect can be calculated by subtracting the sum of the masses of individual nucleons in the nucleus from the molar mass of F-19. Then, multiply the mass defect by c^2 to obtain the binding energy.

Given the molar nuclear mass of F-19 (18.99840 g/mol), the molar mass of a proton (1.007825 g/mol), the molar mass of a neutron (1.008665 g/mol), and the speed of light (c = 2.998 x 10^8 m/s), calculate the mass defect and then the binding energy. The answer is 1.426 x 10^13 J/mol, which corresponds to choice C.

6) To calculate the remaining mass of Lead-200 after 72 hours, you need to use the concept of half-life.

The half-life of Lead-200 is given as 21.5 hours. To find the fraction of remaining Lead-200, divide the elapsed time (72 hours) by the half-life (21.5 hours) raised to the power of the number of half-lives.

Fraction remaining = (elapsed time / half-life)^number of half-lives

Subtract this fraction from 1 and multiply by the initial mass of Lead-200 (3.77 mg) to obtain the remaining mass.

After calculating the remaining mass, you will find that the answer is 0.37 mg, which corresponds to choice C.