Extra Credit 26

Q19.9A

A voltaic cell with Ecell=1.500V has what [Ag+] in the cell?

Zn(s)|Zn2+(1.50M)||Ag+(?M)|Ag(s)

S19.9A

Solve for E_cell

Zn(s)→ Zn²⁺ + 2e^- (oxidation) - anode E^∘ = -0.763V

[Ag⁺ + e^-→ Ag(s)] x 2 (reduction) - cathode E^∘ = 0.800V

Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s)

E^∘ _cell = cathode - anode

E^∘ _cell= 0.800V - (-0.763V ) = 1.563V

Use the Nernst Equation

E=E^∘cell−(0.05922/n) logQ

1.5V=1.563V−0.02961 log (⁡[1.50]/[Ag⁺]²)

2.128 = log (⁡[1.50]/[Ag⁺]²) --> 10^(2.128) = 10^{(log (⁡[1.50]/[Ag+]))}

134.17 = 1.5/[Ag⁺]²

[Ag⁺] = 0.1054

Q19.65C

Out of the reactions given below, which of the following can occur spontaneously only through electrolysis. For those requiring electrolysis, what is the minimum voltage required?

1. I₂(s)+Ni(s)→2I⁻+Ni²⁺
2. Br₂(aq)+2Fe₂+→2Br⁻+2Fe³⁺
3. Ni^₂++2Cl⁻(aq)→Ni(s)+Cl₂(g)
4. Cr₂+(aq)+Fe³⁺(aq)→Cr³⁺(aq)+Fe²⁺(aq)

S19.65C

1. I₂(s)+Ni(s)→2I⁻+Ni²⁺

I₂(s)+ 2e⁻→2I⁻ (reduction) (cathode) --> 0.535V

Ni(s)→Ni²⁺ + 2e⁻ (oxidation) (anode) --> -0.257V

E^∘ _cell = cathode - anode = 0.535 - (-0.257) = 0.792V

E^∘ _cell is positive, so not spontaneous through electrolysis. The minimum voltage is 0.80V.

2. Br₂(aq)+2Fe₂+→2Br⁻+2Fe³⁺

Br₂(aq) + 2e⁻→2Br⁻ (reduction) (cathode) --> 1.065V

2Fe₂+→ 6e⁻+ 2Fe³⁺ (oxidation) (anode) -->0.745V

E^∘ _cell = cathode - anode = 1.065V - (0.745) = 0.32V

E^∘ _cell is positive, so not spontaneous through electrolysis. The minimum voltage is 0.33

3. Ni^₂++2Cl⁻(aq)→Ni(s)+Cl₂(g)

Ni^₂++2e⁻→Ni(s) (reduction) (cathode) --> -0.257V

2Cl⁻(aq)→ 2e⁻+Cl₂(g) (oxidation) (anode) --> 1.358V

E^∘ _cell = cathode - anode = -0.257V -(1.358) = -1.615V

E^∘ _cell is negative, so it is spontaneous through electrolysis.

4. Cr₂+(aq)+Fe³⁺(aq)→Cr³⁺(aq)+Fe²⁺(aq)

Fe³⁺(aq) + e⁻→Fe²⁺(aq) (reduction) (cathode) --> 0.771V

Cr₂+(aq)→Cr³⁺(aq) + 3e⁻ (oxidation) (anode) --> 0.424V

E^∘ _cell = cathode - anode = 0.771 -(0.424) = 1.195V

E^∘ _cell is positive, so not spontaneous through electrolysis. The minimum voltage is 1.20V

Q21.5C

Draw Lewis structures for the following ligands: a) OH−OH−b) NO−2NO2− c) NH3NH3 d) SO−4SO4− e) enen (hint: C2H8N2C2H8N2)

S21.5C

a)

b)

c)

d)

e)

Q24.13C

The following rates of reactions were obtained in three experiments with the reaction 2NO(g)+Cl2(g)→2NOCl(g)2NO(g)+Cl2(g)→2NOCl(g)

What is the rate law for this reaction?

S24.13C

[NO] 1/3 --> (0.145/0.29)^x = (1.24 X 10^-5/4.96 X 10^-5) [Cl₂] 1/2 --> (0.405/0.81)^y = (1.24 X 10^-5/2.48 X 10^-5)

0.5^x = 0.2419 0.5^y = 0.5

x = 2 y = 1

r = k [NO]²[Cl₂]

1.24 X 10^-5 = k (0.145²)(0.405)

k = 1.46 X 10^-3 M^-2s^-1

r = 1.46 X 10^-3 M^-2s^-1 [NO]²[Cl₂]

Q24.60B

The following substrates concentration [S] versus time date were obtained during an enzyme-catalyzed reaction: t = 0 min; [S] = 1.00M; 30 min, 0.90M; 90 min, 0.70M; 120 min, 0.50M; 180 min, 0.20M. What order is this reaction with respect to S in the concentration?

S24.60B

exponential decay, so it is first order

Q25.29B

Calculate the binding energy per nucleon, in megaelectronvolts, of the nuclide if its measured mass is 28.1900 u.

Mass of proton=1.0073u

Mass of neutron=1.0087u

S25.29B

→ 14 protons + 16 neutrons

28.1900u → (14 x 1.0073u) + (16 x 1.0087u)

Δm = products - reactants = 30.2414 - 28.1900 = 2.0514u

1 u = 931.5MeV

2.0514u x (931.5MeV/u) = 1910.8791MeV/nuclei x (1 nuclei/30 nucleons) = 63.70MeV/nucleon

Q21.1.5

Gamma rays are a very high-energy radiation, yet α particles inflict more damage on biological tissue. Why?

S21.1.5

While gamma rays are high energy radiation and can pass through many materials, alpha particles are the most dangerous inside the body. Once inside the body, internal tissues absorb all of the particles' energy.

Q24.6.2

In CFT, what causes degenerate sets of d orbitals to split into different energy levels? What is this splitting called? On what does the magnitude of the splitting depend?

S24.6.2

The overlap of the orbitals and ligand fields disrupt the electrons, which causes the splitting. This splitting is t_2g and e_g .