Exam 3 Review!!

Redox

100

Identify the oxidation states of nitrogen and chromium in the following reaction:

Cr₂O₇²⁻+NO₂^-→Cr³⁺+NO₃^-

Cr: +6 in Cr₂O₇^2- and +3 in Cr³⁺

N: +3 in NO₂^- and +5 in NO₃^-

200

Identify the oxidizing agent and reducing agent, in the following:

Zn+HCl→ZnCl₂+H₂

Oxidizing Agent: HCl

Reducing Agent: Zn

300

Iron reacts with oxygen to form iron(III) oxide (rust). Write the half-reactions for the oxidation and reduction processes that occur in this reaction. Also, calculate the oxidation state change for iron in this reaction.

1. Oxidation Half-Reaction: Fe→Fe³⁺+3e⁻ (Iron loses electrons)
2. Reduction Half-Reaction: O₂+4e⁻→2O²⁻ (Oxygen gains electrons)
3. Iron's oxidation state changes from 0 in Fe to +3 in Fe³⁺

400

Balance the following redox reaction in acidic solution using the half-reaction method: 

Cr₂O₇²⁻+Fe²⁺→Cr³⁺+Fe³⁺

1. Oxidation Half-Reaction: Fe²⁺→Fe³⁺+e⁻ 
2. Reduction Half-Reaction: Cr₂O₇²⁻+6e⁻+14H⁺→2Cr³⁺+7H₂O
3. Multiplying the oxidation half-reaction by 6 and adding it to the reduction half-reaction gives the balanced equation:

6Fe²⁺+Cr₂O₇²⁻+14H⁺→2Cr³⁺+7H₂O+6Fe³⁺

500

Balance the following redox reaction in basic solution:

H₂O₂+CrO₄²⁻→Cr³⁺+O₂

1. Oxidation Half-Reaction: 2OH^-+H₂O₂→O₂+2H₂O+2e⁻ (first balance O, then add H⁺, then add same amount OH^- to both sides)
2. Reduction Half-Reaction: CrO₄²⁻+3e⁻+4H₂O→Cr³⁺+8OH⁻ (first balance O, then add H⁺, then add same amount OH^- to both sides)
3. Multiplying the oxidation half-reaction by 3, multiplying the reduction half-reaction by 2, and adding them together gives the balanced equation:

3H₂O₂+2CrO₄²⁻+8H₂O+6OH^-→3O₂+2Cr³⁺+6H₂O+16OH^-

3H₂O₂+2CrO₄²⁻+2H₂O→3O₂+2Cr³⁺+10OH⁻

Electrochemistry

100

Given the following standard reduction potentials:

Zn²⁺+2e⁻→Zn            E°=−0.76 V 

Ag⁺+e⁻→Ag            E°=+0.80 V

Which metal is more likely to oxidize under standard conditions?

Zn is more likely to oxidize since it has a lower standard reduction potential.

200

Given the following standard reduction potentials:

Br₂+2e⁻→2Br⁻      E°=1.09 V

MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O       E°=1.51 V

Determine the stronger oxidizing agent.

The higher the standard reduction potential, the stronger the oxidizing agent. Therefore, MnO₄⁻ is the stronger oxidizing agent

300

For the reaction:

MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O          E°=+1.51 V

Calculate the equilibrium constant (K) at 25°C.

Nernst Equation:
E = E° − (0.0592/n)*log(Q)

At equilibrium, E=0, Q=K -> 

E^o=(0.0592/n)*log(K)

n = 5 electrons transferred

K = 10^{(1.51*5/0.0592)} = 6.31 x 10¹²

400

A current of 12.5 A is passed through a CuSO₄ solution. How long, in hours, would this current have to be applied to plate out 20 g of copper? The molar mass of copper is 63.55 g/mol. Assume 100% current efficiency.

(20 g / 63.55 g/mol) = 0.315 mol

Cu -> Cu²⁺ + 2e^-

2 mol of electrons=2 Faraday (F)=192970 C/mol

Q = 0.315 mol×192970 C/mol = 60785.55 C

t = Q/I = (60785.55 C) / (12.5 A) = 4862.8 s

1.35 hours

500

A voltaic cell is constructed using the following half-reactions: 

Cu²⁺+2e⁻→Cu       E°=+0.34 V 

Ag⁺+e⁻→Ag       E°=+0.80 V

Calculate the standard cell potential (E°_cell) and the cell potential when the concentration of Cu2+ ions is 0.10 M and the concentration of Ag+ ions is 0.50 M.

E°_cell = E°(cathode) - E°(anode)

E°_cell = (0.80 V) - (0.34 V) = 0.46 V

E_cell = E°_cell - (0.0592 / n)*log(Q)

Q = [Cu²⁺] / [Ag⁺] 

E_cell = 0.46 V - (0.0592 / 2)*log(0.10 M / 0.50 M)

E_cell = 0.4807 V

Thermodynamics

100

We are studying a reaction as our system. The change in enthalpy of the system is positive. Is the reaction endothermic or exothermic?

The reaction absorbs heat and is endothermic.

200

Indicate which substance has the higher entropy value: C₂H₅OH(l) or C₃H₇OH(l).

C₃H₇OH(l)

It is a larger molecule, and so more microstates describing its motions are available.

300

To produce sucrose, the body couples the following two reactions. Write the overall reaction and determine whether it will be spontaneous. 

glucose + fructose <-> sucrose + H₂O (deltaG = 27 kJ/mol)

ATP + H₂O <-> ADP + P_i (deltaG = -30 kJ/mol)

The reaction will be spontaneous.

glucose + fructose + ATP <-> sucrose + ADP + P_i (deltaG = -3 kJ/mol)

400

For the reaction 2A + B <-> C, consider the entropy data below.

Substance       S^o(J/mol K) at 298 K

A                           8.4

B                           12.6

C                           79.4

If the Gibbs free energy change for a reaction is 6.65 kJ/mol, is the reaction endothermic or exothermic?

Endothermic (deltaH = 21.5 kJ/mol)

Steps

1. deltaS = 79.4 - [2(8.4)+12.6] = 50J/mol K

2. Calculate deltaH using deltaG = deltaH - TdeltaS

500

The reaction 2A + B <-> C was run with initial concentrations of [A]=0.275 M, [B]=0.525 M, and [C]=0 M. If [C] = 2.6 x 10^-3 M at equilibrium, determine Gibbs free energy for the reaction at 298K.

deltaG=6.65 kJ/mol

Steps:

1. Fill out an ICE table

-From the table, you will find that [C] at equilibrium is x (x=2.6x10^-3 M)

2. Find equilibrium concentrations from the ICE table and calculate K

3. Use deltaG=-RTlnK

Nuclear Chemistry

100

Which of the following types of nuclear decay decrease the nuclear charge?

A) Positron emission    B) Beta emission

C) Electron capture

Position emission and Electron capture

200

Mn-55 can be prepared by electron capture from which of the following isotopes?

A) Mn-56    B) Fe-55

C) Cr-55     D) Co-57

Fe-55

⁵⁵₂₆Fe + ⁰_-1B  ->  ⁵⁵₂₅Mn

300

A radioactive isotope decays by alpha emission followed by two beta emissions. What is the change in the mass number and atomic number of the original isotope?

The mass number decreases by 4 and the atomic number is unchanged

⁴₂He + 2⁰_-1B

400

You clean out your fridge for the first time in a while and discover some mysterious leftovers. The C-14 radiation level was measured to be 11.1 nuclei/min/C-14. How old are the leftovers? (Assume that for fresh leftovers, C-14 radiation levels are 15.3 nuclei/min/C-14 and t_1/2 = 5730)

2653 years! Better throw them away

ln(rate_t/rate₀) = -0.693t/t_1/2

ln(11.1/15.3) = -0.693t/5730

500

Nb-93 is the only stable isotope of Nb. Determine the primary mode of decay for Nb-97 and Nb-90 and write the balanced nuclear reactions.

⁹⁷₄₁Nb is neutron rich -> decay by beta emission

⁹⁷₄₁Nb -> ⁰_-1B + ⁹⁷₄₂Mo  

⁹⁰₄₁Nb is proton rich -> decay by positron emission (or electron capture)

⁹⁰₄₁Nb -> ⁰₁B + ⁹⁰₄₀Zr