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Ch. 13 Activity

2.28 Exam 2 review

1. What happens at the cathode?  What happens at the anode?  In a galvanic cell, which way to the electrons flow?  In an electrolytic cell?  What is the purpose of a salt bridge?  How does it work?  Do anions or cations flow to the cathode?  To the anode?

Reduction happens at the cathode, which means gain of electrons.  Oxidation happens at the anode, which means loss of electrons.  (Remember Red Cat/An Ox and LEO goes GER.)  In a galvanic, or spontaneous, cell, electrons flow from the anode to the cathode.  In an electrolytic, or non-spontaneous, cell, electrons still flow from the anode to the cathode – the direction of electron flow is what defines the anode and the cathode.  The purpose of the salt bridge is to balance the charge of each half reaction – without it, the reaction would stop much before completion because of the buildup of charge.  It is a soluble salt (which is ionic by definition).  The ionic bonds of the salt break apart as the reaction moves forward.  Cations will flow into the cathode and anions will flow into the anode (makes sense, right?).

2. Balance the following redox reaction in acid:

MnO₄^-(aq) + Fe²⁺(aq) à Mn²⁺(aq) + Fe³⁺(aq)

Which half reaction is oxidation?  Which is reduction?  How many electrons does this process involve?

I.               Split into half reactions:
Fe²⁺ à Fe³⁺ + e- (oxidation)
MnO₄^- à Mn²⁺ (reduction – this makes sense because it is losing oxygen)
The oxidation reaction is balanced but the reduction reaction is not – we must recall rules from chem 142.

II.             Add water to balance oxygens:
MnO₄^- à Mn²⁺ + 4H₂O

III.           Add H⁺ to balance H’s from water:
MnO₄^- + 8H⁺ à Mn²⁺ + 4H₂O

IV.            Balance charge:
MnO₄^- + 8H⁺ + 5e^- à Mn²⁺ + 4H₂O

This process involves 5 electrons, and the total balanced reaction is:

MnO₄^- + 5Fe²⁺ + 8H⁺à Mn²⁺ + 5Fe³⁺ +4H₂O

Now that you know the number of electrons of this redox reaction, you should be able to do calculations such as E_cell, Gibbs energy, etc. (given the necessary additional information).

3. What is the sign of E_cell in a galvanic cell?  In an electrolytic cell?  You should be able to calculate ΔG given E_cell, E_cell given E^o_cell (what does this mean?), equilibrium constant, K, and reaction quotient, Q.  You should be able to use the Nernst equation.  Write down the equations you would need for this.  Know what each variable stands for.

E_cell in a galvanic cell is always positive and negative for an electrolytic cell.
ΔG^o = -nFE^o_cell
E^o_cell = RT/nF * ln(K)
E_cell = E^o_cell – RT/nF * ln(Q)
The last equation, the Nernst equation gives the E_cell for non-standard conditions.  At standard conditions, all concentrations are equal to 1M.

4. Electrolysis of water has been proposed as a solution to the dwindling gasoline resources as a way to store energy in chemical bonds:

2H₂O à 2H₂ + O₂

The fuel made in this reaction is H₂, which is flammable in air.  The reverse reaction of that listed above is highly exothermic.  Electrolysis of water in an electrochemical cell is electrolytic, i.e. non-spontaneous.  The E^o_cell for this reaction is -1.23 V. 

a.) What is the E^o_cell for the hydrogen half reaction, 2H⁺ +2e^- à H₂?  (Hint: you know this already without looking in any tables!) 

E^o_cell for this reaction is 0 V.  This is because the hydrogen half reaction is used as the ‘zero’ for electrochemistry, against which we measure all other half reactions.  It is zero by definition.

b.) Using this information, what is the E^o_cell for the oxygen half reaction?  Balance this half reaction. 

The oxygen half reaction must therefore have an E^o_cell of -1.23 V.  Balance reaction (from steps listed above):
I. 2H₂O à O₂
II. No need to add water
III. 2H₂O à O₂ + 4H⁺
IV. 2H₂O à O₂ + 4H⁺ + 4e^-

c.) How many electrons are involved in this process?  Be sure your overall reaction reduces to the reaction listed above.
This is a four-electron process.  The reaction in part a.) must be multiplied by two and added to the balanced half reaction in part b.).

5. The photoelectric effect was first observed by Hertz in 1887.  Einstein was awarded his only Nobel Prize in 1921 for "for his services to theoretical physics, and especially for his discovery of the law of the photoelectric effect". 

a.) What is the photoelectric effect?  What does it tell us about light?  Why is this significant?  

The photoelectric effect is when light is shined on metal, electrons are ejected.  This tells us that light interacts with matter and therefore has particle properties, in addition to its wave properties.

b.) If intensity of light shining on the metal increases, what changes?  If wavelength of the light increases, what changes?

Increasing the intensity of light increases the number of electrons that are ejected from the metal.  Increasing the wavelength (decreasing energy) will have no effect on the photoelectric effect until a certain cutoff wavelength, above which there is not enough energy to eject the electrons.

6. What is the difference between Ĥ, the Hamiltonian, and E, energy in the Schrödinger equation?  What energy boundary conditions are imposed on the particle in a box?  What is the physical meaning of Ψ*Ψ?  What is Ψ*?

Ĥ is an operator, which tells you to do something to the wave function Ψ (namely, take derivatives) while energy E is a constant, or just a number.  The Schrödinger equation says that when you operate on the wave function with the Hamiltonian, you will get back energy time the wave function: ĤΨ=EΨ.  In particle in a box, the POTENTIAL energy inside the box is 0 and the PE outside the box is infinite.  We do this because it is easy to do math to solve this problem (it can be done by hand and you will learn it in physical chemistry if you take that class).  Ψ*Ψ is the probability of finding an electron in a certain spot.  Places with zero probability are called NODES.  Ψ* is the complex conjugate of the wave function.  That just means that you put a negative in every place you see i, the square root of negative 1.

2.21 Activity

1. Sodium atoms have a characteristic yellow color when excited in a flame.  The color comes from the emission of light if 589.0 nm (this is the same experiment as the flame emission spectroscopy we did in lab 4 with potassium).

What is the frequency of this radiation? What is the energy of this radiation per photon? Per mole of photons?

ν = c/λ = 2.9979*10⁸ m/s / 5.890*10^-7 m = 5.090*10¹⁴ s^-1 (s^-1 is also called Hz – Hertz – and is the SI unit for frequency)

ΔE = hν = 6.626*10³⁴ J*s * 5.090*10¹⁴ s^-1 = 3.373*10^-19 J – This is the energy per photon.  To get ΔE per mole, multiply by Avogadro’s number to get 2.031*10⁵ J/mol.

2. What is the wavelength (in nm) of an electron (mass = 9.11*10^-31 kg) traveling at 5.31*10⁶ m/s?  What is the wavelength (in m) of a tennis ball (mass = 56 g) traveling at 80 km/hr?  What does this tell you about quantum versus classical mechanics?

λ = h/m*v
For the electron: λ = 6.626*10^-34 J*s /(9.11*10³¹ kg * 5.31*10⁶ m/s) = 1.37*10^-10 m = 0.137 nm
For the tennis ball: λ = 6.626*10^-34 J*s /(0.56 kg * 22.22 m/s) = 5.324*10^-35 m

Because the wavelength is so small for the tennis ball, it is not something that could be measured.  This is why we use different equations for quantum mechanics and classical mechanics.

3. Calculate the change in energy corresponding to the relaxation of an electron from the n=4 state to the n=2 state in the hydrogen atom.  What does the sign of this energy change tell you?

ΔE = E_final – E_initial = E₄ – E₂ = -2.178*10^-18 J [Z²/(4²-2²)] = -1.815*10^-19 J (Z=1 here because we are dealing with the hydrogen atom).  The sign tells you that energy is released in this process.

4. For the n=4 energy level, how many electrons can fit, including spin?  What are the possible l values and to what orbitals do they correspond?  What are the m_l values for each l?  What do the m_l values tell you?

For n=4, there are s, p, d, and f orbitals which can fit 2, 6, 10, and 14 electrons, respectively.  Therefore, 32 electrons can fit in n=4.  The possible l values are 1, 2, 3, and 4, corresponding to the s, p, d, and f orbitals, respectively.  The m_l value for s is 0; for p, -1, 0, 1; for d, -2, -1, 0, 1, and 2; for f, -3, -2, -1, 0, 1, 2, 3.  These values are related to the position of the orbital in space relative to the other orbitals.  They are a way to label each orbital.  Each orbital contains two spin-orbitals, which means that each orbital can contain an electron with spin +1/2 and spin -1/2.

5. What is the electron configuration of chromium? What is the configuration in shorthand notation?  How many valence electrons are there?  Load the orbital diagram with electrons.

Cr: 1s²2s²2p⁶3s²3p⁶3d⁵4s¹ or in shorthand: [Ar] 3d⁵4s¹

There are 6 valence electrons.
When you draw the orbital diagram, be sure that all orbitals that are half full have electrons with the same spin.  It is convention to draw the electrons in orbitals that are half full with spin up.
Note that the d orbital and the s orbital are both half full, rather than the d orbital having six electrons and the s orbital having none.  Even though the 4s orbital is lower in energy than the 3d orbital, there is a stabilization energy associated with half-full orbitals.

6. According to periodic trends, order the following atoms in terms of atomic radius, ionization energy, and electron affinity.
a.) Li, B, N, F

Atomic radius: Li>B>N>F

Ionization energy: F>N>B>Li

Electron affinity: F>N>B>Li
b.) Fr, Sr, Ga, P

Atomic radius: Fr>Sr>Ga>P

Ionization energy: P>Ga>Sr>Fr

Electron affinity: P>Ga>Sr>Fr

2.14 Activity

1.31 Activity

1. A system does 3 J of work on the surroundings, and 12 J of work are added to the system.  What is the energy change of the system? 9 J What is the energy change of the surroundings? -9 J

2. A quantity of 0.850 mol of an ideal gas initially at a pressure of 15.0 atm and 300K is allowed to expand isothermally until its final pressure is 1.00 atm.  Calculate the value of the work done if the expansion is carried out (a) against a vaccum, (b) against a constant external pressure of 1.00 atm, and (c) reversibly – use the equation w_rev=-nRT*ln(P₁/P₂).

a.) w=-P_extDV, here P_ext = 0, so w=0

b.) P_ext=1.00 atm.  Find initial and final volumes using the ideal gas law:

      V₁=nRT/P₁; V₂=nRT/P₂; w=-P_extDV, where P_ext is equal to P₂

      DV= V₂-V₁ = nRT({1/P₂}-{1/P₁})

      Therefore, w=-nRTP₂({1/P₂}-{1/P₁}) = -19.5 L*atm = -1.98*10³ J

c.) for an isothermal, reversible process, we use the following equation for work:

     w_rev=-nRT*ln(V₂/V₁)=-nRT*ln(P₂/P₁) = -5.74*10³ J

3. Calculate the values of DE and DH for the heating of 55.40 g of xenon from 300 K to 400 K.  Assume ideal-gas behavior and that the heat capacities at constant volume and constant pressure are independent of temperature.

Xenon is a monatomic “ideal” gas, so C_v=3/2*R and C_p=5/2*R (always true for ideal monoatomic gases).  We have the two following equations (always true):
DE=nC_vDT = 526J; DH=nC_pDT = 877 J

4. Calculate the entropy change when 2.0 mol of an ideal gas are allowed to expand isothermally from an initial volume of 1.5 L to 2.4 L.

DS=nR*ln(V₂/V₁) = 7.8 J/K

5. A quantity of 0.50 mol of an ideal gas at 20^oC expands isothermally against a pressure of 2.0 atm from 1.0 L to 5.0 L.  Calculate DS_sys, DS_surr, and DS_univ.

Use the ideal gas law to find initial pressure = 12 atm.  Use the equation:

DS_sys = nR*ln(V₂/V₁) = 6.7 J/K

DS_surr = q_surr/T; DE=0 since DT=0 (it is an isothermal process).  Therefore, w=-q.  We can calculate w=-PDV, and q_sys=-q_surr, so q_surr=w.  We get DS_surr = -2.8 J/K
DS_univ=DS_sys+DS_surr=3.9J/K

6. Supercooled water is liquid water that has been cooled below its normal freezing point.  This state is thermodynamically unstable and tends to freeze into ice spontaneously.  Suppose we have 2.0 mol of supercooled water turning to ice at -10^oC and 1.0 atm.  Calculate DS_sys, DS_surr, and DS_univ for this process (set up a thermodynamic cycle).  C_p water = 75.3 J/K*mol and C_p ice = 37.7 J/K*mol; the molar heat of fusion of water is 6.01 kJ/mol. (DS_fus=n*DH_fus/T, q_surr=n*DH_fus)

Step 1: Water at -10 à Water at 0:

DS = nC_p*ln(T_f/T_i) = 5.6 J/K

Step 2: Water freezes at 0:

DS_fus=DH_fus/T = -44.0 J/K

Step 3: Ice at 0 à Ice at -10:

DS = nC_p*ln(T_f/T_i) = -2.8 J/K

Therefore, DS_tot=DS_sys =5.6 J/K - 44.0 J/K - 2.8 J/K = -41.2 J/K

To find DS_surr, we must find q for each step, add those numbers together and plug into the equation DS_surr = q_rev/T.

Step 1: q_surr=-nC_PDT = -1.5 kJ
Step 2: q_surr=n*DH_fus = 12 kJ
Step 3: q_surr=-nC_pDT = 754 J

q_surr(tot) = 11 kJ, DS_surr=41.8 J/K

DS_univ = DS_sys + DS_surr = 0.6 J/K

7. A 0.590 mol sample of an ideal gas initially at 300 K and 1.50 bar is compressed isothermally to a final pressure of 6.90 bar.  Calculate the change in Gibbs energy for this process.

For an ideal gas, DG = nRT*ln(P₂/P₁) = 2.23 kJ

1.23 Activity

Chem 152 B Activity 1/24/12 – answers will be posted at chem152B.blogspot.com

1. For a chemical reaction involving only the gas phase, entropy is related to the total number of moles on either side of the equation.  A decrease means lower entropy; an increase means higher entropy.

For the reaction

2H₂(g) + O₂(g) à 2H₂O(g)

do you expect the change in entropy to be positive or negative?  Why?  Do you expect the change in enthalpy to be positive or negative?  Why?  (Think about your background knowledge of this reaction.)  What might this tell you about the spontaneity of the reaction?

You would expect the change in entropy to be negative because there are more moles of gas in the reactants than the products.  You would expect the change in enthalpy be negative because we know this reaction to be exothermic – hydrogen is flammable in air.  DS_univ = DS_sys + DS_surr = DS_sys  - DH/T.  DH is negative so we have a negative number minus a negative number.  When temperature is small, DH will outweigh DS and the reaction will be spontaneous (i.e. DS_univ will be positive).  When temperature is large, DS will outweigh DH and the reaction will not be spontaneous (i.e. DS_univ will be negative).

2. For a chemical reaction involving different phases, the production of gas will, in general, increase the entropy much more than the increase in the number of moles of a liquid or solid.

For the reaction

2HNO₃(aq) + Na₂CO₃(s) à 2NaNO₃(aq) + H₂O(l) + CO₂(g)

do you expect the change in entropy to be positive or negative?  Why?

You would expect the change in entropy to be positive since there are more moles of gas in the products than the reactants.