# Practice Problems for Exam 4

Practice problems for exam 4 are listed below. This exam will cover sections 10.1 - 20.9

## Topic: Solubility and Precipitation

1. #### Estimate Ksp for compound in basic solution

A saturated solution of Hg(OH)2 in water has a pOH = 4.50. Estimate Ksp for this compound.
Ksp = 1.6 x 10-14
##### Explanation
The equilibrium equation is:

Hg(OH)2 <-> Hg2+ + 2OH-

while the equilibrium expression is:

Ksp = [Hg2+][OH-]2

The pOH value gives the [OH-] = 10-pOH
[OH-] = 10-4.50 = 3.16 x 10-5 M
The [Hg2+] = (1/2)[OH-] = 1.58 x 10-5 M
Ksp = (1.58 x 10-5)(3.16 x 10-5)2
Ksp = 1.6 x 10-14
2. #### Solubility of Fe(II) hydroxide

The Ksp for Fe(OH)2 is 8.0 x 10-16. Determine the solubility of Fe(OH)2 in units of g/L.
5.3 x 10-4 g/L
##### Explanation
The equilibrium dissociation reaction is written as:

Fe(OH)2 <-> Fe2+ + 2OH-

The equilibrium expression is:

Ksp = [Fe2+][OH-]2

The equilibrium concentration can be found by assuming only slight dissociation of the iron hydroxide:
 Fe(OH)2 <-> Fe2+ + 2OH- init xs 0 0 change -- +x +2x final xs x 2x
Ksp = [Fe2+][OH-]2
8.0 x 10-16 = (x)(2x)2
8.0 x 10-16 = 4x3
x = 5.85 x 10-6 M
MW(Fe(OH)2) = 89.85 g/mol
solubility = (x)(MW) = (5.85 x 10-6 M)(89.85 g/mol)
solubility = 5.3 x 10-4 g/L
3. #### Solubility of Fe(II) with Common Ion

The Ksp for Fe(OH)2 is 8.0 x 10-16. Determine the free Fe2+ concentration, in units of g/L, in a solution having pH = 13.30.
[Fe2+] = 1.1 x 10-12 g/L
##### Explanation
The equilibrium dissociation reaction is written as:

Fe(OH)2 <-> Fe2+ + 2OH-

The equilibrium expression is:

Ksp = [Fe2+][OH-]2

The equilibrium concentration can be found by assuming only slight dissociation of the iron hydroxide. However, in this case, a common ion, OH- is present. The concentration of this ion is:
[OH-] = 10-pOH = 10-(14-pH)
[OH-] = 10-0.7 = 0.20 M
 Fe(OH)2 <-> Fe2+ + 2OH- init xs 0 0.20 M change -- +x 0.20 M + 2x final xs x 0.20 M + 2x
8.0 x 10-16 = (x)(0.20 M + 2x)2
can assume x is small compared to 0.20 M, since Ksp is so small
8.0 x 10-16 = (x)(0.20 M)2
x = [Fe2+] = 2.0 x 10-14 M
MW(Fe) = 55.6 g/mol
[Fe2+] = (2.0 x 10-14 M)(55.6 g/mol)
[Fe2+] = 1.1 x 10-12 g/L
4. #### Precipitation of Fe(II) hydroxide

The Ksp for Fe(OH)2 is 8.0 x 10-16. Determine the [OH-] that must be exceeded in a 4.6 x 10-6 M FeCl2 solution to precipitate Fe(OH)2.
[OH-] = 1.3 x 10-5 M
##### Explanation
Since FeCl2 is the salt of a strong acid, it will dissociate completely in water at low molal concentration
FeCl2 -> Fe2+ + 2Cl-
Therefore, [Fe2+] = 4.60 x 10-6 M.
For the equilibrium reaction
Fe(OH)2 -> Fe2+ + 2OH-
to lead to solid precipitate, Q > K (equilibrium will shift to reactants)
Q = [Fe2+][OH-]2 > 8.0 x 10-16
(4.60 x 10-6 M)[OH-]2 > 8.0 x 10-16
[OH-]2 > 1.73 x 10-10
[OH-] = 1.3 x 10-5 M

## Topic: Enthalpies

1. #### Fundamentals of Thermodynamic Systems

The following statements discuss different thermodynamic systems. Determine whether each is true or false.
• Energy is not a state function
• There is heat transfer in a constant-pressure process
• A closed system must be in a steady state
• False
• True
• False
##### Explanation
• Energy only depends on the initial and final conditions of the system, and independent of pathway. Therefore, energy is a state function.
• A system under constant pressure can be involved in heat transfer. The heat transfered under constant pressure is called enthalpy.
• A closed system can exchange heat and work (but not matter). Therefore, it does not have to be in a steady state in reference to energy.
2. #### Transfer of Matter, Heat, and Work Across System Boudaries

The following statements pertain to the transfer of heat, matter, and work across a system boundary. Determine whether each statement is true or false.
• Work is able to cross the boundary of a closed system
• Matter is able to cross the boundary of an open system
• Heat is able to cross the boundary of an isolated system
• True
• True
• False
##### Explanation
• A closed system will allow heat and work to pass, but not matter.
• An open system will allow heat, work and matter to pass.
• In an isolated system, nothing can pass the boundary.
3. #### Determine Internal Energy Change

Calculate the change in the internal energy for a process in which the system releases 546 J of heat to the surroundings and does 11.0 J or work on the surroundings
-557 J
##### Explanation
DE = Q + W
The sign of Q is negative if heat is released to the surroundings
The sign of W is negative if work is done on the surroundings
DE = (-546 J) + (-11 J)
DE -557 J
4. #### Work Done On or By System

From the conditions and given definitions of the system, determine whether there is work done on the system, work done by the system, or no work done.
• One mole of ideal gas in a piston chamber expands to double its volume (system = piston chamber).
• Gaseous dichlorofluoromethane, a refrigerant, is compressed in the compressor of an air conditioner, to try an liquify it (system = compressor).
• A can of spray paint is discharged in air (system = spray).
• The space shuttle's cargo bay doors are opened while in orbit, releasing a little bit of residual atmosphere into space (system = cargo bay).
• A balloon expands as a small piece of dry ice inside the balloon sublimes (system = air surrounding the balloon).
• Work done by system
• Work done by system
• Work done by system
• No work done
• Work done on system
##### Explanation
• As the piston chamber expands, it is performing work on the surroundings. Therefore, work is done by the system.
• The compressor is working the liquify the gas. Since the system is the compressor, work is being done by the system.
• The spray is doing work against the atmosphere. Work is therefore being done by the system, since the system is defined as the spray.
• No work is done when gas expands into a vacuum.
• The balloon is expanding against the surrounding atmosphere. Since the surrounding air is the system, work is being done on it.
5. #### Specific Heat Calculation

It requires 2816 J of energy to change the temperature of 23.5 g of a substance from 95.9 oF to 136.2 oF. Determine the specific heat of the substance.
5.35 J/g-K
##### Explanation
Specific heat is the amount of energy needed to raise 1 gram of a substance 1 degree in temperature. Units are J/g-K.
Specific heat = Q/(mass * DT)
DT = T2 - T1
T1 = (5/9)(95.9 - 32.0) = 35.5 oC
T2 = (5/9)(136.2 - 32.0) = 57.9 oC
DT) = 57.9 - 35.5 = 22.4 K
Note: temperature differences are the same in K and oC
Specific heat = (2816 J)/(23.5 g * 22.4 K) = 5.35 J/g-K
6. #### Primer on Enthalpies

Consider the following statements regarding enthalpy and enthalpies of formation. Determine whether each statement is true or false.
• The change in enthalpy of any system cannot be measured.
• Enthalpy is a state function.
• The change in enthalpy is equal to the heat added to a system under constant volume conditions.
• The standard enthalpy of formation for the most stable form of any element is always greater than zero.
• False
• True
• False
• False
##### Explanation
• The value DH can be measured from the heat absorbed or released by a system under constant pressure conditions.
• Enthalpy is independent of pathway, and therefore is a state function.
• The change in enthalpy is the heat added under constant pressure conditions.
• The standard enthalpy of formation for the most stable form of any element is equal to zero.
7. #### Reaction Enthalpies

Consider the followng statements regarding reaction enthalpies. Determine whether each statement is true, false, or sometimes true.
• An exothermic reaction is a spontaneous reaction.
• The enthalpy of reaction depends on the state of the reactants and products.
• A reaction is endothermic when the reaction enthalpy is less than zero.
• The sign of the reaction enthalpy is the same for both the forward and reverse directions of the chemical reaction.
• The magnitude of DHrxn is related to the amount of reactant consumed.
• Sometimes true
• True
• False
• False
• True
##### Explanation
• The spontaneity of a reaction will depend both on the change of enthalpy and entropy of the reaction.
• The enthalpy change for a reaction depends on the state of the reactants and products.
• Endothermic reactions have DHrxn > 0.
• The sign of the reaction enthalpy will change with the direction of the reaction, while the magnitude will remain the same.
• DHrxn is an extensive property, therefore, directly proportional to the amount of reactant consumed.
8. #### Determine Enthalpy of Formation for Aluminium Oxide

For the reaction:

Al2O3(s) + HCl(g) -> H2O(g) + AlCl3(s)

Balance the equation and determine DHof Al2O3, given DHrxn is 86.94 kJ per one mole of Al2O3 consumed. The following thermodynamic data may be useful:
DHof HCl(g) = -92.30 kJ/mol
DHof H2O(g) = -241.82 kJ/mol
DHof AlCl3(s) = -705.60 kJ/mol
Al2O3(s) + 6HCl(g) -> 3H2O(g) + 2AlCl3(s)
DHof Al2O3 = -1670 kJ/mol
##### Explanation
The balanced reaction is:
Al2O3(s) + 6HCl(g) -> 3H2O(g) + 2AlCl3(s)
The ethalpy of reaction can be determined using Hess's Law.
DHrxn = S n DHf(products) - S m DHf(reactants)
where m and n are the stoichiometric coefficients of the reactants and products, respectively, of the chemical reaction.
DHrxn = 3 * DHf H2O(g) + 2 * DHf AlCl3(s) - (1 * DHf Al2O3(s) + 6 * DHf HCl(g))
86.94 kJ = 3 mol * (-241.82 kJ/mol) + 2 * (-705.60 kJ/mol) - {1 mol * (? kJ/mol) + 6 * ( -92.30 kJ/mol) }
DHrxn = -1670 kJ/mol
9. #### Determine Heat Released in Combustion of Methane

The combustion of methane, CH4, in oxygen results in the formation of liquid water and carbon dioxide. How much heat is released if 13.0 g of methane is burned in excess oxygen? The following thermodynamic data may be useful:
DHof CH4(g) = -74.80 kJ/mol
DHof H2O(l) = -285.83 kJ/mol
DHof CO2(g) = -393.5 kJ/mol
DHof O2(g) = 0 kJ/mol
723 kJ
##### Explanation
The balanced reaction is:

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)

The enthalpy of reaction can be determined using Hess's Law.
DHrxn = S nDHf(products) -S mDHf(reactants)
where m and n are the stoichiometric coefficients of the reactants and products, respectively, of the chemical reaction.
DHrxn =1 * DHf CO2(g) +2 * DHf H2O(l) -(1 * DHf CH4(g) +2 * DHf O2(g))
DHrxn =1 mol * (-393.5 kJ/mol) + 2 * (-285.83 kJ/mol) - {1 mol * (-74.80 kJ/mol) + 2 * ( 0 kJ/mol) }
DHrxn = -890.36 kJ/1 mol CH4 consumed
Amount of CH4 consumed = (13.0 g)/(16.0 g/mol) = 0.8125 mol
Amount of heat released = (-890.36 kJ/1 mol CH4)*(0.8125 mol CH4)
Amount of heat released = 723 kJ

## Spontaneity

1. #### Thermodynamic Quantities and Their Relation to Spontaneity

Consider the following statements regarding a spontaneous reaction at constant pressure and temperature. Determine whether each statement is true, false, or sometimes true.
• DSrxn > 0
• DSsurr > 0
• DGrxn = 0
• DHrxn < 0
• DStot = 0
• Sometimes true
• Sometimes true
• False
• Sometimes true
• False
##### Explanation
• For a spontaneous reaction, DStot = DSsurr + DSsys > 0. DSrxn is equivalent to DSsys and can take any value for a spontaneous reaction as long as DStot > 0.
• For a spontaneous reaction, DStot = DSsurr + DSsys > 0. DSsurr and can take any value for a spontaneous reaction as long as DStot > 0.
• DGrxn < 0 for a spontaneous reaction.
• Exothermic reactions will be spontaneous when the expression DG = DH -TDS is less than zero.
• The total entropy must increase for a spontaneous process, as defined by the second law of thermodynamics.
2. #### Thermodynamics of a Gliding Hockey Puck

An isolated system consists of a hockey puck on an ice rink. The hockey puck glides slowly on the ice until it comes to a complete stop. Consider the following statements regarding the change from initial to final state of the system. Determine whether each statement is true or false.
• The process is not spontaneous
• The total entropy of the system has not changed
• The kinetic energy of the puck was converted to heat.
• The total energy of the system has decreased.
• The process is reversible.
• False
• False
• True
• False
• False
##### Explanation
• Since the puck came to rest with no external work performed, the process is spontaneous.
• The total entropy has increased. Since the system is isolated, and the process spontaneous, the entropy change must be positive.
• The puck came to rest due to friction between the surface of the puck and surface of the ice. The kinetic energy of the puck is transformed to heat energy in the ice molecules.
• The total energy of the isolated system must be conserved. As noted above the kinetic energy of the puck was transformed to heat energy.
• The process is not reversible since the entropy change is greater than zero.

## Topic: Entropy

1. #### Primer on the Second Law of Thermodynamics

Consider the following statements regarding entropy and the second law of thermodynamics. Determine whether each statement is true, false, or sometimes true.
• For an irreversible process, DSuniv < 0.
• For a reversible process, DSsurr < 0
• The change in the entropy of a system undergoing a reversible process is equal to that of the same system undergoing an irreversible process, as long as the initial and final states are the same.
• Entropy is a measure of the disorder of a system.
• In an isolated system, the change in entropy is a negative value.
• False
• Sometimes true
• True
• True
• False
##### Explanation
• For an irreversible process, DSuniv must be greater than zero.
• A reversible process requires that DSuniv = 0. Since DSuniv = DSsys + DSsurr, there are some instances when a reversible process will have DSsurr < 0.
• Since entropy and change in entropy are state functions, the change in entropy for a given system is independent of the pathway. This is true if the pathway is reversible or irreversible.
• Entropy is associated with the extent of randomness of a system.
• For an isolated system, the entropy change must be positive. The change in entropy must increase for an isolated system based on the second law.
2. #### Determine Relative Entropy Between Two Systems

For each of the following pairs, determine the relative standard entropy between the two systems. Fill in the blank with the word greater, lower, or same.
• 1 mol SO3(g) at STP has a ... entropy than 1 mol O2(g) at STP.
• 1 mol Br2(l) has a ... entropy than 1 mol Br2(s).
• 1 mol Cl2(g) at 200 K and 1 atm has a ... entropy than 1 mol Cl2(g) at STP.
• 1 mol N2(g) in 22.4 L has a ... entropy than 1 mol N2(g) in 224 L.
• Greater
• Greater
• Lower
• Lower
##### Explanation
• The SO3 molecule has more bonds (3) than the O2 molecule (2), there it has more degrees of freedom of motion and greater entropy.
• The intermolecular interactions between neighboring molecules in the liquid phase are weaker, therefore, the Br2(l) system will have a greater entropy.
• The Cl2(g) molecules at the lower temperature will have a lower average kinetic energy, and hence a lower entropy.
• In the first system, the N2(g) molecules are held is a smaller container. Therefore, their translational degrees of freedom are more limited, and the system has a lower entropy.
3. #### Entropy Change for Reversible Expansion of an Ideal Gas

The volume of 0.100 mol of He(g) at 27 oC is increased isothermally from 2.0 L to 5.0 L. Assuming the gas to be ideal, calculate the entropy change for the process.
DS = 0.76 J/K
##### Explanation
The entropy change for a reversible process is defined as:
DS = qrev/T
for constant temperature.
Since the process is defined as isothermal (constant temperature)
DS = qrev/T
The heat released from the expansion of an ideal gas is:
qrev = -wrev = nRT ln(Vfinal/Vinitial)
qrev = (0.100mol)(8.314 J/mol-K)(300 K)ln(5.0 L/2.0 L)
qrev = +228.5 J
DS = (228.5 J)/(300 K)
DS = 0.76 J/K
4. #### Predict the Sign of the Entropy Change for a Set of Reactions

For each of the following reactions, predict whether DS is positive or negative.
• O(g) + NO2(g) -> NO3(g)
• 2Na(s) + Cl2(g) -> 2NaCl(s)
• H2O(g) -> H2O(l)
• 2POCl3(g) -> 2PCl3(g) + O2(g)
• 3O2(g) -> 2O3(g)
• Negative
• Negative
• Negative
• Positive
• Negative
##### Explanation
• The number of moles of gaseous species decreases from reactants to products, therefore, the entropy change of the system should be negative.
• Reactants in the solid and gaseous phases are converted to product in the solid phase only. The entropy change should therefore be negative.
• The phase change from gaseous to liquid form should lead to a negative entropy change.
• The number of moles of gaseous species increases from reactants to products, therefore, the entropy change of the system should be positive.
• The number of moles of gaseous species decreases from reactants to products, therefore, the entropy change of the system should be negative.
5. #### Calculate Entropy Change of System Associated with Phase Change

Calculate the entropy change when 1 mol of liquid mercury is transformed into a solid at a constant pressure of 1 atm. The following thermodynamic data may be useful:
DHfus = 2.289 kJ/mol
DHvap = 58.99 kJ/mol
Tf = -38.8 oC
Tb = 356.7 oC
-9.77 J/K
##### Explanation
DSrev = qrev/T
DSrev = DHfus/T
DHfus = -2.289 kJ/mol = -2289 J/mol
DSrev = (1 mol) (-2289 J/mol)/(-38.8 + 273.15)
DSrev = (1 mol) (-2289 J/mol)/(234.35 K)
DSrev = -9.77 J/K

## Topic: Free Energy and Equilibrium

1. #### Primer on Gibb's Free Energy

Consider the following statements related to the Gibbs free energy. Decide whether each statement is true or false.
• The term DS in the expression for the free energy is associated with the change in entropy of the system.
• The free energy of a gas depends on the absolute amount of gas, but not on its pressure.
• If DG for a reaction is zero, it occurs with the same tendency in both directions.
• At constant temperature and pressure, a reaction is always spontaneous if the change in the Gibbs free energy for the reaction is negative.
• A reaction which is spontaneous in the forward direction can never become spontaneous in the reverse direction, regardless of the relative concentrations of reactants and products.
• True
• False
• True
• True
• False
##### Explanation
• The term DS represents the entropy change of the system.
• The free energy under standard conditions for gases is 1 atm pressure. Deviations from standard conditions will lead to a change in the free energy.
• When DG = 0, the chemical reaction is in equilibrium. Therefore, the rates of the forward and reverse reactions are equal.
• If DG < 0, the reaction in the forward direction is spontaneous.
• Non-spontaneous reactions can become spontaneous if an external force is applied. A change in reactant or product concentration can cause such a disturbance.
2. #### Free Energy Change Associated with Combustion of Methane

Estimate the standard free energy change for the combustion of methane

CH4(g) + O2(g) -> H2O(g) + CO2(g)

assuming 1 mole of methane consumed. The following thermodynamic data may be useful: DHof CH4(g) = -74.80 kJ/mol
So CH4(g) = 186.3 J/K-mol
DGof CH4(g) = -50.8 kJ/mol
DHof O2(g) = 0 kJ/mol
So O2(g) = 205.0 J/K-mol
DGof O2(g) = 0 kJ/mol
DHof H2O(g) = -241.8 kJ/mol
So H2O(g) = 188.8 J/K-mol
DGof H2O(g) = -228.6 kJ/mol
DHof CO2(g) = -393.5 kJ/mol
So CO2(g) = 213.6 J/K-mol
DGof CO2(g) = -394.4 kJ/mol
DGo = -800.7 kJ
##### Explanation
The balanced reaction is:

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)

The standard free energy change for the reaction can be determined using
DGorxn = S nDGof(products) -S mDGof(reactants)
where m and n are the stoichiometric coefficients of the reactants and products, respectively, of the chemical reaction.
DGorxn =1 * DGof CO2(g) +2 * DGof H2O(g) -(1 * DGof CH4(g) +2 * DGof O2(g))
DGorxn =1 mol * (-394.4 kJ/mol) + 2 * (-228.6 kJ/mol) - {1 mol * (-50.8 kJ/mol) + 2 * ( 0 kJ/mol) }
DGorxn = -800.8 kJ/1 mol CH4 consumed
3. #### Free Energy Change Associated with Combustion of Methane at T = 6590 K

Estimate the free energy change for the combustion of methane

CH4(g) + O2(g) -> H2O(g) + CO2(g)

at T = 6590 K, assuming 1 mole of methane consumed. The following thermodynamic data may be useful:
DHof CH4(g) = -74.80 kJ/mol
So CH4(g) = 186.3 J/K-mol
DGof CH4(g) = -50.8 kJ/mol
DHof O2(g) = 0 kJ/mol
So O2(g) = 205.0 J/K-mol
DGof O2(g) = 0 kJ/mol
DHof H2O(g) = -241.8 kJ/mol
So H2O(g) = 188.8 J/K-mol
DGof H2O(g) = -228.6 kJ/mol
DHof CO2(g) = -393.5 kJ/mol
So CO2(g) = 213.6 J/K-mol
DGof CO2(g) = -394.4 kJ/mol
DG = -769.0 kJ
##### Explanation
Since the temperature is not 298 K, need to apply the relation:
DG = DH - TDS
to estimate the free energy change for the reaction.
The values DH and DS can be calculated for standard conditions. The balanced reaction is:

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)

The enthalpy of reaction can be determined using Hess's Law.
DHrxn = S nDHf(products) -S mDHf(reactants)
where m and n are the stoichiometric coefficients of the reactants and products, respectively, of the chemical reaction.
DHrxn =1 * DHf CO2(g) +2 * DHf H2O(g) -(1 * DHf CH4(g) +2 * DHf O2(g))
DHrxn =1 mol * (-393.5 kJ/mol) + 2 * (-241.8 kJ/mol) - {1 mol * (-74.80 kJ/mol) + 2 * ( 0 kJ/mol) }
DHrxn = -802.3 kJ for 1 mol CH4 consumed
The entropy change of reaction can be determined from the expression.
DSrxn = S nSf(products) -S mSf(reactants)
where m and n are the stoichiometric coefficients of the reactants and products, respectively, of the chemical reaction.
DSrxn =1 * Sf CO2(g) +2 * Sf H2O(l) -(1 * Sf CH4(g) +2 * Sf O2(g))
DHrxn =1 mol * (213.6 J/mol-K) + 2 * (188.83 J/mol-K) - {1 mol * (186.3 J/mol-K) + 2 * (205.0 J/mol-K) }
DSrxn = -5.04 J/K for 1 mol CH4 consumed
DG = DH - TDS
DG = (-802.3 kJ) - (6590 K)(-0.00504 kJ/K)
DG = -769.0 kJ
4. #### Dependency of Free Energy on Temperature

Consider the following statements regarding the dependency of free energy on temperature. Determine whether each statement is true or false.
• As temperature is increased, endothermic processes will go faster.
• The heat transfer to the surroundings is more important at low temperatures.
• The change in entropy of the system is more important at lower temperatures.
• A reaction which is spontaneous at a low temperature but becomes non-spontaneous at a higher temperature is " entropy driven "".
• An exothermic process in which solids turn into gases is spontaneous at all temperatures.
• True
• True
• False
• False
• True
##### Explanation
The questions are based on the free energy relation
DG = DH- TDS.
• An increase in temperature will increase the rate constant for both an endothermic and exothermic reaction,therefore, the reactions will go faster if the temperature is raised.
• At lower temperatures, the enthalpy term will dominate the temperature - free energy relationship. Hence the exothermic nature of the reaction is more important at lower temperatures.
• Since the free energy depends on the product of temperature and the entropy change, the change of entropy of the system is more important at higher temperatures.
• An entropy driven reaction is one that is non-spontaneous at low temperatures, and spontaneous at higher temperatures (where the entropy term will dominate).
• The process of solids turning to gases will have a positive entropy change, while the exothermic reaction has DH < 0. Therefore, DG will be less than zero for all values of temperature.
5. #### Free Energy and Equilibrium Relations

The relation between free energy and equilibrium is discussed in each statement below. Decide which statements are true or false.
• At equilibrium, the reaction quotient (Q) equals the equilibrium constant (K).
• DGo depends on initial reactant concentrations.
• When using K, which was calculated from DGo, the concentration of gaseous species can be reported in either atmospheres or molarity.
• DGo depends on temperature.
• A reaction cannot proceed without application of external work if DG is positive.
• True
• False
• False
• True
• True
##### Explanation
• Q=K at equilibrium.
• DGo represents the free energy change under standard conditions, and is therefore independent of reactant concentration.
• Gas concentrations must be reported in atm.
• DGo can be calculated for other temperatures
• If DG > 0, the reaction is not spontaneous. To make the reaction spontaneous, external work must be applied to the system.
6. #### Standard Free Energy Change Associated with Solubility Equilibrium

A sample of Mn(OH)2, a slightly soluble salt, is placed in pure water at room temperature (25.0 oC). When the solution reaches saturation, the water has a pOH = 4.14. Determine the standard free energy change associated with the equilibrium dissociation process.
DGo = 73 kJ/mol
##### Explanation
The equilibrium equation is:
Mn(OH)2 <-> Mn2+ + 2OH- while the equilibrium expression is:
Ksp = [Mn2=][OH-]2 The pOH value gives the [OH-] = 10-pOH
[OH-] = 10-4.14 = 7.24 x 10-5
The [Mn2+] = (1/2)[OH-] = 3.62 x 10-5
Ksp = (3.62 x 10-5)(7.24 x 10-5)2
Ksp = 1.89 x 10-13 Since the reaction is at equilibrium, DG = 0 and
DGo = -RT ln Ksp
DGo = -(8.314 J/mol-k)(298 K) ln (1.89 x 10-13)
DGo = +73 kJ/mol
7. #### Free Energy Change and Combustion of Methane at Equilibrium

Estimate the free energy change for the combustion of methane

CH4(g) + O2(g) -> H2O(g) + CO2(g)

at equilibrium, assuming 1 mole of methane consumed. The following thermodynamic data may be useful:
DHof CH4(g) = -74.80 kJ/mol
So CH4(g) = 186.3 J/K-mol
DGof CH4(g) = -50.8 kJ/mol
DHof O2(g) = 0 kJ/mol
So O2(g) = 205.0 J/K-mol
DGof O2(g) = 0 kJ/mol
DHof H2O(g) = -241.8 kJ/mol
So H2O(g) = 188.8 J/K-mol
DGof H2O(g) = -228.6 kJ/mol
DHof CO2(g) = -393.5 kJ/mol
So CO2(g) = 213.6 J/K-mol
DGof CO2(g) = -394.4 kJ/mol
DG = 0 kJ/mol
##### Explanation
For any process in equilibrium, the free energy change is zero.

## Topic: Balancing Redox Reactions

1. #### Assigning Oxidation Numbers in Redox Reactions

Consider the following statements regarding the redox reaction

Xe(g) + 2F2(g) -> XeF4(g)

Determine whether each statement is true or false
• The oxidation number of F in XeF4(g) is -1.
• The oxidation number of Xe in XeF4(g) is 0.
• In this reaction, Xe is oxidized.
• The oxidation number of Xe in Xe(g) is -8.
• The oxidation number of F in F2 is +7.
• True
• False
• True
• False
• False
##### Explanation
• Since F is a halogen, it will attain an oxidation number -1 in the molecule XeF4.
• The oxidation number of Xe in XeF4 will be +4.
• The oxidation number of Xe changes from +0 to +4 when going from reactants to products. Since the oxidation number of this element increases, it is oxidized.
• The oxidation number of any element is 0.
• The oxidation number of any element is 0.
2. #### Balancing a Redox Reaction in Acidic Solution

Complete and balance the following reaction that occurs in acidic solution.
MnO4-(aq) + CH3OH(aq) -> Mn2+ + HCO2H(aq)
12H+ + 4MnO4- + 5CH3OH -> 11H2O + 4Mn2+ + 5HCO2H
##### Explanation

MnO4-(aq) + CH3OH(aq) -> Mn2+ + HCO2H(aq)

Step 1: Identify the species that are oxidized and reduced:
Mn is reduced from Mn7+ in MnO4- to Mn2+
C is oxidized from C2- in CH3OH to C2+ in HCO2H
MnO4- + 5e- -> Mn2+
CH3OH -> HCO2H + 4e-
Step 2: Balance the number of electrons required for redox process:
4*[MnO4- + 5e- -> Mn2+]
5*[CH3OH -> HCO2H + 4e-]
4MnO4- + 20e- -> 4Mn2+
5CH3OH -> 5HCO2H + 20e-
Step 3: Add the half reactions:
4MnO4- + 5CH3OH -> 4Mn2+ + 5HCO2H
Step 4: Balance the charges by adding H+ as required (since acidic solution):
4 negative charges on reactant side
8 positive charges on the product side
Add 12 positive charges on the reactant side
12H+ + 4MnO4- + 5CH3OH -> 4Mn2+ + 5HCO2H
Step 5: Balance the H and O atoms by addition of water as required:
32 H and 21 O on reactant side
10 H and 10 O on product side
Add 11 water molecules on the product side
12H+ + 4MnO4- + 5CH3OH -> 11H2O + 4Mn2+ + 5HCO2H
Reaction is now complete and balanced.
3. #### Balancing a Redox Reaction in Basic Solution

Complete and balance the following reaction that occurs in basic solution.

H2O2(aq) + Cl2O7(aq) -> ClO2-(aq) + O2(g)

2OH- + Cl2O7 + 4H2O2 -> 5H2O + 2ClO2- + 4O2
##### Explanation

H2O2(aq) + Cl2O7(aq) -> ClO2-(aq) + O2(g)

Step 1: Identify the species that are oxidized and reduced:
Cl is reduced from Cl7+ in Cl2O7 to Cl3+ in ClO2-
O is oxidized from O1- in H2O2 to O0 in O2
Cl2O7 + 8e- -> 2ClO2-
H2O2 -> O2 + 2e-
Step 2: Balance the number of electrons required for redox process:
1*[Cl2O7 + 8e- -> 2ClO2-]
4*[H2O2 -> O2 + 2e-]

Cl2O7 + 8e- -> 2ClO2-
4H2O2 -> 4O2 + 8e-
Step 3: Add the half reactions:
Cl2O7 + 4H2O2 -> 2ClO2- + 4O2
Step 4: Balance the charges by adding OH- as required (since basic solution):
0 charges on reactant side
2 negative charge on the product side
Add 2 negative charge on the reactant side
2OH- + Cl2O7 + 4H2O2 -> 2ClO2- + 4O2
Step 5: Balance the H and O atoms by addition of water as required:
10 H and 17 O on reactant side
0 H and 12 O on product side
Add 5 water molecules on the product side
2OH- + Cl2O7 + 4H2O2 -> 5H2O + 2ClO2- + 4O2
Reaction is now complete and balanced

## Topic: Voltaic Cells

1. #### Voltaic Cell Composed of Cu and Zn Electrodes

Consider the following statements regarding a standard voltaic cell with copper (+0.34 V) and zinc (-0.76 V) electrodes. Determine whether each statement is true or false
• The zinc electrode undergoes oxidation.
• In the external circuit, negative ions flow from the copper electrode to the zinc electrode.
• The zinc electrode serves as the positive pole of the cell.
• The copper electrode serves as the cathode.
• In the salt bridge, negative ions flow from the copper electrode to the zinc electrode.
• True
• False
• False
• True
• True
##### Explanation
The voltaic cell will be a spontaneous redox reaction, where the cell potential is greater than 0. Given the two electrodes, a positive cell potential will be reduction occurs at the Cu electrode, and oxidation at the Zn electrode:

Cu2+ + 2e- -> Cu
Zn -> Zn2+ + 2e-

• As shown above, the Zn electrode will undergo oxidation.
• Electrons will flow in the external circuit from the Zn electrode to the Cu electrode.
• Since the source of electrons is the Zn electrode, it will serve as the negative pole of the cell.
• Reduction occurs at the cathode, therefore, the Cu electrode is the cathode.
• Since negative ions flow from the Zn electrode to the Cu electrode in the external circuit, they must flow from the Cu electrode to the Zn electrode in the salt bridge.
2. #### Voltaic Cell Composed of Cu and Ag Electrodes

Consider the following statements regarding a standard voltaic cell with copper (+0.34 V) and silver (+0.80 V) electrodes. Determine whether each statement is true or false.
• The copper ions undergo reduction.
• The cell potential is -0.46 V.
• In the external circuit, electrons flow from the anode to the cathode.
• The silver electrode serves as the anode.
• The copper electrode serves as the negative pole of the cell.
• False
• False
• True
• False
• True
##### Explanation
The voltaic cell will be a spontaneous redox reaction, where the cell potential is greater than 0. Given the two electrodes, a positive cell potential will be reduction occurs at the Ag electrode, and oxidation at the Cu electrode:

Ag+ + 1e- -> Ag
Cu -> Cu2+ + 2e-

• The Cu electrode will undergo oxidation, producing Cu2+ ions.
• The cell potential cannot be negative, since a voltaic cell is spontaneous. The potential is +0.46 V.
• Electrons will flow in the external circuit from the electrode undergoing oxidation (anode) to that undergoing reduction (cathode).
• Silver ions are being reduced at the silver electrode. Therefore, this electrode is the cathode.
• The Cu electrode is experiencing oxidation, and is the source of electrons in the external circuit. It is therefore the negative pole of the cell.
3. #### Cell Potential Calculation for Voltaic Cell

What would be the emf of a voltaic cell which consists of the two standard half cells below:

Ti4+(aq) + 1e- -> Ti3+(aq)     Eo = 0.00 V
Sn2+(aq) + 2e- -> Sn(s)    Eo = -0.14 V

Ecell = 0.14 V
##### Explanation
Ecell = Eored(cathode) - Eored(anode)
For a voltaic cell, Ecell must be greater than zero.
Therefore, the Ti reduction will occur at the cathode, and Sn(s) will be oxidized at the anode.
Ecell = 0.0 V - (-0.14 V)
Ecell = 0.14 V

## Topic: Cell Potential

1. #### Cell Potential and Free Energy Calculation

Calculate DGo for the reaction:

2Ti4+(aq) + Sn(s) -> Sn2+(aq) + 2Ti3+

given the half-cell reduction reactions

Ti4+(aq) + 1e- -> Ti3+(aq)     Eo = 0.00 V
Sn2+(aq) + 2e- -> Sn(s)    Eo = -0.14 V

DGo = -27.0 kJ/mol
##### Explanation
The reaction as written:

2Ti4+(aq) + Sn(s) -> Sn2+(aq) + 2Ti3+

has Ti4+ undergoing reduction and Sn(s) being oxidized.
The cell potential is:
Ecell = Eored(cathode) - Eored(anode)
Ecell = 0.0 V - (-0.14 V)
Ecell = 0.14 V
The standard free energy change can be calculated from the cell potential using the relation
DGo = -nFEo
DGo = -(2 mol e)(96485 J/V-mol)(0.14 V)
DGo = -27000 J
2. #### Predict the Spontaneity of a Set of Reactions

Which of the following reactions are spontaneous?
• 2Cl-(aq) + I2(s) -> Cl2(g) + 2I-(aq)
• 2LiF(s) -> 2Li(s) + F2(g)
• 2Ag(s) + Zn(NO3)2(aq) -> 2AgNO3(aq) + Zn(s)
• S(s) + 2Mn2+(aq) + 3H2O(l) -> H2SO3(aq) + Mn(s) + 4H+(aq)
• Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)
• Cu(s) + 2H+(aq) -> Cu2+(aq) + H2(g)
The only reaction that is spontaneous is:
Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)
##### Explanation
For an redox reaction to be spontaneous, the value Ecell > 0. A table of half-cell potential is available in the textbook on p. 1128.
2Cl-(aq) + I2(s) -> Cl2(g) + 2I-(aq)
2Cl- -> Cl2 + 2e-     Eored = +1.359 V
I2 + 2e- -> 2I-     Eored = +0.536 V
Eocell = Eored (cathode) - Eored (anode)
Eocell = (+0.536 V) - (+1.359 V) = -0.823 V
2LiF(s) -> 2Li(s) + F2(g)
2Li+ + 2e- -> 2Li(s)     Eored = -3.05 V
2F- -> F2 + 2e-     Eored = +2.87 V
Eocell = Eored (cathode) - Eored (anode)
Eocell = (-3.05 V) - (+2.87 V) = -5.92 V
2Ag(s) + Zn(NO3)2(aq) -> 2AgNO3(aq) + Zn(s)
Ag -> Ag+ + 1e-     Eored = +0.799 V
Zn2+ + 2e- -> Zn     Eored = -0.763 V
Eocell = Eored (cathode) - Eored (anode)
Eocell = (-0.763 V) - (+0.799 V) = -1.535 V
S(s) + 2Mn2+(aq) + 3H2O(l) -> H2SO3(aq) + Mn(s) + 4H+(aq)
S + + 3H2O -> H2SO3 + 4H+ + 4e-     Eored = +0.45 V
Mn2+ + 2e- -> Mn     Eored = -1.18 V
Eocell = Eored (cathode) - Eored (anode)
Eocell = (-1.18 V) - (+0.45 V) = -1.63 V
Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)
Zn(s) -> Zn2+ + 2e-     Eored = -0.763 V
Cu2+ + 2e- -> Cu     Eored = +0.337 V
Eocell = Eored (cathode) - Eored (anode)
Eocell = (+0.337 V) - (-0.763 V) = +1.10 V
Cu(s) + 2H+(aq) -> Cu2+(aq) + H2(g)
Cu -> Cu2+ + 2e-     Eored = +0.337 V
2H+ + 2e- -> H2     Eored = 0 V
Eocell = Eored (cathode) - Eored (anode)
Eocell = (0 V) - (+0.337 V) = -0.337 V
3. #### Predict the Spontaneity of a Set of Reactions

Which of the following reactions are spontaneous?
• 2Cl-(aq) + I2(s) -> Cl2(g) + 2I-(aq)
• 2LiF(s) -> 2Li(s) + F2(g)
• 2Ag(s) + Zn(NO3)2(aq) -> 2AgNO3(aq) + Zn(s)
• S(s) + 2Mn2+(aq) + 3H2O(l) -> H2SO3(aq) + Mn(s) + 4H+(aq)
• Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)
• Cu(s) + 2H+(aq) -> Cu2+(aq) + H2(g)
The only reaction that is spontaneous is:
Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)
##### Explanation
For an redox reaction to be spontaneous, the value Ecell > 0. A table of half-cell potential is available in the textbook on p. 1128.
2Cl-(aq) + I2(s) -> Cl2(g) + 2I-(aq)
2Cl- -> Cl2 + 2e-     Eored = +1.359 V
I2 + 2e- -> 2I-     Eored = +0.536 V
Eocell = Eored (cathode) - Eored (anode)
Eocell = (+0.536 V) - (+1.359 V) = -0.823 V
2LiF(s) -> 2Li(s) + F2(g)
2Li+ + 2e- -> 2Li(s)     Eored = -3.05 V
2F- -> F2 + 2e-     Eored = +2.87 V
Eocell = Eored (cathode) - Eored (anode)
Eocell = (-3.05 V) - (+2.87 V) = -5.92 V
2Ag(s) + Zn(NO3)2(aq) -> 2AgNO3(aq) + Zn(s)
Ag -> Ag+ + 1e-     Eored = +0.799 V
Zn2+ + 2e- -> Zn     Eored = -0.763 V
Eocell = Eored (cathode) - Eored (anode)
Eocell = (-0.763 V) - (+0.799 V) = -1.535 V
S(s) + 2Mn2+(aq) + 3H2O(l) -> H2SO3(aq) + Mn(s) + 4H+(aq)
S + + 3H2O -> H2SO3 + 4H+ + 4e-     Eored = +0.45 V
Mn2+ + 2e- -> Mn     Eored = -1.18 V
Eocell = Eored (cathode) - Eored (anode)
Eocell = (-1.18 V) - (+0.45 V) = -1.63 V
Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)
Zn(s) -> Zn2+ + 2e-     Eored = -0.763 V
Cu2+ + 2e- -> Cu     Eored = +0.337 V
Eocell = Eored (cathode) - Eored (anode)
Eocell = (+0.337 V) - (-0.763 V) = +1.10 V
Cu(s) + 2H+(aq) -> Cu2+(aq) + H2(g)
Cu -> Cu2+ + 2e-     Eored = +0.337 V
2H+ + 2e- -> H2     Eored = 0 V
Eocell = Eored (cathode) - Eored (anode)
Eocell = (0 V) - (+0.337 V) = -0.337 V

## Topic: Nerst Equation

1. #### Potential of a Concentration Cell

Calculate the cell potential at T = 25 oC, assuming ideal solutions.

Fe | Fe2+ (1.43 x 10-5 M) || Fe2+ (5.00 x 10-4 M) | Fe

E = +0.046 V
##### Explanation
The cell potential can be determined from the Nernst equation
E = Eo - (0.0592/n) log Q at 298 K, where
Q is the reaction quotient
For reaction is written in shorthand notation, as discussed in lecture. Two half cell reactions are given, separated by a double line, which represents a salt bridge. The half cell reaction to the left is the oxidation of Fe(s) to Fe2+. The half cell reaction to the right is the reduction of Fe2+ to Fe(s):
Fe(s) -> Fe2+ (1.43 x 10-5 M) + 2e-
Fe2+ (5.00 x 10-4 M) + 2e- -> Fe(s)
The overall reaction is:
Fe(s) + Fe2+ (5.00 x 10-4 M) -> Fe2+ (1.43 x 10-5 M) + Fe(s)
The standard cell potential is Eo = 0, since the half-cell reactions are identical.
Q = [Fe2+ (1.43 x 10-5 M)]/[Fe2+ (5.00 x 10-4 M)]
Q = (1.43 x 10-5 M)/(5.00 x 10-4 M)
Q = 0.0286
E = +0.00 V - (0.0592/2) log (0.0286)
E = +0.046 V
2. #### EMF for a Coupled Hydrogen Electrodes at Different pH Values

A voltaic cell is constructed with two hydrogen electrodes, one with pH = 0.00 and the other with pH = 6.60, both under 1.00 atm of H2(g). Calculate the emf of the cell at room temperature.
E = +0.391 V
##### Explanation
The cell potential can be determined from the Nernst equation,
E = Eo - (0.0592/n) log Q at 298 K, where
Q is the reaction quotient
For the reaction as written, with the SHE is both the anode and cathode:
2H+ + 2e- -> H2
H2 -> 2H+ + 2e-
Since it is a voltaic cell, it has to be spontaneous, with E > 0. This will occur is the cell with the lower [H+] is the anode, since the H2 gase will be oxidized to form more [H+].
2H+(pH=0.00) + 2e- -> H2
H2 -> 2H+(pH=6.60) + 2e-
Q = [H+ (pH=6.60)]2/[H+ (pH=0.00)]2
Q = [10-6.60]2/[10-0.00]2
Q = 6.31 x 10-14
E = +0.00 V - (0.0592/2) log (6.31 x 10-14)
E = +0.391 V
3. #### Free Energy Change with Non Standard Voltaic Cell

When a copper electrode in 1.00 M Cu(NO3)2 solution is coupled to a standard hydrogen electrode at 25 oC, the cell potential is 0.34 V, with the SHE being the anode. Determine the free energy change when the Cu2+ concentration is reduced to 5.40 x 10-4 M.
DG = -46.9 kJ
##### Explanation
The free energy change can be determined from
DG = -nFE, where
n is the number of mol of electrons transferred
F is 96,485 Coulombs per mole of electroncs transferred
E is the cell potential
The cell potential can be calculated using the Nernst equation
E = Eo - (0.0592/n) log Q at 298 K, where
Q is the reaction quotient
For the reaction as written, with the SHE as the anode:
Cu2+ + 2e- -> Cu(s)
H2 -> 2H+ + 2e-
Cu2+ + H2 -> Cu(s) + 2H+
Q = [H+]2/[Cu2+]
E = +0.34 V - (0.0592/2) log 1/(5.4 x 10-4 M)
E = +0.243 V
DG = -nFE, where
DG = -(2 mol)(96485 J/V-mol)(+0.243 V)
DG = -46.9 kJ

## Topic: Electrolysis

1. #### Electrolysis Cell with C and Ag Electrodes

Consider the following statements regarding an electrolysis cell for coating silver with copper. The cell consists of graphite (chemically inert at standard conditions) and silver (+0.80 V) electrodes in a 1 M CuI2 solution. The standard reduction potential of I2 is +0.54 V and of Cu2+ is +0.34 V. Determine whether each statement is true or false.
• During the plating process, the copper ion concentration decreases because metallic copper is deposited at the cathode.
• The silver electrode serves as the positive pole of the cell.
• The minimum potential needed to drive this process is 0.46 V.
• The copper ions undergo reduction.
• The graphite electrode serves as the cathode.
• True
• False
• False
• True
• False
##### Explanation
An electrolysis cell is not spontaneous, and work is required to make the process proceed. Since Cu will be plated on the silver electrode, reduction is occuring at the silver electrode, and oxidation at the graphite electrode:

Cu2+ + 2e- -> Cu
2I- -> I2 + 2e-

• Copper ions are being reduced at the cathode.
• Cu is begin reduced at the Ag electrode making it the cathode. The cathode is the negative terminal in an electrolytic cell.
• The minimum potential needed is the potential difference between the iodine and copper half cells, which is 0.20 V.
• The copper ions are undergoing reduction to form Cu metal at the silver electrode.
• Oxidation is happening at the graphite electrode, therefore, this electrode is the anode.
2. #### Electrolysis Cell with Cu and Ag Electrodes

Consider the following statements regarding an electrolysis cell for coating silver with copper. The cell consists of copper (+0.34 V) and silver (+0.80 V) electrodes in a 1 M CuI2 solution. The standard reduction potential of I2 is +0.54 V and of Cu2+ is +0.34 V. Determine whether each statement is true or false.
• The copper electrode serves at the anode.
• During the plating process, copper ions are formed at the anode and metallic copper is deposited at the cathode.
• The minimum potential needed to drive this process is 0.46 V.
• The copper electrode serves as the positive pole of the cell.
• The copper electrode undergoes oxidation.
• True
• True
• False
• False
• True
##### Explanation
An electrolysis cell is not spontaneous, and work is required to make the process proceed. Since Cu will be plated on the silver electrode, reduction is occuring at the silver electrode, and oxidation at the copper electrode:

Cu2+ + 2e- -> Cu
Cu -> Cu2+ + 2e-

Note that the second reaction does not involve oxidation of the I- in solution, since the reaction has a greater reduction potential than that of Cu2+.
• Oxidation is occurring at the copper electrode, therefore, it is the anode.
• The copper electrode is being oxidized, producing Cu2+ ions. These ions are then being reduced at the silver cathode, forming copper metal. Note that the copper ion concentration does not change during the plating process.
• The minimum potential needed is the potential difference between the copper and copper half cells, which is 0 V.
• The copper electrode is the location of the oxidation, and will therefore be thenegative pole of the electrolysis cell.
• Oxidation is happening at the copper electrode, and copper ions are being released into solution.
3. #### Electrolysis of LiF in Aqueous Medium

What elements are produced in the electrolysis of LiF(aq)?
hydrogen gas is produced at the cathode
oxygen gas is produced at the anode
##### Explanation
The electrolysis of LiF will usually lead to the production of Li metal and molecular flourine
2LiF -> 2Li + F2
where the half cell reactions, with reduction potentials, are
Li+ + 1e- -> Li, Eored = -3.05 V
2F- -> F2 + 2e-, Eored = +2.87 V
However, the presence of water will provide competing reactions
2H20 + 2e- -> H2 + 2OH-, Eored = -0.83 V
2H20 -> O2 + 4H+ + 4e-, Eored = +1.23 V
Since the reduction of water to form H2 has a smaller reduction potential than the reduction of Li+, hydrogen gas will be formed at the cathode.
Since the oxidation of water to form O2 has a smaller reduction potential than the oxidation of F-, oxygen gas will be formed at the anode.
4. #### Electrolysis of LiF and Production of Lithium

Calculate the mass of Li produced in 8.8 hr by the electrolysis of molten LiF if the electrical current is 6.7 A?
15.3 g
##### Explanation
The electrolysis of molten LiF will produce Li metal and molecular fluorine
2LiF -> 2Li + F2
During the 8.8 hr, the number of coulombs of electrical charge that are passed into the cell is:
Coulombs = Amps x seconds
Coulombs = (6.7 A)(8.8 hr)(3600 s/hr) = 2.12 x 105
The number of moles of electrons that pass into the cell are:
mol e = (2.12 x 105 C)/(96485 C/mol e) = 2.20
The number of moles of Li produced is in a one-to-one ratio with the number of moles of electrons introduced, since the Li reduction half cell is:
Li+ + 1e- -> Li
mass of Li = (mol Li)(MW) = (2.20 mol)(6.941 g/mol) = 15.3 g

## Topic: Previous Exams (Spring 2010)

1. #### Voltaic Cell

A voltaic cell is constructed with a Cu electrode in a CuSO4 solution and a Ag electrode in a Ag2SO4 solution. The standard reduction potential, Eored, for Cu2+ is 0.337 V and the standard reduction potential for Ag+ is 0.799 V. On the sketch below clearly indicate the cathode and the anode and the reactions occurring at each one. Indicate how many electrons are transferred per reaction and the value of Eocell
V = 0.462 V
Anode rxn: Cu(s) -> Cu2+ + 2e-
Cathode rxn: 2Ag+ + 2e- -> 2Ag(s)
two electrons transferred
Content
2. #### Equivalence point

A 0.1 M solution of a weak base, A-, is titrated with HCl. At the equivalence point the pH of the solution was determined to be 5.4. Determine the Ka of the conjugate acid HA. (Ignore any volume changes when adding the HCl).
1.58 x 10-10
##### Explanation
At equivalence point, all A- is converted to HA.
 HA <=> H+ + A- i 0.1 M 0 0 c -x +x +x e 0.1 - x x x
Assume x is small
Ka = [H+][A-]/[HA] = x2/0.1 = [3.98x10-6]2/0.01
Ka = 1.58 x 10-10
3. #### Gibbs Free Energy

For the reaction

N2O4 <-> 2NO2

at what temperature will the P(NO2) = 0.667 atm and P(N2O4) = 0.333 atm? The following thermodynamic data may be useful: Hof (N2O4) = 9.66 kJ/mol, So (N2O4)= 304.3 J/Kmol, Hof (N2O) = 33.84 kJ/mol, So (N2O)= 240.45 J/Kmol
333 K
##### Explanation
ΔHo = 2*33.84 kJ/mol - 9.66 kJ/mol = 58.02 kJ/mol
ΔSo = 2*240.45 J/Kmol) - 304.3 J/Kmol = 176.6 J/Kmol
at equilibrium Δ = 0
ΔGo = ΔHo - TΔSo = -RTlnK
ΔHo/T - ΔSo = -RlnK
ΔHo/T = -RlnK + ΔSo
T = ΔHo/(-RlnK+ ΔSo)
T = 58020 J/mol/ (-8.3145 J/Kmol * ln (0.6672/0.333) + 176.6 J/Kmol)
T = 58020 J/mol / 174.2 J/Kmol
T = 333 K
4. #### Entropy

Determine whether each of the following statements is true or false.
1. According to the third law of thermodynamics, the entropy of a substance at 0 K is defined to be 10 J.
2. The entropy of the universe increases in a spontaneous process
3. Enthalpy and entropy are not state functions
4. For a system at equilibrium ΔG is negative
5. Spontaneous processes are irreversible
1. False
2. True
3. False
4. False
5. True
##### Explanation
1. The entropy at 0 K is 0.
2. This is the definition of a spontaneous process
3. Enthalpy and entropy are state functions
4. At equilibrium ΔGo is 0.
5. #### Solubility

Significant environmental damage can be caused by acidic mine drainage in both coal mining areas in the East and hard rock mining operations in the West. An illustrative example of the reactions in mine tailings concerns the oxidation of pyrite (fool’s gold).

4FeS2 (s) + 15O2 (g) + 2H2O(l) -> 4 Fe3+ (aq) + 4H+ (aq) + 8 SO42- (aq).

The resulting acidic conditions leach Fe3+ and other heavy metals into the surrounding watersheds. As the pH of the Fe3+ solution increases Fe(OH)3 will precipitate (Ksp = 2.7 x 10-39). What is the molar solubility of Fe(OH)3? What is the molar solubility if the pH of the solution is raised to 9?
##### Answwer
Solubility = 1 x 10-10
at pH = 9, Solubility = 2.7 x 10-24
##### Explanation
 Fe(OH)3 <=> Fe3+ + 3OH- i - 0 0 c - +x +3x e - x 3x
Ksp = [Fe3+][OH-]3 = [x][3x]3 = 27x4
2.7x10-39 = 27x4
molar solubility = 1 x 10-10

 Fe(OH)3 <=> Fe3+ + 3OH- i - 0 1x10-5 c - +x +3x e - x 1x10-5 + 3x
assume x is small
Ksp = 2.7x10-29 = [x][1x10-5]3 x = 2.7x10-24
6. #### Voltaic Cell

Determine whether each of the following statements regarding a voltaic cell constructed from a copper electrode in a CuSO4 (Eored = 0.33 V) solution and a Zn electrode in a ZnSO4 solution (Eored = -0.76 V) is true or false.
1. The reduction of Zn2+ occurs at the cathode.
2. In the external circuit the electrons flow from the cathode to the anode.
3. The cell potential is 1.09 V.
4. The redox reaction that occurs in this voltaic cell is non-spontaneous
5. The maximum obtainable work from this redox reaction does not depend on the cell potential.
1. False
2. False
3. True
4. False
5. False
##### Explanation
1. Zinc is oxidized at the anode
2. Electrons flow from the anode to the cathode
3. Eocell = 0.33 - (-0.76) V = 1.09
4. Since Eocell is positive the reaction is spontaneous
5. The obtainable work from the cell does depend on the concentration according to wmax = nFE
7. #### Concentration Cell

Between heartbeats the potential across the heart cell membrane must be at least –85 mV to prevent the risk of arrhythmia. The potential is maintained through the use of ion pumps in the membrane which selectively pump K+ into the cell and can be represented with the following equilibrium:

K+ (dilute, outside cell) <-> K+ (concentrated, inside cell)

Assume you can approximate the potential across the heart cell membrane as a concentration cell involving K+ ions using the given equilibrium equation. If the concentration of K+ ions inside the cell is 0.155 M, what concentration of K+ ions would you need outside the heart cell to maintain a potential of -0.085 V?
0.0057 M
##### Explanation
E = Eo - 0.0592/n * logQ
n = 1
Eo = 0, for a concentration cell
E = -0.0592/1 * log [K+,inside cell]/[K+, outside cell]
-0.085 = -0.0592 * log [0.155 M] / [x]
log [0.155 M]/[x] = 1.436
0.155 M/x = 27.278
x = 0.0057 M

## Topic: Previous Exams (Spring 2011)

1. #### Voltaic cell

A voltaic cell is constructed with a Cu electrode in a solution of CuSO4 and an Au electrode in a solution of Au3+ (aq). The standard reduction potential for Cu2+ is Eored = 0.34 V and for Au3+ is Eored = 1.50 V. Indicate the cathode and the anode reactions, how many electrons are transferred per reaction, and the value of Eocell
Anode: 3Cu(s) -> 3Cu2+(aq) + 6 e-
Cathode: 2 Au3+(aq) + 6e- -> 2Au(s)
Eocell = 1.16 V
six electrons transferred
##### Explanation
For a voltaic cell the cell potential has to be greater than 0 V. Based on the standard reduction potentials Au3+ is a better oxidizing agent (it will be reduced) than Cu2+. Therefore Au3+ will be reduced and Cu(s) will be oxidized making Cu the anode and Au the cathode.
Eocell = 1.5 - 0.34 = 1.16 V
Six electrons are transferred based on balancing the numcer of electrons released by the oxidation reaction and consumed by the reduction reaction.
2. #### Redox reactions

For the following redox reactions below indicate what is being oxidized and what is being reduced.
1. Xe (g) + 2 F2 (g) → XeF4 (g)
2. I2O5 (s) + 5 CO (g) → I2 (s) + 5 CO2 (aq)
1. Oxidized: Xe(g), Reduced: F2 (g)
2. Oxidized: CO, Reduced: I2O5
##### Explanation
1. Xe is changes from a 0 to a +4 oxidation state while the F atoms in F2 change from a 0 to a -1 oxidation state
2. The C in CO changes from a +2 to a +4 oxidation state while the I atoms in I2O5 changes from a +7 to a 0 oxidation state
3. #### Gibbs Free Energy

Answer the following questions concerning spontaneity as true or false.
1. For a spontaneous process the enthalpy is always negative.
2. A spontaneous process is reversible.
3. In a voltaic cell, a spontaneous movement of electrons from anode to cathode will occur if Eocell is negative.
4. A concentration cell is always at equilibrium since Eocell is always 0.
1. False
2. False
3. False
4. False
##### Explanation
1. The spontaneity of a process depends on the enthalpy and the entropy
2. All spontaneous processes are irreversible
3. If Eocell the process is not spontaneous
4. If the concentrations of the dilute and concentration solutions are not the same in a concentration cell it is not at equilibrium.
4. #### Entropy

Complete the following statements concerning entropy comparisons with the words greater than, less than, or equal to.
1. The entropy of 1 mol H2 at 0 K is ______________ the entropy of 1 mol CH4 at 0 K.
2. The entropy of 1 mol H2 (g) at STP is ______________ the entropy of 2 mol H2 (g) at STP.
3. The entropy of 1 mol SO3(g) at STP is ____________ the entropy than 1 mol O2(g) at STP
4. The entropy of 1 mol H2O (g) is _______________ the entropy of 1 mol H2O (l).
1. equal to
2. less than
3. greater than
4. greater than
##### Explanation
1. A 0 K the entropy of a substance is 0
2. The entropy increases with the number of molecules
3. The entropy increases with the number of atoms in a molecule
4. The entropy of a gas is higher than the entropy of a liquid
5. #### Corrosion

Answer the following questions concerning the corrosion process with either true or false. The following information may be useful: Eored (Cu2+) = 0.34 V, Eored (Fe2+) = -0.44 V , Eored (Sn2+) = -0.14 V, Eored (Zn2+) = -0.77 V
1. Corrosion is a redox reaction.
2. Zn is not an acceptable sacrificial anode to protect Fe.
3. Sn is not an acceptable sacrificial anode to protect Fe.
4. The cell potential corresponding to the rusting of Fe can be used to determine the rate of Fe oxidation.
1. True
2. False
3. True
4. False
##### Explanation
1. Corrosion is an electrochemical process. In the Fe example, Fe(s) -> Fe2+ + 2e-
2. The standard reduction potential of Zn is lower than for Fe indicating that it will oxidize before Fe and serve as a sacrificial anode.
3. The standard reduction potential of Sn is higher than for Fe indicating that Fe will oxidize before Sn so Sn wil not serve as a sacrificial anode.
4. The cell potential is a thermodynamic propertty and can't be used to determine reaction rates.
6. #### Electrolysis

Consider the following statements regarding an electrolysis cell for coating silver with copper. The cell consists of graphite (chemically inert at standard conditions) and silver (+0.80 V) electrodes in a 1 M CuI2 solution. The standard reduction potential of I2 is +0.54 V and of Cu2+ is +0.34 V. Answer true or false.
1. During the plating process the copper ion concentration decreases.
2. The minimum potential needed to drive this reaction is 0.46 V.
3. The copper ions undergo reduction
4. The graphite electrode serves as the cathode
1. True
2. False
3. True
4. False
##### Explanation
1. If copper is to be plated out of solution its concentration will decrease.
2. The cell potential for this reaction is Ecell = 0.34 - 0.54 = -0.2 V
3. Copper is reduced from an oxidation state of +2 to an oxidation state of 0.
4. The silver electrode is the cathode and the graphite electrode is the anode.
7. #### Ksp

A galvanic cell is constructed using a standard hydrogen electrode (SHE) half cell and a half cell containing a silver electrode and a solution in which Ag+ (aq), Cl- (aq), and AgCl (s) are in equilibrium. The [Cl-] is 1.00 x 10-3 M. The two half reactions occurring in the galvanic cell are:

Anode: H2 (g) + 2 H2O (l) -> 2 H+ (aq) + 2 e-
Cathode: 2 Ag+ (aq) + 2e- -> 2 Ag(s)

The cell potential is 0.397 V and the standard cell potential is 0.80 V and the overall reaction is

H2 (g, 1 atm) + 2 H2O (l) + 2 Ag+ (aq) → 2 H+ (aq, 1 M) + 2 Ag(s)

Calculate the value of Ksp for AgCl at 25 oC. Hint: Write the Ksp expression for AgCl and use the electrochemical reaction to solve for any unknown concentrations.
1.56 x 10-10
##### Explanation
AgCl (s) <=> Ag+ (aq) + Cl - (aq)
Ksp = [Ag+][Cl-]
Ecell = Eocell - 0.0592/n * log Q
0.397 = 0.8 - 0.0592/2 * log Q
log Q = 13.6
Q = 4.12 x 1013
Q = [H+]2/[H2][Ag+]2
Q = 1 / 1*[Ag+]2
[Ag+]2 = 2.43 x 10-14
[Aq+] = 1.56 x 10-7
Ksp = [1.56 x 10-7][1 x 10-3] = 1.56 x 10-10
8. #### Gibbs Free energy

Carbon monoxide is toxic because it binds more strongly to hemoglobin (Hgb) than O2 (g). Consider the following two reactions and standard Gibbs free energy changes.

Hgb (aq) + O2 (g) → HgbO2 (aq), ΔGo = -70 kJ
Hgb (aq) + CO (g) → HgbCO (aq), ΔGo = -80 kJ

What is the equilibrium constant for the following reaction at 25 oC.

HbgO2 (aq) + CO (g) → HbgCO (aq) + O2 (g), ΔGo = -10 kJ