Unit B2 FR Practice #2

Unit B2 FR Practice #2 Electrochemistry

Name_____________________Per___

1. A student is given a standard galvanic cell, represented above, that has a Cu electrode and a Sn electrode. As current flows through the cell, the student determines that the Cu electrode increases in mass and the Sn electrode decreases in mass.

(a) Identify the electrode at which oxidation is occurring. Explain your reasoning based on the student’s observations.

(b) As the mass of the Sn electrode decreases, where does the mass go?

(c) In the expanded view of the center portion of the salt bridge shown in the diagram below, draw and label a particle view of what occurs in the salt bridge as the cell begins to operate. Omit solvent molecules and use arrows to show the movement of particles.

(d) A nonstandard cell is made by replacing the 1.0 M solutions of Cu(NO3)2 and Sn(NO3)2 in the standard cell with 0.50 M solutions of Cu(NO3)2 and Sn(NO3)2 . The volumes of solutions in the nonstandard cell are identical to those in the standard cell.

Save this question (i) for Unit C

(i) Is the cell potential of the nonstandard cell greater than, less than, or equal to the cell potential of the standard cell?

Justify your answer.

(ii) Both the standard and nonstandard cells can be used to power an electronic device. Would the nonstandard cell power the device for the same time, a longer time, or a shorter time as compared with the standard cell? Justify your answer.

(e) In another experiment, the student places a new Sn electrode into a fresh solution of 1.0 M Cu(NO3)2.

(i) Using information from the table to the right, write a net-ionic equation for the reaction between the Sn electrode and the Cu(NO3)2 solution that would be thermodynamically favorable. Justify that the reaction is thermodynamically favorable.

Save this question (e ii) for Unit C

(ii) Calculate the value of ∆G° for the reaction. Include units with your answer.

Half Reaction Eº (V) Cu⁺ + e⁻ → Cu(s) 0.52 Cu²⁺ + 2 e⁻ → Cu(s) 0.34 Sn⁴⁺ + 2 e⁻ → Sn²⁺ 0.15 Sn²⁺ + 2 e⁻ → Sn(s) −0.14

1.0 M Cu(NO3)2

Voltmeter

KNO3 Salt Bridge

Sn

1.0 M Sn(NO3)2

Cu

Unit B FR Practice #2 Electrochemistry

(page # of # ) 2 14

ANSWER 1

(a) Remember if sometimes you feel at a loss while trying to answer a question, go back and read any extra information given, and be sure and look at the diagrams. Note also that the question ask you to “explain your reasoning based on the student’s OBSERVATIONS.”

Since the electrode is losing mass (student observed), the Sn atoms must be forming Sn²⁺ ions and this process is oxidation.

Alternatively you might say...

Since the cell is operating (student observed), the Eº must be positive and based on the Eº values given in the table later in the question, we can see that the Sn must be oxidized in the tin and copper combination.

#

(b) The atoms on the Sn electrode are going into the solution as Sn²⁺ ions.

It is important to note that the loss of mass dos NOT go to the copper electrode and certainly NOT because electrons are transferred to the copper electrode, which would be a dramatic overestimation of the contribution of electrons toward the mass of a substance.

(c) In the diagram shown above, you must show charges on the potassium and nitration ions, and you must NOT show any electrons, such as using the symbol e⁻ since electrons do NOT travel through the salt bridge.

Save this question (f) for Unit C

(d i) The voltage is the same. In the nonstandard reaction the reaction quotient, Q # has not changed because the concentrations of Sn²⁺ and Cu²⁺ are equal to each other in both cases. Compared to the standard cell, Q is not any further from or closer to equilibrium, K

Q is mathematically tied to the voltage through the Nernst Equation; # If the value of Q remains unchanged, then Ecell will remain the same. More specifically in this situation, when Q = 1 as it is in the standard and in the nonstandard situation, then log(1) = 0, causing Ecell = Eºcell.

(d ii) The nonstandard cell would power the device for a shorter time because there are less Cu²⁺ ions which would be exhausted more quickly.

Alternatively you could take an equilibrium angle and say...

The nonstandard cell would power the device for a shorter time because the reaction will reach Ecell = 0 more quickly.

(e i) Remember that in spite of the equations presented in the table, you MUST use the reactants given in the question. Cu²⁺, from the Cu(NO3)2 solution and Sn metal. This requires that you combine the second equation as given, and flip the last equation to an oxidation.

Cu²⁺ + Sn → Cu + Sn²⁺ and for this reaction E will be positive, and thus the reaction is thermodynamically stable.

#

OR you could say....

Cu²⁺ + Sn → Cu + Sn²⁺ and the cell observations from earlier parts of the question indicate that the reaction actually does occur when tin metal is oxidized, and Cu²⁺ ions are reduced, thus the Ecell must be positive.

Saving (e ii) for unit C

(e ii) # 


E_cell^o = Ered o + Eox

o + 0.48 = +0.34

( )

+ 0.14

( )

Q= ⎡⎣Sn²⁺⎤⎦

Cu²⁺

⎡⎣ ⎤⎦

⎛

⎝⎜ ⎞

⎠⎟

E_cell= Ecell

o −0.0592 n logQ

⎛⎝⎜ ⎞

⎠⎟

E_cell^o = E_red^o + E_ox^o + 0.48 = +0.34

( )

+ 0.14

( )

∆ Gº= −nFE_cell^o

∆ Gº= −2mol e⁻

mol_Rxn ×96, 485C

1mol e⁻ ×0.48J

1C = −93,000J / mol_Rxn= −93kJ / mol_Rxn

K⁺ NO₃⁻

Unit B2 FR Practice #2 Electrochemistry

Name_____________________Per___

2 Al(s) + 3 Zn2+(aq) → 2 Al3+(aq) + 3 Zn(s)

2. Respond to the following statements and questions that relate to the species and the reaction represented above.

You should have learned (a) and (b) in first year…try them.

(a) Write the complete electron configuration (e.g., 1s²2s² . . .) for Zn²⁺.

(b) Which species, Zn or Zn²⁺, has the greater ionization energy? Justify your answer.

(c) Identify the species that is oxidized in the reaction.

The diagram to the right shows a galvanic cell based on the reaction. Assume that the temperature is 25°C.

(d) The diagram includes a salt bridge that is filled with a saturated solution of KNO3. Describe what happens in the salt bridge as the cell operates.

(e) Determine the value of the standard voltage, E°, for the cell.

Save this question (f) for Unit C

(f) Indicate whether the value of the standard free-energy change, ΔG°, for the cell reaction is positive, negative, or zero. Justify your answer.

(g) If the concentration of Al(NO3)3 in the Al(s)/Al3+(aq) half-cell is lowered from 1.0 M to 0.01 M at 25°C, does the cell voltage increase, decrease, or remain the same? Justify your answer.

1.0 M Al(NO3)3

Voltmeter

Salt Bridge

Zn

1.0 M Zn(NO3)2

NO3− K⁺ Al

NO3−

NO3−

NO3−

Al³⁺

NO3−

NO3−

Zn²⁺

Standard Reduction

Potentials (V)

at 298 K Zn²⁺ + 2 e⁻ → Zn(s) -0.76

Al³⁺ + 3 e⁻ → Al(s) -1.66

Unit B FR Practice #2 Electrochemistry

(page # of # ) 4 14

ANSWER 2

(a)

1s

²

2s

²

2p

⁶

3s

²

3p

⁶

3d

¹⁰

(b) While we haven’t gotten to unit H that covers ionization energy yet, you should remember this from first year chem. Before answering this question, you’ll need to remember that ionization energy is the energy needed to remove an electron.

Zn²⁺ has a higher ionization energy, because the electron being removed from Zn²⁺ would be in the third energy level − closer to the nucleus than the electron removed from a Zn atom which would be in the 4th energy level, 4s2. The effective nuclear charge felt by the electron removed would be greater for the electron in the 3rd energy level electron in Zn²⁺.

(c) Al

(d) As the cell operates, NO3− ions flow toward the Al half cell and K⁺ ions flow toward the Zn half cell in order to maintain the balance of charge.

(e) Eºcell = Eºred + Eºox Eºcell = (−0.76) + (+1.66 V) = 0.90 V

Save question (f) for unit C

(f) You can justify the sign of the ∆Gº by one of two different methods.

∆Gº is negative because the reaction is spontaneous.

OR you can justify with;

∆Gº is negative because Eº is positive and = −nF Eº will result in positive ∆Gº since n and F are both positive.

(g) You can mathematically justify with the Nernst Equation;

[Al³⁺] is a product of the reaction. Compared to standard conditions for which Q = 1, a reduction in the [Al³⁺] makes starting conditions presented in the problem, represented by Q, # less than 1, thus the log term in the Nernst

Equation, will become negative, which will cause Ecell to become more positive, thereby increasing the voltage.

OR you can justify with Q’s position relative to K (equilibrium)

[Al³⁺] is a product in the reaction. Compared to standard conditions for which Q = 1, a reduction in the [Al³⁺] makes the starting conditions presented in the problem, represented by Q, # , smaller than 1 and thus further from the

“dead battery,” or the equilibrium position, K, which is a number larger than 1 for a thermodynamically favorable reaction.

Therefore, the driving force for the reaction, represented by the cell voltage, Ecell , is increased.

Q= ⎡⎣Al³⁺⎤⎦² Zn²⁺

⎡⎣ ⎤⎦³

⎛

⎝⎜

⎜

⎞

⎠⎟

⎟ E_cell= E_cell^o −0.0592

n logQ

⎛⎝⎜ ⎞

⎠⎟

Q= ⎡⎣Al³⁺⎤⎦² Zn²⁺

⎡⎣ ⎤⎦³

⎛

⎝⎜

⎜

⎞

⎠⎟

⎟

Unit B FR Practice #2 Electrochemistry

(page # of # )5 14 5 Fe2+(aq) + MnO4−(aq) + 8 H+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(L)

3. A galvanic cell and the balanced equation for the spontaneous cell reaction are shown above. The two reduction half-reactions for the overall reaction that occurs in the cell

are shown in the table below.

(a) On the diagram to the right, clearly label the cathode.

(b) Calculate the value of the standard potential, E°, for the spontaneous cell reaction.

(c) How many moles of electrons are transferred when 1.0 mol of MnO4-(aq) is consumed in the overall cell reaction?

Save question (d) for Unit C

(d) Calculate the value of the equilibrium constant, Keq, for the cell reaction at 25°C. Explain what the magnitude of Keq tells you about the extent of the reaction.

Three solutions, one containing Fe2+(aq), one containing MnO4−(aq), and one containing H+(aq), are mixed in a beaker and allowed to react. The initial concentrations of the species in the mixture are 0.60 M Fe2+(aq), 0.10 M MnO4−(aq), and 1.0 M H⁺

(aq).

(e) When the reaction mixture has come to equilibrium, which species has the higher concentration, Mn2+(aq) or MnO4-(aq)? Explain.

(f) When the reaction mixture has come to equilibrium, what are the molar concentrations of Fe2+(aq) and Fe3+(aq)?

2010 AP^® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)

© 2010 The College Board.

Visit the College Board on the Web: www.collegeboard.com.

GO ON TO THE NEXT PAGE.

-7-

5 Fe²⁺(aq) + MnO₄^!(aq) + 8 H⁺(aq) → 5 Fe³⁺(aq) + Mn²⁺(aq) + 4 H₂O(l)

2. A galvanic cell and the balanced equation for the spontaneous cell reaction are shown above. The two reduction half-reactions for the overall reaction that occurs in the cell are shown in the table below.

Half-Reaction E° (V) at

298 K Fe³⁺(aq) + e^! " Fe²⁺(aq) + 0.77 MnO₄^!(aq) + 8 H⁺(aq) + 5 e^! →###Mn²⁺(aq) + 4 H₂O(l) +1.49

(a) On the diagram, clearly label the cathode.

(b) Calculate the value of the standard potential, E°, for the spontaneous cell reaction.

(c) How many moles of electrons are transferred when 1.0 mol of MnO₄^!(aq) is consumed in the overall cell reaction?

(d) Calculate the value of the equilibrium constant, K_eq, for the cell reaction at 25°C. Explain what the magnitude of K_eq tells you about the extent of the reaction.

Three solutions, one containing Fe²⁺(aq), one containing MnO₄^!(aq), and one containing H⁺(aq), are mixed in a beaker and allowed to react. The initial concentrations of the species in the mixture are 0.60 M Fe²⁺(aq) , 0.10 M MnO₄^!(aq) , and!1.0 M H⁺(aq) .

(e) When the reaction mixture has come to equilibrium, which species has the higher concentration, Mn²⁺(aq) or MnO₄^!(aq)? Explain.

(f) When the reaction mixture has come to equilibrium, what are the molar concentrations of Fe²⁺(aq) and Fe³⁺(aq)?

2010 AP^® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)

© 2010 The College Board.

Visit the College Board on the Web: www.collegeboard.com.

GO ON TO THE NEXT PAGE.

-7-

5 Fe²⁺(aq) + MnO₄^!(aq) + 8 H⁺(aq) → 5 Fe³⁺(aq) + Mn²⁺(aq) + 4 H₂O(l)

2. A galvanic cell and the balanced equation for the spontaneous cell reaction are shown above. The two reduction half-reactions for the overall reaction that occurs in the cell are shown in the table below.

Half-Reaction E° (V) at

298 K Fe³⁺(aq) + e^! " Fe²⁺(aq) + 0.77 MnO₄^!(aq) + 8 H⁺(aq) + 5 e^! →###Mn²⁺(aq) + 4 H₂O(l) +1.49

(a) On the diagram, clearly label the cathode.

(b) Calculate the value of the standard potential, E°, for the spontaneous cell reaction.

(c) How many moles of electrons are transferred when 1.0 mol of MnO₄^!(aq) is consumed in the overall cell reaction?

(d) Calculate the value of the equilibrium constant, K_eq, for the cell reaction at 25°C. Explain what the magnitude of K_eq tells you about the extent of the reaction.

Three solutions, one containing Fe²⁺(aq), one containing MnO₄^!(aq), and one containing H⁺(aq), are mixed in a beaker and allowed to react. The initial concentrations of the species in the mixture are 0.60 M Fe²⁺(aq) , 0.10 M MnO₄^!(aq) , and!1.0 M H⁺(aq) .

(e) When the reaction mixture has come to equilibrium, which species has the higher concentration, Mn²⁺(aq) or MnO₄^!(aq)? Explain.

(f) When the reaction mixture has come to equilibrium, what are the molar concentrations of Fe²⁺(aq) and Fe³⁺(aq)?

Unit B FR Practice #2 Electrochemistry

(page # of # ) 6 14

ANSWER 3

(a) The electrode on the the right is the cathode.

(b) Eºcell = Eºred + Eºox Eºcell = (+1.49 V) + (−0.77 V) = 0.72 V

(c) 5 moles of electrons are transferred when 1 mole of MnO4−(aq) is consumed.

Save question (d) for Unit C

(d) At equilibrium, ∆G = 0 and since ∆G = −RT lnK and ∆G = −nF Eº, thus −RT lnK = −nF Eº, solve for lnK

#

This very large K tells us that the reaction goes essentially to completion.

(e) [Mn²⁺] will be greater than [MnO4−] because as indicated by (d), the reaction essentially goes to completion and there is more than enough Fe²⁺ and H⁺ to react completely with the MnO4− which will be essentially zero when equilibrium is reached.

(f) At equilibrium, we can assume essentially all of the MnO4− is used, thus we can stoich over to how much Fe²⁺ is used and a simple subtraction will tell us how much Fe²⁺ is left. The the stoichiometry will tell us much Fe³⁺ is produced.

##

ln K= nFEº RT =

5 mol of e⁻

mol of reaction× 96,500 coul

mol of e⁻× 0.72Joule coul

8.31 Joule

mol of reaction K× 298K ln K= 140.3 K = 8.42 ×10⁶⁰

0.10M MnO⎡⎣ ₄⁻⎤⎦reacted× 5Fe³⁺

1MnO₄⁻= 0.50M Fe⎡⎣ ³⁺⎤⎦produced

0.10M MnO⎡⎣ ₄⁻⎤⎦reacted× 5Fe²⁺

1MnO₄⁻= 0.50M Fe⎡⎣ ²⁺⎤⎦reacted 0.60M Fe⎡⎣ ²⁺⎤⎦initial− 0.50M Fe⎡⎣ ²⁺⎤⎦reacted= 0.10M Fe⎡⎣ ²⁺⎤⎦leftOver

Unit B FR Practice #2 Electrochemistry

(page # of # )7 14 4. Answer the following questions about the diagram to the right.

(a) Write a balanced net ionic equation for the spontaneous reaction that takes place in the cell.

(b) Calculate the standard cell potential, Eº, for the reaction in part (a).

(c) In the diagram above,

(i) label the anode and cathode on the dotted lines provided, and

(ii) indicate, in the boxes below the half-cells, the concentration of AgNO3 and the concentration of Zn(NO3)2 that are needed to generate Eº.

(d) How will the cell potential be affected if KI is added to the silver half-cell? Justify your answer.

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Standard Reduction

Potentials (V)

at 298 K Ag⁺ + e⁻ → Ag(s) +0.80 Zn²⁺ + 2 e⁻ → Zn(s) -0.76

Unit B FR Practice #2 Electrochemistry

(page # of # ) 8 14

ANSWER 4

(a) In spite of the order of questions (a) and (b), the only way to know which metal is oxidized, and which metal is reduced, is by looking at the reduction potential values.

#

Thus we can tell the Zn gets oxidized: Zn + 2 Ag⁺ → Zn²⁺ + 2 Ag

(b) Eºcell = Eºred + Eºox Eºcell = (+0.80 V) + (+0.76 V) = +1.56 V

(c i) Zn is the anode and the Ag is the cathode.

(c ii) In part (b) we are asked for the standard cell potential, Eº, thus;

#

Save this question (d) for Unit C

(d) You can justify with Q’s position relative to K (equilibrium)

A precipitate will form as I⁻ ions react with [Ag⁺] ions in solution in the cathode compartment. [Ag⁺] is a reactant in the reaction. Compared to standard conditions for which Q = 1, a reduction in the [Ag⁺] makes the starting conditions presented in the problem, represented by Q # , greater than 1 and closer to the final equilibrium position, K, which is a number larger than 1.


Therefore, the driving force of the reaction, represented by the cell voltage, Ecell , is decreased.

OR you can mathematically justify with the Nernst Equation;

[Ag⁺] is a reactant in the reaction. Compared to standard conditions for which Q = 1, a reduction in the [Ag⁺] makes the starting conditions presented in the problem, represented by Q, # greater than 1, thus the log term in the

Nernst Equation, will be positive, which will reduce Ecell , thereby decreasing the voltage.

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Page 13

Q= ⎡⎣Zn²⁺⎤⎦

Ag⁺

⎡⎣ ⎤⎦²

⎛

⎝⎜

⎜

⎞

⎠⎟

⎟

Q=⎡⎣Zn²⁺⎤⎦

Ni²⁺

⎡⎣ ⎤⎦

⎛

⎝⎜ ⎞

⎠⎟ E_cell= E_cell^o −0.0592

n logQ

⎛⎝⎜ ⎞

⎠⎟

Unit B FR Practice #2 Electrochemistry

(page # of # )9 14 5. Answer the following questions about electrochemistry.

(a) Several different electrochemical cells can be constructed using the materials and the reduction potential table shown below. Write the balanced net-ionic equation for the reaction that occurs using the cell that would have the greatest positive value of Eºcell

(b) Calculate the standard potential, Eºcell, for the reaction written in part (a)

(c) A cell is constructed based on the reaction in part (a) above. Label the metal used for the anode on the cell shown in the figure to the right.

(d) Of the compounds NaOH, CuS, and NaNO3, which one is appropriate to use in a salt bridge? Briefly explain your answer, and for each of the other compounds, include a reason why it is not appropriate?

(e) Another standard cell is based on the following reaction.

Zn + Pb²⁺ → Zn²⁺ + Pb

If the concentration of [Zn²⁺] is decreased from 1.0 M to 0.25 M, would the cell potential increase, decrease, or remain the same? Justify your answer.

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Page 10

Standard Reduction

Potentials (V)

at 298 K Cu²⁺ + 2 e⁻ → Cu(s) +0.34

Fe²⁺ + 2 e⁻ → Fe(s) -0.44 Al³⁺ + 3 e⁻ → Al(s) -1.66

Unit B FR Practice #2 Electrochemistry

(page # of # ) 10 14

ANSWER 5

(a) Remember that one of the reactions must be flipped to an oxidation to result in the largest positive Eºcell.

#

(b) Eºcell = Eºred + Eºox Eºcell = (+0.34 V) + (+1.66 V) = +2.00 V

(c) Oxidation occurs at the anode, thus the metal will be aluminum.

(d) NaNO3 is appropriate for a salt bridge. NaNO3 is soluble in water and neither the cation or the anion will react with the any of the ions in the anode or cathode compartments.

NaOH is not appropriate. The anion, OH⁻, would migrate towards the anode compartment and would react with Al³⁺ in the solution.

CuS is not appropriate. CuS is insoluble in water, so there would be no free ions available to migrate to the anode or cathode compartments and balance the charge.

(e) You can mathematically justify with the Nernst Equation

[Zn²⁺] is a product of the reaction. Compared to standard conditions for which Q = 1, a reduction in the [Zn²⁺] makes the

# less than 1, thus the log term in the Nernst Equation, # will be negative, which

will cause Ecell to become more positive, thereby increasing the voltage.

OR you can you can justify with Q’s position relative to K (equilibrium)

[Zn²⁺] is a product in the reaction. Compared to standard conditions for which Q = 1, a reduction in the [Zn²⁺] makes the starting conditions presented in the problem, represented by Q # , smaller than 1 and further from the final equilibrium position, K which will be a number larger than 1.

Therefore, the driving force for the reaction, represented by the cell voltage, Ecell , is increased.

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Page 9

Q=⎡⎣Zn²⁺⎤⎦

Pb²⁺

⎡⎣ ⎤⎦

⎛

⎝⎜ ⎞

⎠⎟ E_cell= Ecell

o −0.0592 n logQ

⎛⎝⎜ ⎞

⎠⎟

Q=⎡⎣Zn²⁺⎤⎦

Pb²⁺

⎡⎣ ⎤⎦

⎛

⎝⎜ ⎞

⎠⎟

Unit B FR Practice #2 Electrochemistry

(page # of # )11 14 6. The diagram below shows the experimental setup for a typical electrochemical

cell that contains two standard half-cells. The cell operates according to the reaction represented by the following equation.

Zn(s) + Ni2+(aq) → Ni(s) + Zn2+(aq)

(a) Identify M and M²⁺ in the diagram to the right.

(b) specify the initial concentration for M²⁺ in solution.

(c) Indicate which of the metal electrodes is the cathode. Write the balanced equation for the reaction that occurs in the half- cell containing the cathode.

(d) What would be the effect on the cell voltage if the concentration of Zn²⁺ was reduced to 0.100 M in the half-cell containing the Zn electrode?

(e) Describe what would happen to the cell voltage if the salt bridge was removed. Explain.

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Unit B FR Practice #2 Electrochemistry

(page # of # ) 12 14

ANSWER 6

(a) The diagram indicates the electrons are flowing from left to right, from the anode to the cathode. Oxidation is occurring at the anode, and reduction is occurring at the cathode. According to the reaction given, the zinc is being oxidized, thus Metal, M must be zinc, and the M²⁺ solution must be Zn²⁺. Further, since we are told the cell contains two standard half cells, we know that the molarity must be 1 M. Standard cells are always 1 M at the start of the reaction.

(b) Metal X is the cathode where the reduction occurs, which according to the reaction is nickel.

The half reaction is: Ni²⁺ + 2e⁻ → Ni(s)

(c) You can justify with Q’s position relative to K (equilibrium)

[Zn²⁺] is a product in the reaction. Compared to standard conditions for which Q = 1, a reduction in the [Zn²⁺] makes the starting conditions presented in the problem, represented by Q # , smaller than 1 and further from the final equilibrium position, K which is a number larger than 1.

Therefore, the driving force for the reaction, represented by the cell voltage, Ecell , is increased.

OR you can mathematically justify with the Nernst Equation;

[Zn²⁺] is a product of the reaction. Compared to standard conditions for which Q = 1, a reduction in the [Zn²⁺] makes the starting conditions presented in the problem, Q, # smaller than 1 , thus the log term in the Nernst Equation,

will become negative, which will cause Ecell to become more positive, thereby increasing the voltage.

(d) The cell voltage will drop to zero when the salt bridge is removed. This happens because the salt bridge is needed to allow the charge to balance in the solutions. In the absence of the salt bridge, ions cannot flow into the anode compartment to balance the build up of cations and cannot flow into the cathode compartment to balance the reduction of cations (build up of negative charge).

Q=⎡⎣Zn²⁺⎤⎦

Ni²⁺

⎡⎣ ⎤⎦

⎛

⎝⎜ ⎞

⎠⎟

Q=⎡⎣Zn²⁺⎤⎦

Ni²⁺

⎡⎣ ⎤⎦

⎛

⎝⎜ ⎞

⎠⎟ E_cell= Ecell

o −0.0592 n logQ

⎛⎝⎜ ⎞

⎠⎟

Unit B FR Practice #2 Electrochemistry

(page # of # )13 14 7. Answer the following questions that refer to the galvanic cell shown in the

diagram to the right.

(a) Identify the anode of the cell and write the half reaction that occurs there.

(b) Write the net ionic equation for the overall reaction that occurs as the cell operates and calculate the value of the standard cell potential, Eºcell.

(c) Indicate how the value of Eºcell would be affected if the concentration of Ni(NO3)2(aq) was changed from 1.0 M to 0.10 M and the concentration for Zn(NO3)2(aq) remained at 1.0 M. Justify your answer.

(d) Specify whether the value of Keq for the original cell reaction is less than 1, greater than 1, or equal to 1. Justify your answer.

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Standard Reduction

Potentials (V)

at 298 K Ni²⁺ + 2 e⁻ → Ni(s) -0.25 Zn²⁺ + 2 e⁻ → Zn(s) -0.76

Unit B FR Practice #2 Electrochemistry

(page # of # ) 14 14

ANSWER 7

(a) In spite of the order of questions (a) and (b), the only way to know which metal is oxidized, and which metal is reduced, is by looking at the reduction potential values.

The metal that gets oxidized will be the one with the more negative reduction potential anode. Thus the Zn will be the anode, the electrode on the right. The half reaction that occurs there will be Zn(s) → Zn2+(aq) + 2e⁻

(b) Zn(s) + Ni2+(aq) → Ni(s) + Zn2+(aq)

Eºcell = Eºred + Eºox Eºcell = (−0.25 V) + (+0.76 V) = +0.51 V

Save this question (c) for Unit C

(c) You can mathematically justify with the Nernst Equation;

[Ni²⁺] is a reactant in the reaction. Compared to standard conditions for which Q = 1, a reduction in the [Ni²⁺] makes the starting conditions presented in the problem, represented by Q # greater than 1 and the log term in the Nernst

Equation, will become positive, which will reduce Ecell , thereby decreasing the voltage.

OR you can justify with Q’s position relative to K (equilibrium)

[Ni²⁺] is a reactant in the reaction. Compared to standard conditions for which Q = 1, a reduction in the [Ni²⁺] makes the starting conditions presented in the problem, represented by Q # , greater than 1 and closer to the final equilibrium position, K, which is a number larger than 1.

Therefore, the driving force for the reaction, represented by the cell voltage, Ecell , is decreased.

Save this question (d) for Unit C

(d) K > 1 because Eºcell is positive, indicating the reaction is spontaneous, resulting in a K value larger than 1.

Q=⎡⎣Zn²⁺⎤⎦

Ni²⁺

⎡⎣ ⎤⎦

⎛

⎝⎜ ⎞

⎠⎟ E_cell= E_cell^o −0.0592

n logQ

⎛⎝⎜ ⎞

⎠⎟

Q=⎡⎣Zn²⁺⎤⎦

Ni²⁺

⎡⎣ ⎤⎦

⎛

⎝⎜ ⎞

⎠⎟