electrochemicalcells ii supplementalworksheet...

Name:____________________

Revised CR 1/16/14 © LaBrake & Vanden Bout 2014

Electrochemical Cells II – Supplemental Worksheet KEY All the electrochemical cells on this worksheet are the same ones on the first Electrochemical Cells worksheet. To make the work on this worksheet easier, refer to the work you did on the previous Electrochemical Cells worksheet. Experimental Observations of Electrochemical Cells

1. Consider the voltaic cell that contains standard Co2+/Co and Au3+/Au electrodes. The following experimental observations were noted: (1) Metallic gold plates out on one electrode and the gold ion concentration decreases around that electrode, and (2) The mass of the cobalt electrode decreases and the cobalt (II) ion concentration increases around that electrode.

a. Recall the diagram, overall balanced redox reaction, and the standard cell potential for this cell?

<Co(s)|Co2+(aq)||Au3+(aq)|Au(s)> E°cell = E°cathode – E°anode = 1.50 V – (–0.28 V) = 1.78 V

b. What is the standard reaction free energy for this cell?

€

ΔG˚= –nFE˚

ΔG˚= – 6 mol e –

mol rxn⎛ ⎝ ⎜ ⎞

⎠ ⎟ 96485 C

mol e –⎛ ⎝ ⎜ ⎞

⎠ ⎟ 1.78 J

C( )ΔG˚= –1030.46 kJ

mol rxn

Name:____________________


c. Calculate the equilibrium constant for the overall redox reaction of this cell at 25°C.

€

ΔG˚= –RT lnKΔG˚= –1030.46 kJ

mol rxn = –1030460 Jmol rxn

lnK = – ΔG˚RT

= −–1030460 J

mol rxn

(8.314 Jmol rxn⋅K )(298.15K)

= 415.7

K = e415.7

This is a very large K, this reflects the spontaneous nature of this reaction.

d. Calculate the emf at 25°C of this cell in which the concentration of Co2+ ions is 0.30 mol/L and that of the Au3+ ions is 0.0010 mol/L. Here we use the Nernst Equation. Since we are finding the emf at 25°C we can use the version of the Nernst Equation that includes the room temperature value into its constant:

€

Ecell = E˚− 0.05916n

logQ

Q =[Co2+]3

[Au3+]2=[0.30M]3

[0.0010M]2= 27000

Ecell =1.78V −0.059166mol e–

log(27000)

Ecell =1.74V

e. Assume you use this cell as a battery to power one of your electric devices

that draws 1 mA of current. If you ran the battery at a constant 1 mA for 3000 hours, how many grams of Co would be consumed? To figure this problem out, we will do a series of conversions: current and time ! charge ! moles of e– ! moles of Co ! grams of Co

€

1mA = 0.001Cs

C used = current × time =0.001C1s

⎛

⎝ ⎜

⎞

⎠ ⎟ ×

3600s1hr

⎛

⎝ ⎜

⎞

⎠ ⎟ × 3000hr( )

C used =10800C

mass Co = (10800C) ×1mol e–

96485C

⎛

⎝ ⎜

⎞

⎠ ⎟ ×

1mol Co2mol e–⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

59g Co1mol Co

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

mass Co = 3.30gCo

So 3.30 grams of cobalt would be consumed.

Name:____________________


2. Consider the electrolysis of molten calcium chloride with inert electrodes. The following experimental observations were noted: (1) Bubbles of pale green chlorine gas are produced at one electrode, and (2) Silvery white molten metallic calcium is produced at the other electrode.


<Pt(s)|Cl-‐(aq)|Cl2(g)||Ca2+(aq)|Ca(s)|Pt(s)> E°cell = E°cathode – E°anode = –2.76 V – (1.36 V) = –4.12 V


€

ΔG˚= –nFE˚

ΔG˚= – 2 mol e –


⎠ ⎟ 96485 C

mol e –⎛ ⎝ ⎜ ⎞

⎠ ⎟ –4.12 J

C( )ΔG˚= 795 kJ

mol rxn

c. Calculate the equilibrium constant for the overall redox reaction of this cell at

25°C.

Name:____________________


€

ΔG˚= –RT lnKΔG˚= 795 kJ

mol rxn = 795000 Jmol rxn

lnK = – ΔG˚RT

= −795000 J

mol rxn

(8.314 Jmol rxn⋅K )(298.15K)

= –320.7

K = e–320.7

This is a very small K, this reflects the non-‐spontaneous nature of this reaction.

d. How many hours are required to plate 12.00 g of metallic calcium from 1.00 M CaCl2 (aq) by using a current of 3.00 A? To figure this problem out, we will do a series of conversions: grams of Ca ! moles of Ca ! moles of e– ! charge ! charge and current ! time

€

time = 12.00gCa( ) ×1mol Ca40.1g Ca

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

2mol e–

1mol Ca

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

96485C1mol e–⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

1s3C⎛

⎝ ⎜

⎞

⎠ ⎟ ×

1hr3600s⎛

⎝ ⎜

⎞

⎠ ⎟

time = 5.347hr

It would require 5.3 hours to plate 12.00 g of metallic calcium from 1.00 M CaCl2 (aq) by using a current of 3.00 A.

e. Determine the volume (in liters, at STP) of chlorine gas that can be produced in this cell by using a current of 7.30 mA for 2.11 hours. To figure this problem out, we will do a series of conversions: current and time ! charge ! moles of e-‐ ! moles of Co ! liters of Cl2

€

current = 7.30mA = 0.0073 Cs

time = 2.11hr × 3600s1hr

⎛

⎝ ⎜

⎞

⎠ ⎟ = 7596s


⎛

⎝ ⎜

⎞

⎠ ⎟ × (7596s)

C used = 55.45C

volume Cl2 = (55.45C) ×1mol e–

96485C

⎛

⎝ ⎜

⎞

⎠ ⎟ ×

1mol Cl22mol e–

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

22.4L Cl21mol Cl2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

volume Cl2 = 6.44 ×10−3L

In this situation 6.44x10-‐3L or 6.44 mL Cl2 would be produced.

Name:____________________


Short Hand Notation of Electrochemical Cells 3. Consider the following cell <Ni(s)|Ni2+(aq)||Ag+(aq)|Ag(s)>


E°cell = E°cathode – E°anode = 0.80 V – (–0.23 V) = 1.03 V


€

ΔG˚= –nFE˚

ΔG˚= – 2 mol e –


⎠ ⎟ 96485 C

mol e –⎛ ⎝ ⎜ ⎞

⎠ ⎟ 1.03 J

C( )ΔG˚= −198.76 kJ

mol rxn

c. Calculate the equilibrium constant for the overall redox reaction of this cell at

25°C.

€

ΔG˚= –RT lnKΔG˚= −198.76 kJ

mol rxn = −198760 Jmol rxn

lnK = – ΔG˚RT

= −−198760 J

mol rxn

(8.314 Jmol rxn⋅K )(298.15K)

= 80.18

K = e80.18 = 6.65 ×1034

This is a very large K, this reflects the spontaneous nature of this reaction.

Name:____________________


d. What is the concentration of Ni2+ ions if the emf at 25°C of this cell is +1.68 V and the concentration of Ag+ ions is 0.0280 mol/L? Here we use the Nernst Equation. Since we are finding the emf at 25°C we can use the version of the Nernst Equation that includes the room temperature value into its constant:

€

Ecell = E˚− 0.05916n

logQ

Q =[Ni2+][Ag+]2

=x

[0.0280M]2

1.68 =1.03V −0.059162mol e–

log x[0.0280M]2⎛

⎝ ⎜

⎞

⎠ ⎟

log x[0.0280M]2⎛

⎝ ⎜

⎞

⎠ ⎟ = −21.97

x[0.0280M]2

=1.06 ×10−22

x = [Ni2+] = 8.32 ×10−26M

e. How many hours would it take for this galvanic cell to plate 30.00 g of

metallic silver from 1.00 M solutions of Ni2+(aq) and Ag+(aq) assuming it produces a constant current of 2.25 A? To figure this problem out, we will do a series of conversions: grams of Ag ! moles of Ag ! moles of e-‐ ! charge ! current ! time

€

2.25A = 2.25 Cs

time = 30.00g Ag( ) × 1mol Ag107.87g Ag

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

1mol e–

1mol Ag

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

96485C1mol e–⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

1s2.25C⎛

⎝ ⎜

⎞

⎠ ⎟ ×

1hr3600s⎛

⎝ ⎜

⎞

⎠ ⎟

time = 3.3hours

It would require 3.3 hours to plate 30.00 g of metallic silver from 1.00 M solutions of Ni2+(aq) and Ag+(aq) assuming it produces a constant current of 2.25 A.

Name:____________________


4. Consider the following cell <Pt(s)| Ce3+(aq)|Ce4+(aq)||Cu2+(aq) |Cu(s)> a. Recall the diagram, overall balanced redox reaction, and the standard cell

potential for this cell?

E°cell = E°cathode – E°anode = 0.34 V – (1.70 V) = –1.36 V


€

ΔG˚= –nFE˚

ΔG˚= – 2 mol e –


⎠ ⎟ 96485 C

mol e –⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1.36 J

C( )ΔG˚= 262.44 kJ

mol rxn


€

ΔG˚= –RT lnKΔG˚= 262.44 kJ


lnK = – ΔG˚RT

= −262440 J

mol rxn

(8.314 Jmol rxn⋅K )(298.15K)

= –105.87

K = e–105.87 ≈1.0 ×10−46


Name:____________________


d. What is the concentration of Ce4+ ions if the emf at 25°C of this cell is –1.20 V, the concentration of Cu2+ ions is 0.60 mol/L and the concentration of Ce3+ ions is 0.30 mol/L? Here we use the Nernst Equation. Since we are finding the emf at 25°C we can use the version of the Nernst Equation that includes the room temperature value into its constant:

€

Ecell = E˚− 0.05916n

logQ

Q =[Ce4+]2

[Cu2+][Ce3+]2=

x 2

[0.60M][0.30M]2=

x 2

0.054

−1.20V = –1.36V −0.059162mol e–

log x 2

0.054⎛

⎝ ⎜

⎞

⎠ ⎟

log x 2

0.054⎛

⎝ ⎜

⎞

⎠ ⎟ = −5.409

x 2

0.054= 3.90 ×10−6

x = [Ce4+] = 4.59 ×10−4M

e. Determine the mass (in grams) of metal copper that can be produced in this

cell by using a current of 5.0 A for 2.7 days. To figure this problem out, we will do a series of conversions: current and time ! charge ! moles of e– ! moles of Cu ! grams of Cu

€

5.0A = 5.0 Cs


⎛

⎝ ⎜

⎞

⎠ ⎟ ×

3600s1hr

⎛

⎝ ⎜

⎞

⎠ ⎟ ×

24hr1day⎛

⎝ ⎜

⎞

⎠ ⎟ × 2.7days( )

C used =1166400C

mass Cu = (1166400C) ×1mol e–

96485C

⎛

⎝ ⎜

⎞

⎠ ⎟ ×

1mol Cu2mol e–⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

63.55g Cu1mol Cu

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

mass Cu = 384gCu

So 384 g Cu would be produced.

Name:____________________


Electrochemical Cell Diagrams 5. Consider the following cell:


<Cu(s)|Cu2+(aq)||Zn2+(aq)|Zn(s)> E°cell = E°cathode – E°anode = –0.76 V – (0.34 V) = –1.10 V


Name:____________________


€

ΔG˚= –nFE˚

ΔG˚= – 2 mol e –


⎠ ⎟ 96485 C

mol e –⎛ ⎝ ⎜ ⎞

⎠ ⎟ −1.10 J

C( )ΔG˚= 212.27 kJ

mol rxn


€

ΔG˚= –RT lnKΔG˚= 212.27 kJ


lnK = – ΔG˚RT

= −212270 J

mol rxn

(8.314 Jmol rxn⋅K )(298.15K)

= –85.63

K = e–85.63 ≈ 6.5 ×10−38


d. What is the potential for this cell if the Cu2+ concentration is 5.8 x 10-‐3M and the Zn2+ concentration is 1.3 x 10-‐1 M ? Here we use the Nernst Equation. Since we are finding the emf at 25°C we can use the version of the Nernst Equation that includes the room temperature value into its constant:

€

Ecell = E˚− 0.05916n

logQ

Q =[Cu2+][Zn2+]

=[5.8 ×10−3M][1.3 ×10−1M]

= 0.0446

Ecell = –1.10V −0.059162mol e–

log 0.0446( )

Ecell = −1.06V

e. Determine the mass (in grams) of zinc metal that can be produced in this cell

by using a current of 6.0 A for 1.5 days. To figure this problem out, we will do a series of conversions: current and time ! charge ! moles of e– ! moles of Zn ! grams of Zn

Name:____________________


€

6.0A = 6.0 Cs


⎛

⎝ ⎜

⎞

⎠ ⎟ ×

3600s1hr

⎛

⎝ ⎜

⎞

⎠ ⎟ ×

24hr1day⎛

⎝ ⎜

⎞

⎠ ⎟ × 1.5days( )

C used = 777600C

mass Zn = (777600C) ×1mol e–

96485C

⎛

⎝ ⎜

⎞

⎠ ⎟ ×

1mol Zn2mol e–⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

65.38g Zn1mol Zn

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

mass Zn = 263.46gZn

So 263.5 g Zn would be produced 6. Consider the following cell:


<Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s)> E°cell = E°cathode – E°anode = 0.34 V – (–0.76 V) = 1.10 V

Name:____________________



€

ΔG˚= –nFE˚

ΔG˚= – 2 mol e –


⎠ ⎟ 96485 C

mol e –⎛ ⎝ ⎜ ⎞

⎠ ⎟ 1.10 J

C( )ΔG˚= −212.27 kJ

mol rxn


€

ΔG˚= –RT lnKΔG˚= −212.27 kJ

mol rxn = −212270 Jmol rxn

lnK = – ΔG˚RT

= −−212270 J

mol rxn

(8.314 Jmol rxn⋅K )(298.15K)

= 85.63

K = e85.63 ≈1.5 ×1037

This is a very large K, this reflects the spontaneous nature of this reaction. d. Calculate the emf at 25°C of this cell in which the concentration of Cu2+ ions is

0.1250 mol/L and that of the Zn2+ ions is 0.0020 mol/L. Here we use the Nernst Equation. Since we are finding the emf at 25°C we can use the version of the Nernst Equation that includes the room temperature value into its constant:

€

Ecell = E˚− 0.05916n

logQ

Q =[Zn2+][Cu2+]

=[0.0020M][0.1250M]

= 0.016

Ecell = –1.10V −0.059162mol e–

log 0.016( )

Ecell = −1.15V

e. How many hours would it take for this galvanic cell to plate 15.00 g of

metallic copper from 1.00 M solutions of Cu2+(aq) and Zn2+(aq) assuming it produces a constant current of 5.0 A? To figure this problem out, we will do a series of conversions: grams of Cu ! moles of Cu ! moles of e-‐ ! charge ! current ! time

€

5.0A = 5.0 Cs

time = 15.00g Cu( ) × 1mol Cu63.55g Cu

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

2mol e–

1mol Cu

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

96485C1mol e–⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ ×

1s5.0C⎛

⎝ ⎜

⎞

⎠ ⎟ ×

1hr3600s⎛

⎝ ⎜

⎞

⎠ ⎟

time = 2.53hours

It would require 2.53 hours to plate 15.00 g of metallic copper from 1.00 M solutions of Cu2+(aq) and Zn2+(aq) assuming it produces a constant current of 5.0 A.

electrochemicalcells ii supplementalworksheet...

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