Experiment 4 Electrochemical Cells PDF

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EXPERIMENT 4: GALVANIC CELLS, THE NERNST EQUATION

De Vera, Ar Jay B. Gonzales, Joshua C. Lucas, Ana Lowela L. Sison, Herschel Annicole S. Velaso, Louisse CpE B12

General Chemistry Instructor (CM011L): Sir Rommel Galvan

Mapua University

Abstract Electrochemistry deals with the study of the relationship between electricity and chemical processes. Electricity can be created by the consecutive movement of electrons in a chemical reaction which is called the oxidation-reduction or redox reaction. Electrochemical cells are used to control the movement of electrons in a reaction. It is divided into two types, the galvanic cell and electrolytic cell, both utilizing the principle of oxidation–reduction or redox reactions. The objective of this experiment is to identify the effect of concentration changes on cell potential and determine the unknown concentration of an element using the Nernst equation. The group was able to obtain the results accurately with precision by using various laboratory instruments and following multiple instructions to conduct the experiment.

Experiment 4 Galvanic Cells, the Nernst Equation Electrochemical cells are of two types, galvanic and electrolytic, both employing the principle of oxidation–reduction (redox) reactions. In galvanic or voltaic cells, redox reactions occur spontaneously as is common with all portable batteries of which we are very familiar. Electric cars, flashlights, watches, and power tools operate because of a specific spontaneous redox reaction. Electrolytic cells are driven by nonspontaneous redox reactions, reactions that require energy to occur. The recharging of batteries, electroplating and refining of metals, and generation of various gases all require the use of energy to cause the redox reaction to proceed. Experimentally, when copper wire is placed into a silver ion solution, copper atoms spontaneously donate electrons, when copper atoms are oxidized, to the silver ions which are reduced. Silver ions migrate to the copper atoms to pick up electrons and form silver atoms at the copper metal–solution interface; the copper ions that form then move into the solution away from the interface.

Methods Materials and Procedures Materials used in the experiment: 1. Collect the electrodes, solutions, and equipment. Obtain four small 50 mL beakers and fill them three-fourths full of the 0.1 M solutions. Share these solutions with other chemists or groups of chemists in the laboratory. Polish strips of copper, zinc, magnesium, and iron metal with steel wool or sandpaper, rinse briefly with dilute 0.1 M of HNO3, and rinse with deionized water. These polished metals, used as electrodes, should be bent to extend over the lip of their respective beakers. Check out a multimeter or a voltmeter with two electrical wires preferably a red and black wire attached to alligator clips. 2. Set up the copper–zinc cell. Place a Cu strip electrode in the CuSO4 solution and a Zn strip electrode in the Zn(NO3)2 solution. Roll and flatten a piece of filter paper; wet the filter paper with a 0.1 M KNO3 solution. Fold and insert the ends of the filter paper into the solutions in the two beakers. Set the multimeter to the 2000-mV range or as appropriate. Connect one electrode to the negative terminal of the multimeter and the other to the positive terminal. 3. Determine the copper–zinc cell potential. If the multimeter reads a negative potential, reverse the connections to the electrodes. Read and record the positive cell potential. Identify the metal strips that serve as the cathode positive terminal and the anode. Write

an equation for the half-reaction occurring at each electrode. Combine the two halfreactions to write the equation for the cell reaction. 4. Repeat for the remaining cells. Determine the cell potentials for all possible galvanic cells that can be constructed from the four redox couples. Refer to the Report Sheet for the various galvanic cells. Prepare a new salt bridge for each galvanic cell. 5. Determine the relative reduction potentials. Assuming the reduction potential of the Zn2+(0.1 M)/Zn redox couple is –0.79 V, calculate the reduction potentials of all other redox couples. 6. Determine the reduction potential of the unknown redox couple. Place a 0.1 M solution and electrode obtained from your instructor in a small beaker. Determine the reduction potential, relative to the Zn2+(0.1 M)/Zn redox couple, for your unknown redox couple. Effect of Concentration Changes on Cell Potential 1. Effect of different molar concentrations. Set up the galvanic cell shown in Figure 32.5, using 1 M CuSO4 and 0.001 M CuSO4 solutions. Immerse a polished copper electrode in each solution. Prepare a salt bridge to connect the two half-cells. Measure the cell potential. Determine the anode and the cathode. Write an equation for the reaction occurring at each electrode. 2. Effect of complex formation. Add 2–5 mL of 6 M NH3 to the 0.001 M CuSO4 solution until any precipitate redissolves. Observe and record any changes in the halfcell and the cell potential. 3. Effect of precipitate formation. Add 2–5 mL of 0.2 M Na2S to the 0.001 M CuSO4 solution now containing the added NH3. What is observed in the half-cell and what happens to the cell potential? Record your observations.

The Nernst Equation and an Unknown Concentration 1. Prepare the diluted solutions. Prepare solutions 1 through 4 as shown in Figure 32.6 using a 1-mL pipet and 100-mL volumetric flasks.4 See Pre-laboratory Assignment, question 3. Be sure to rinse the pipet with the more concentrated solution before making the transfer. Use deionized water for dilution to the mark in the volumetric flasks. Calculate the molar concentration of the Cu2+ ion for each solution and record. 2. Measure and calculate the cell potential using small 50 mL beakers. 3. Measure and calculate the cell potentials. Repeat Part C.2 with solutions 3 and 2, respectively. A freshly prepared salt bridge is required for each cell. See data from Part A.3 for the potential of Solution 1. 4. Plot the data. Plot

Ecell , expt

and

Ecell , calc

(ordinate) versus pCu (abscissa) on the

same piece of linear graph paper or by using appropriate software for the four concentrations of CuSO4. Have your instructor approve your graph. 5. Determine the concentration of the unknown. Obtain a CuSO4 solution with an unknown copper ion concentration from your instructor and set up a like galvanic cell. Determine Ecell

as in Part C.2. Using the graph, determine the unknown copper(II) ion

concentration in the solution.

Calculations Prelaboratory 1. In a galvanic cell, a) oxidation occurs at the (name of electrode)

anode

b) the cathode is the (sign) electrode

positive

c) cations flow in solution toward the (name of electrode)

cathode

d) electrons flow from the (name of electrode) to (name of electrode)

anode to cathode

2. a. What is the purpose of a salt bridge? Explain. Answer: The purpose of the salt bridge is to make the half-cells electrically neutral within the internal circuit. Also, it prevents half-cells from reaching equilibrium at faster rate. b. How is the salt bridge prepared in this experiment? Answer: The salt bridge is prepared by twisting a piece of filter paper and dipping it in a 0.1M KNO3 solution. Both ends of the filter paper are then inserted into the two beakers. 3. Experimental Procedure, Part C.1. A 1-mL pipet is used to transfer 1.0 mL of a 0.10 M CuSO4 solution to a 100-mL volumetric flask. The volumetric flask is then filled to the mark with deionized water. See Figure 32.6. What is the molar concentration of the diluted solution? Show calculations expressing the concentration with the correct number of significant figures. Answer:

(0.1 M CuSo 4)(1 mL ) 100 mL

= 0.1 M

4. Refer to Figure 32.2 and equations 32.8 and 32.9.

a. What is the value of the cell constant (y-intercept)? Answer:

E ° cell = constant – 0.0592/2 (pCu) 0.95 = constant – 0.0592/2 (6) Constant = 1.1276

b. What is the [Cu2+] if the measured cell potential is 1.05 V? Answer: E ° cell = constant + 0.0592/2 (log[Cu2+]) 1.05 V = 1.1276 +0.0592/2 (log[Cu2+]) [Cu2+] = 2.39 mol/L c. What should be the cell potential if the [Cu2+] is 1.0 x 10–6 mol/L? Answer: E ° cell = constant + 0.0592/2 (log[Cu2+]) 1.1276 + 0.0592/2 (log 1.0 x 10-6 mol/L) = 1.04V

5. Consider a galvanic cell consisting of the following two redox couples: −¿ ¿ E° = +0.80 V e —› Ag(s) −¿ ° Cr3+(0.010 M) + 3 e¿ —› Cr(s) E = –0.74 V a. Write the equation for the half-reaction occurring at the cathode. Ag+(0.010 M) +

Cathode: Ag  Ag+ + eb. Write the equation for the half-reaction occurring at the anode. Anode: Cr  Cr3+ + 3ec. Write the equation for the cell reaction. 3 Ag+ + Cr  3 Ag + Cr3+

d. What is the standard cell potential,

°

Ecell

for the cell?

E °

Ag

E °

Cr=

E °

cell

=+0.80V -0.74V

= +0.80V – (- 0.74V) = 1.54 V

e. Realizing the nonstandard concentrations, what is the actual cell potential,

Ecell ,

for the cell? See equation 32.6. Hint: What is the value of n in the Nernst equation? Answer:

Ecell =

Ecell = 1.54V -

Ecell

-

0.0592 log Q n

Cr 3+ ¿ 3 ( Ag) log ¿ ] 0.0592 ¿ 2 0.010 M 3 (0.010 M ) ] 0.0592 ¿ 2

log Ecell = 1.54V -

Ecell = 1.4216 V 6. The extent of corrosion in the steel reinforcing rods (rebar) of concrete is measured by the galvanic cell shown in the diagram of the instrument. The half-cell of the probe is usually a AgCl/Ag redox couple: AgCl +

−¿ ¿ e

Ag + Cl– (1.0 M)

E

°

= +0.23 V

Corrosion is said to be severe if the cell potential is measured at greater than 0.41 V. Under

these conditions, what is the iron(II) concentration on the rebar? See equation

32.6. −¿ ¿ 2+¿+2 e ¿ Fe

Answer:

Fe

E

°

= -0.44 V



Fe

−¿ ¿ 2+¿+2 e ¿ Fe

Fe + 2 AgCl 

2+¿ Fe¿

+

2+¿ +2Cl – 2 Ag¿

¿ (Ag)2 log ¿ ] 0.0592 ¿ n

Fe 2+ Ecell =

Ecell -

¿ (1 x 10−5 )2 log ¿ ] 0.0592 ¿ 2

Fe 2+ 0.41 = 0.67 -

Fe2+ = 0.067 mol/L

Report Sheet Galvanic Cells, the Nernst Equation A. Reduction Potentials of Several Redox Couples Galvanic

Ecell Measured

Anode

Equation for Anode HalfReaction

Cathode

0.81V

Zn

Zn  Zn2+ + 2e-

Cu

Fe

Fe  Fe + 2e

Zn

Zn  Zn + 2e

Cell Cu–Zn

Equation for Cathode HalfReaction

Cu2+ + 2e--> Cu (s)

Cu–Fe

Zn–Fe

1.02V

0.2V

2+

2+

-

-

Cu

Cu2+ + 2e-

Fe

--> Cu (s) Fe2+ + 2e--> Fe

1. Write balanced equations for the six cell reactions. Cu–Zn: Cu + Zn2+  Zn + Cu2+ Cu-Fe: Fe + Cu2+  Fe2+ + Cu Zn-Fe: Zn + Fe2+  Zn2+ + Fe

2. Complete the table as follows: Galvanic Cell

Ecell Measured

For the Redox Couple

Cu–Zn

0.81V

Zn-Fe

0.2V

Zn-Zn

0

2+¿/Cu Cu¿ 2+¿/ Fe Fe¿ 2+¿/ Zn Zn¿

Reduction Potential (experimental ) 0.02

Reduction Potential (theoretical)

% Error

0.31

93.55%

-0.59

-0.47

25.53%

-0.79 V

-0.79

0.00

Calculation for the Reduction Potential (experimental) EM2+/M = Ecell Measured + (-0.79 V) Cu – Zn  0.81V + (-0.79V) = 0.02V Zn – Fe  0.2V + (-0.79V) = -0.59V % Error Calculation

Cu – Zn  % Error =

% Error =

experimental−theoretical theoretical 0.02 V −0.31 V 0.31 V

x 100

x 100

% Error = |- 93.55%| % Error = 93.55% Zn – Fe  % Error =

% Error =

experimental−theoretical theoretical

x 100

−0.59−(−0.47 V ) x 100 −0.47 V

% Error = |25.53 %| % Error = 25.53%

B. Effect of Concentration Changes on Cell Potential 1. Cell potential of concentration cell

0.02V

Anode half-reaction

[1M CuSO4 ] Cu(s)  Cu2+ (aq) + 2e-

Cathode half-reaction

[0.001 M CuSO4] Cu2+ (aq) + 2e-  Cu(s)

Explain why a potential is recorded. Answer: A potential is recorded because there is a difference in molar concentrations. The experiment done shows how the one with the higher concentrations gives away electrons to the one who has lesser concentrations. The result will make the cell potential to be positive.

C. The Nernst Equation and an Unknown Concentration

1. Complete the following table with the concentrations of the Cu (No3)2 the measured cell potentials,

Ecell , expt . Use equation 32.9 to determine

solutions and Ecell , calc .

Solution Number

Concentration of Cu (No3)2

Ecell , experimental

1

0.1 mol/L

0.965

2

0.001 M

0.896

-3

9.01

3

0.00001 M

0.842

-5

8.54

4

0.0000001 M

0.812

-7

8.30

2+¿ ], Cu¿ pCu -1

–log [

Ecell ,calculated

9.65

2. Account for any significant difference between the measured and calculated Ecell values.

Laboratory Questions 1. The filter paper salt bridge is not wetted with the 0.1 M

KNO3

solution. As a result,

will the measured potential of the cell be too high, too low, or unaffected? Explain. Answer: The result of the measured potential of the cell would be too low. If the filter paper salt bridge is not wetted, this charge difference would prevent further flow of electrons, because electrons leave one half of a galvanic cell and flow to the other, a difference in charge is established. On the other hand, A salt bridge maintain a balance in charge between reduction and oxidation vessels while keeping the contents of each, separately. With the balance of charge difference, electrons can flow once again, and the reduction and oxidation reactions can proceed.

Therefore, keeping the two cells separately is preferable form the point of view of eliminating variables from an experiment. 2. A positive potential is recorded when the copper electrode is the positive electrode. Is the copper electrode the cathode or the anode of the cell? Explain. Answer: The positive electrode is the cathode, wherein the reduction occurs, and electrons are gained. Therefore, it is indeed the cathode. 3. The measured reduction potentials are not equal to the calculated reduction potentials. Give two reasons why this might be observed. Answer: It might be because of inaccurate measuring equipment or the Energy lost or accepted from surroundings. 4. Would the cell potential be higher or lower if the

NH 3

(aq) had been added to the 1 M

Cu SO 4 solution instead of the 0.001 M Cu SO 4 solution of the cell? Explain. Answer: The cell potential would be lower if the

NH 3

(aq) had been added instead of the

0.001 M Cu SO 4 solution of the cell since, the 1 M Cu SO 4

has a higher concentration

compared to 0.001M Cu SO 4 . 5. The cell potential increased (compared to Part B.2) with the addition of the Na2S solution to the 0.001 M CuSO4 solution. Explain. Answer: The cell potential increased with the addition of Na2S solution because more electrons travel from anode to cathode when the 0.001 M CuSO4 solution discharges electrons. 6. As the concentration of the copper(II) ion increased from solution 4 to solution 1, did the measured cell potentials increase or decrease? Explain why the change occurred.

Answer: The measured potentials increased as the concentration of copper (II) ion increased since, electrons released from the 0.001 M CuSO4 solution flow from anode to cathode. 7. Suppose the 0.1 M

2+¿ solution had been diluted (instead of the Zn¿

2+¿ Cu¿

solution),

Would the measured cell potentials have increased or decreased? Explain why the change occurred. Answer: Suppose that Zn had been diluted instead of Cu, the measured cell potential would have decreased. This is due to Zn having significantly lower standard electrode potential as compared to Cu, hence the lower measured potential of the cell. 8. How would you increase or decrease the

decrease the

2+¿ Zn¿

2+¿ Cu¿

concentration and/or increase or

concentration to maximize the cell potential? Explain how the

change for each ion would maximize the cell potential. Answer: To maximize the cell potential, increasing the amount of solute would mean a higher concentration. On the other hand, diluting it with water results to lower concentration.

Results and Discussion This will tackle the results of the reduction potentials of several redox couples. There are 3 galvanic cells in this experiment. The first galvanic cell is Cu-Zn, the value of the measured Ecell of it is 0.81V. Its anode is the element, Zn and the equation for anode half-reaction is Zn

 Zn2+ + 2e-. While the cathode is Cu and the equation for its half-reaction is Cu 2+ + 2e- --> Cu (s). The second galvanic cell is Cu-Fe, the measure

Ecell

is 1.02V. Its anode is Fe and the

equation for half reaction of it is Fe  Fe2+ + 2e-. While the cathode is also the Cu and the

equation of half reaction of it is just the same to the first one. Last galvanic cell is Zn-Fe, the measured

Ecell

is 0.2V. The anode is Zn and the equation for the half-reaction is Zn  Zn2+ +

2e- while the cathode is Fe and the equation of its half-reaction is Fe2+ + 2e- --> Fe. This is discussion about the redox couple of the said galvanic cells, its reduction potential which is experimental and theoretical, and the percent error. The galvanic cell Cu-Zn, the redox couple of it is Cu2+ /Cu, the experimental reduction potential of it is 0.02 while the theoretical is 0.3. However, the percent error is 93.55% which is obviously higher value. The second galvanic cell is Zn-Fe the redox couple of it is Fe2+ ¿ Fe , the experimental reduction potential of it -0.59 while the theoretical is -0.47. The percent error it this is lower compared to the first one, it is 25.53% only. The last galvanic cell is Zn-Zn, the measured

Ecell

is 0. The redox couple is

Zn2+/Zn, the experimental reduction potential of it is -0.79 which is the same with theoretical. The percent error of this galvanic cell is 0.

Conclusion The first galvanic cell is Cu-Zn, its anode is the element, Zn and the equation for anode half-reaction is Zn  Zn2+ + 2e-, while the cathode is Cu and the equation for its half-reaction is Cu2+ + 2e- --> Cu (s). The second galvanic cell is Cu-Fe, Its anode is Fe and the equation for half reaction of it is Fe  Fe2+ + 2e- , while the cathode is also the Cu and the equation of half reaction of is the same as the first cell. The las...