Lab report 16 PDF

Preview

Summary

CHEM 106 LAB REPORT 16...

Description

1

Instructor Yue 28 April 2020 Lab Report 16 Galvanic Cells Data Analysis Part 1: Constructing Galvanic Cells Fe (s) | Fe2+ (aq, 1M) || Cu2+ (aq, 1M) | Cu (s) Experimental: 0.696 V Theoretical: E°cell = E°red + E°ox E°cell = 0.34 V + 0.44 V = 0.78 V Zn (s) | Zn2+ (aq, 1M) || Cu2+ (aq, 1M) | Cu (s) Experimental: 1.093 V Theoretical: E°cell = E°red + E°ox E°cell = 0.76 V + 0.34 V = 1.10 V Zn (s) | Zn2+ (aq, 1M) || Fe2+ (aq, 1M) | Fe (s) Experimental: 0.373 V Theoretical: E°cell = E°red + E°ox E°cell = 0.76 V + -0.44 V = 0.32 V Part 2: Effect of Concentration on Cell Potential Zn (s) | Zn2+ (aq, 0.1M) || Cu2+ (aq, 0.1M) | Cu (s) Experimental: 1.091 V Theoretical: Ecell = E°cell - RT/nF ln Q Ecell =1.1 V - 8.314*298K/2*96485 C/mol electrons ln(0.1M/0.1M) = 1.1 V

2

Zn (s) | Zn2+ (aq, 1M) || Cu2+ (aq, 0.1M) | Cu (s) Experimental: 1.081 V Theoretical: Ecell = E°cell - RT/nF ln Q Ecell =1.1 V - 8.314*298K/2*96485 C/mol electrons ln(1M/0.1M) = 1.0704 V Zn (s) | Zn2+ (aq, 0.1M) || Cu2+ (aq, 1M) | Cu (s) Experimental: 1.104 V Theoretical: Ecell = E°cell - RT/nF ln Q Ecell =1.1 V - 8.314*298K/2*96485 C/mol electrons ln(0.1M/1M) = 1.1296 V Zn (s) | Zn2+ (aq, 1M) || Zn2+ (aq, 1M) | Zn (s) Experimental: 0 V Theoretical: Ecell = E°cell - RT/nF ln Q Ecell = 0 V - 8.314*298K/2*96485 C/mol electrons ln(1M/1M) = 0 V Zn (s) | Zn2+ (aq, 0.1M) || Zn2+ (aq, 1M) | Zn (s) Experimental: 0.026 V Theoretical: Ecell = E°cell – RT/nF ln Q Ecell =0 V - 8.314*298K/2*96485C/mol electrons ln(0.1M/1M) = 0.0296 V Cu (s) | Cu2+ (aq, 0.1M) || Cu2+ (aq, 1M) | Cu (s) Experimental: 0.013 V Theoretical: Ecell = E°cell – RT/nF ln Q Ecell =0 V - 8.314*298K/2*96485 C/mol electrons ln(0.1M1M)= 0.0296 V Part 3: Effect of Temperature on Cell Potential Fe (s) | Fe2+ (aq, 1M) || Cu2+ (aq, 1M) | Cu (s)

3

Theoretical ΔH° = [(0) + (-89.1)] - [(64.8) + (0)] = -153.9 kJ/mol ΔS° = [(-137.7) + (33.2)] - [(-99.6) + (27.3)] = -32.2 J/mol*K ΔG° = [(0) + (-78.9)] - [(65.6) + (0)] = -144 kJ/mol Experimental ΔH° = y- intercept * -nF/10^6 ΔH° = 677* - 2 *96485C/mol electrons/10^6 ΔH°= -130.64 kJ/mol ΔS° = slope * nF/1000 ΔS° = 0.982 * 2 * 96485 C/mol electrons/1000 ΔS°=189.496 J/mol * K ΔG° = ΔH° - TΔS° ΔG°= - 130.64 kJ/mol – 298 * 189.496 J/mol *K/1000J = -187.110 kJ/mol Focus Questions

4

Part 1 1. What are the cell potentials for the three galvanic cells you obtain in the lab? Answer: Fe (s) | Fe2+ (aq, 1M) || Cu2+ (aq, 1M) | Cu (s)

E°cell = 0.696 V

Zn (s) | Zn2+ (aq, 1M) || Cu2+ (aq, 1M) | Cu (s)

E°cell = 1.093 V

Zn (s) | Zn2+ (aq, 1M) || Fe2+ (aq, 1M) | Fe (s)

E°cell = 0.373 V

2. Do your experimental/laboratory values match the theoretical? Answer: Yes, the experimental and the theoretical values matched because the values were very close to one another. Part 2 1. How does concentration affect the cell potential? (look at a,b, and c) Answer: The concentration changes affect the cell potential because with an increase in concentration of anodes, there will be a cell potential decrease. However, a greater concentration of cathodes will increase the cell potential. 2. Do two identical half-cells constitute a galvanic cell? (look at e and f) Answer: Two identical half-cells do not constitute a galvanic cell because in (e) and (f) the concentration varied. However, in (d) the concentration and half cells were identical as the cell potential was 0 V. 3. Do the same type of cell but different concentration constitute a galvanic cell? (look at d and e) Answer: The same type of cell but different concentration may constitute a galvanic cell because for (d) the concentration and half cells were identically yielded with a cell potential of 0. For (e) the concentration changes, which yielded a cell potential of 0.026 V for experimental. Because it’s close to 0, it may constitute as a galvanic cell.

5

Part 3 1. How does temperature affect the cell potential of a) the iron and copper (II) reaction and b) the zinc and copper (II) reaction? Answer: The temperature is directly proportional for iron and copper (II) reaction. 2. What are your experimental values of ΔH°, ΔS°, and ΔG° for these reactions? Are they in agreement with the theoretical values? Discuss any sources of experimental errors. Answer: The experimental values for iron and copper (II) reaction was ΔH° = -130.64 kJ/mol, ΔS° = 189.496 J/mol * K, ΔG°= -187.110 kJ/mol. ΔH° was the closest when compared to the theoretical values, however, ΔS° wasn’t. Some experimental errors that could have occurred was constructing the galvanic cell wrong or mixing the concentrations up or having the battery faced in a wrong way which can hinder the cathode  anode pathway. References 

Smeureanu, G., & Geggier, S. (2019). General Chemistry Laboratory- CHEM 106 at Hunter College, Fall 2019.

Post Lab Questions 1. A voltaic cell is constructed with an Ag/Ag+ half-cell and a Ni/Ni2+ half-cell. a. Which metal will serve as the anode and which as the cathode? Justify your answer quantitatively. Answer: Ni would be the anode and Ag would be the cathode. If we switched it around, the cell potential would be and a non spontaneous reaction which means the experiment would not work. b. Write a cell reaction for this cell. Write the half reactions and label them as oxidation and reduction.

6

Answer: Oxidation: Ni (s) → Ni2+ (aq) + 2eReduction: Ag+ (aq) + e- → Ag (s) c. Write the line notation for this cell Answer: Ni (s) | Ni2+ (aq) || Ag+ (aq) | Ag (s) d. Determine the standard cell potential Answer: Ni (s) → Ni2+ (aq) + 2e-

E° = 0.25 V

2Ag+ (aq) + 2e- → 2Ag (s) E° = 2 (0.3998 V) = 0.7996 V E°cell = 0.7996 V + 0.25 V = 1.0496 V e. If [Ag+] = 0.001M and [Ni2+] = 0.01M, determine the cell potential. Answer: E = E° - 0.0592/n logQ E =1.0496 V - 0.0592/2 log 0.01M/0.001M E = 1.0496V - 0.0296V E = 1.02 V 2. Write a balanced equation from each line notation: a. Ag(s) | Ag+(aq) || Cr3+ (aq) | Cr(s) Answer: 3Ag (s) + Cr3+ (aq) → Cr (s) + 3Ag+ (aq) b. Pb(s) | Pb2+ (aq) || MnO2 (aq) | Mn2+ (aq) | Pt (s) Answer: Pb (s) +MnO2 (aq) → Pb2+ (aq) + Mn2+ (aq) + Pt (s) 3. In the lead storage battery, lead serves as the anode, and lead coated with lead oxide serves as the cathode. Both electrodes dip into an electrolyte solution of sulfuric acid. The electrode reactions are Pb (s) + HSO4- (aq) → PbSO4 (s) + H+ (aq) + 2e-

E°ox = +0.35V

PbO2 (s) + HSO4- (aq) + 3H+ + 2e- → PbSO4 (s) + 2H2O (l)

E°red = +1.69V

7

Cell reaction: Pb (s) + PbO2 (s) + 2HSO4- (aq) + 2H+ (aq) → 2PbSO4 (s) + 2H2O (l) a. Calculate the standard cell potential, E°, for this cell at 25°C. Answer: E°cell = E°ox + E°red E°cell = 0.35 V + 1.69 V = 2.04 V b. The concentration of sulfuric acid in a car battery at 6M, when the battery is fully charged. Use the Nernst equation to determine the cell potential, E. Remember that sulfuric acid is a strong acid (use only the first proton). Answer: Ecell = E°cell – RT/nF ln Q Ecell = 2.04 - 8.314*298/2*96485 ln(6) Ecell = 2.02 volts...