Galvanic Cell, the Nernst Equation Report

General Chemistry II/Lab Experiment 32: Galvanic Cell, the Nernst Equation Hypothesis: If the relative reduction potentials for a number of redox couples are measured, then it is possible to estimate the concentration of ions in a solution using the Nernst Equation. Data: Numerical figures derived from the experiment The experiment was conducted by Reduction Potentials of Several Redox Couples Table 1 Ecell Anode Equation for Anode Measure Reaction d (V) Galvanic Cell Cu-Zn 1.10 Zn Zn (s)--- Zn 2+ +2e- Cathode Equation for Cathode Reaction Cu Cu2++2e-----Cu(s) Cu2++2e----- Cu(s) Cu2++2e- ---- Cu (s) Mg2++2e- ----Mg(s) Fe2++2e- ---- Fe(s) Fe2++2e------ Fe(s) Cu-Mg Cu-Fe 2.70 0.78 Mg Fe Mg (s) --- Mg2+ +2eFe (s) ---- Fe2+ +2e- Cu Cu Zn-Mg 1.60 Mg Zn(s) ---- Zn2+ +2e- Zn Fe-Mg Zn-Fe 1.92 0.32 Mg Zn Mg-----Mg2++ 2eZn (s)---- Zn2+ +2e- Fe Fe TABLE 2 Galvanic Cell Cu-Zn Zn-Fe Zn-Zn Zn-Mg Ecell Measured (V) 1.10 0.32 0 1.60 For the redox Reduction Potential V couple (experiment) Reduction Potential (theoretical) % Error Cu2+/Cu Fe2+/Fe Zn2+/Zn Mg2+/Mg (+) 0.31 (-) 0.47 (-) 0.79 (-) 2.39 + 0.31 - 0.47 -0.79 -2.39 (+)0.32 (-)0.44 (-)0.77 (-)2.45 Grady 2 X2+/X 0.65 Znunknown, X Table 3 Solution Number 1 2 3 4 unknown Concentratio n of Cu(NO3)2 0.1 mol/L 0.01 0.001 0.0001 0.9656 M (-)0.12 (-) 0.14 Ecell experimental (-)log[Cu2+], p[Cu] 1.10 V 1.04V 0.98V 0.92V 1.01V 1 2 3 4 -0.14 Ecell, calculated using equation 32.9 and [Zn2+]=0.1M 1.1 1.0104 0.9208 0.8312 Graph 1 Solution 1 Solution 2 Solution 3 Solution 4 Ecell 1.10 1.04 0.98 0.92 pCu 1 2 3 4 Ecell versus pCu 1.15 1.1 f(x) = − 0.06 x + 1.16 1.05 Ecell 1 0.95 0.9 0.85 0.8 1 2 3 pCu Series1 Linear (Series1) 4 Grady 3 Calculations: Ecell = E°Cell – 0.0592/2 log [Zn2+] +0.0592/2 log [Cu] Solution 2: Ecell= 1.04 – 0.0592/ log [0.1] +0.0592log [0.01] = 1.0104 Y= -0.06x +1.16= Slope from graph 1.01=-0.06x+1.16 x= 2.5 102.5= 3.162 x 10-3M Ecell= 1.01 -0.0592/2log [0.1] +0.0592/2log [3.162 x 10-3]2 = 0.9656 M = Concentration of unknown Discussion: In part A of the experiment, six cell reactions were observed; Cu-Zn, Cu-Mg, Cu-Fe, ZnMg, Fe-Mg, and Zn-Fe—their cell potentials were recorded on the experimental sheet. In part B of the experiment, the effect of concentration changes on cell potential was observed. Firstly, two different concentrations of copper (II) Sulfate were being looked at; 0.001M and 1M. A galvanic cell was made from these two solutions, and a potential was recorded because concentration does indeed affect cell potential. When NH 3 was added, the cell potential increased because diluting an already dilute solution increases the potential greatly because NH 3 can form complexes with the copper ions more easily. When Na 2S is added to the solution, the cell potential once again increases because it acts in the same manner as the NH3. Overall, diluting Grady 4 solutions, especially solutions that already have low concentrations, results in a higher cell potential than that of a concentrated solution. There were several sources of experimental error, specifically in part C of the experiment. There should have been a trend in the values for part C, where different concentrations of Cu were being observed, but there was no set trend. The values were inconsistent, and it may have been due to faulty equipment or contaminated stock solutions because the values being observed for different concentrations should have presented the conclusion that as concentration increases, cell potential decreases. But from the results, no such conclusion could have been made. This conclusion was reached with help from manipulation of the Nernst equation. Conclusion: The objective of this lab was to measure the relative reduction potentials for a number of redox couples, develop an understanding of the movement of electrons, anions, and cations in a galvanic cell, and to study factors affecting cell potentials and estimate the concentration of ions in solution using the Nernst equation. A multimeter was set to 2000-mV and used to measure the potential of each metal in each solution. A piece of filter paper dipped in 0.1 M KNO3 solution was used as a salt bridge to connect both solutions that were being tested. There were 4 cells made, Zinc-cooper, Zinc-iron, Copper-ion and Zinc-unknown. Each solution had their respective metal placed in it and the multimeter was attached to each metal to read to the voltage and determine the cathode and anode of the cells. Part C, four different cells were set up, three of which had different concentrations of copper and one with Zinc. The cell potential was measured Grady 5 for each as zinc remained constant. From that, the concentration of copper was found and recorded as 0.1, 0.01, 0.001, 0.0001, and 0.9656 (Beran, 2014). Pre-lab Questions: Part A Question 1. Write Balanced Equations for the six cell reactions: 1. Cu-Zn Zn(s) + Cu 2+ →Zn 2+ + Cu(s) 2. Cu-Mg Mg(s) + Cu 2+ →Mg 2+ + Cu(s) 3. Cu-Fe Fe(s) + Cu 2+ →Fe 2+ + Cu(s) 4. Zn-Mg Mg(s) + Zn 2+ →Mg 2+ + Zn(s) 5. Fe-Mg Mg(s) + Fe 2+ →Mg 2+ + Fe(s) 6. Zn-Fe Zn(s) + Fe 2+ →Zn 2+ + Fe(s) Question 2. What is the oxidizing agent in the Zn-Mg cell? The overall reaction of Zn-Mg cell is Mg (s) + Zn 2+ (aq) → Zn (s) + Mg 2+ (aq). Zn 2+ has been reduces to Zn (s), and Mg (s) has been oxidized to Mg 2+. Therefore, ZnCl2 is the oxidizing agent. Question 3. Compare the sum of the Cu-Zn and Zn-Mg cell potentials with the Cu-Mg cell potential. Explain Cell potential of Zn-Mg: 0.462 V. Cell potential of Cu-Mg: 0.889 V. Sum of the Zn-Mg and Cu-Mg cell potentials = 0.462 + 0.889= 1.351 V. Cell potential of Zn-Cu: 0.455 V. The sum value is higher than the Zn-Cu cell potential by 0.896. Question 4. Compare the sum of the Zn-Fe and Zn-Mg cell potentials with the Fe-Mg cell potential. Explain. Grady 6 Considering the Zn-Fe cell, in this case Fe is cathode and Zn is anode so Eo cell = Eo cathode - Eo anode. Eo cell = -0.44 + 0.76; Eo cell = 0.32 V. So, cell potential for Zn-Fe cell is 0.32 V. Considering Fe-Mg cell, Fe is cathode and Mg is anode. Eo cell = Eo cathode - Eo anode; Eo cell = -0.44 + 2.37; Eo cell = 1.93. So, the cell potential for Fe-Mg cell is 1.93 V. The sum of Zn-Fe and Zn-Mg cell potentials = 0.32 + 1.61 = 1.93 V. The sum of Zn-Fe and Zn-Mg cell potentials = cell potential for Fe-Mg cell. Question 6. Reduction Potential of the unknown redox couple is: - -(0.12) Part C Question 3. Ecell for the solution of unknown concentration is: -0.9656 Post Laboratory Questions: 1. Part A.3. The filter paper salt bridge is not wetted with the 0.1M KNO3 solution. As a result, will the measured potential of the cell be too high, low low, or unaffected? Explain. The measured potential of the cell would be too low because there would be no flow of electrons from one solution to the other. 2. Part A.3. 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. Grady 7 The measured cell potential would decrease because the increase in number of ions will result in the reaction moving closer to the equilibrium. 3. Part A.5. The measured reduction potentials are not equal to the calculated reduction potentials. Give two reasons why this might be observed. We will be moving the reaction further away from equilibrium which will increase the potential if we diluted the Zn2+. 4. Part C. 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. As the concentration increases, the cell potential also increases. E = E* - (0.0592/n) * log [1/(Cu+2)]. If (Cu+2) increases ; then E will also increase from the above equation.