Electrochemcial Cells

Electrochemical Cells

Laayla Muhammad
Partners: Arlee Vang, Lee Her
March 21, 2007

Purpose: To construct a series of microscale electrochemical half-cells and use the Nernst equation to experimentally determine the voltage of a Zn/Cu cell, the Ksp of AgCl, and the Kf of Cu(NH3)42+. We will basically be constructing a table of relative electrode potentials and be changing the concentration of one of the solutions to see the affects. Then we’d use the electrical potential of a cell containing Ag and AgCl along with the Nernst equation to determine the solubility product of AgCl. Finally, creating Cu (NH3)42+, the potential and the Nernst equation can be used to calculate the formation constant of the tetramminecopper (II) complex ion using microscale techniques.


Procedure:
Determining the Reduction Potentials:
First, a test cell needs to be prepared to measure the voltage of the copper and zinc half cells by putting 2 mL of 1.0 M Zn(NO3)2 solution in one of the center wells and putting 2 mL of 1.0 M Cu(NO3)2 in an adjacent one. Then taking a small strip of filter paper, soaked in KNO3 solution, and draping it across the two wells, as a salt bridge. Then we polished strips of Zn and Cu metal and placed them in appropriate well containing the solutions of the ions. Then using a voltmeter, the potential difference between the two half-cells was measured, making sure the reading was positive. We made sure to note which electrode played the role of an anode and a cathode. The electrode connected to the positive terminal is the cathode and is undergoing reduction, while oxidation occurred at the electrode connected to the negative terminal, which was the anode. Next step was to prepare half cells in other wells by pouring some 1.0 M solution of the following ions in different wells and polishing the metals with sandpaper or steel wood so they’re shiny and inserting them into the well containing the ion of same metals (fresh trips of filter paper soaked in 1.0 M potassium nitrate were used as salt bridges): Ag|Ag+, Cu|Cu2+, Fe|Fe3+, Mg|Mg2+, Pb|Pb2+, and Zn|Zn2+. We recorded the data using a table to help organize our findings.
Reduction Equation for each Ion Arranged in Decreasing Order of Potential:
Then we composed reduction equations for each metal ion and arranged the equations in decreasing order of measured potential in an “Eo” type of table. We recorded the standard potentials using the hydrogen electrode as standard, and calculated the difference between two values using another data table to stay organized.
Measure Cell Potentials:
Next we measured the potential difference between at least 4 combinations of various electrodes and used the table of electrode potentials to predict the voltage and which half cell will be the anode and cathode. Next step involves comparing predicted and measured potentials and using a table to record the data.
Changing Ion Concentration:
We diluted 1.0 M Cu(NO3)2 by adding 2 drops of the solution into 18 drops of deionized water in a small test tube. We mixed well and did this two times more to decrease the concentration to 0.0010 M. Then we poured some of this solution in one of the wells and added a polished copper wire to measure the voltage against the standard zinc electrode. We recorded the data and next step includes writing a net ionic equation for the reaction occurring in the cell and using the Nernst equation to calculate the expected voltage to be compared to the measured one.
Finding the Solubility:
We poured 10 mL of 1.0 M NaCl solution into a beaker and added 1.0 M AgNO3 and stirred. Then we poured some of the solution into one of the wells and added an Ag metal electrode and measured the potential difference versus this half-cell and the zinc half-cell. Next step involves writing out the net ionic equation for the reaction and using Nernst equation to calculate the concentration of the Ag+ ion and the value of the solubility of AgCl, to compare the calculated value with the measured.
Finding the Formation Constant of Cu(NH3)4:
We found the volume of one drop of 1.0 M Cu(NO3)2 solution by counting the number of drops in 1 mL, which was 17 drops for us. Then we put 10 mL of 6.0 M NH3 in a beaker and added 1 drop of 1.0 M Cu(NO3)2 solution. Then we poured some of the solution in one of the wells and added a Cu metal electrode to measure the potential difference vs. the Zn electrode, to write a balanced net ionic equation for the reaction occurring in the cell. The Cu(NH3)4 concentration is calculated by assuming that all of the Cu2+ in solution is present as the complex ion. The uncomplexed Cu2+ concentration is calculated from the cell potential using the Nernst equation. Last step includes finding the formation constant, the equilibrium constant for the equation and comparing the calculated value from the measured.


Data:
Determining Reduction Potentials

Voltage of each half-cell versus the zinc electrode:

Cell Voltage Anode Cathode
Zn vs. Ag 1.231 Ag Zn
Zn vs. Cu .801 Cu Zn
Zn vs. Fe .396 Fe Zn
Zn vs. Mg -.625 Zn Mg
Zn vs. Pb .391 Pb Zn
Zn vs. Zn 0.00 (assumed)





Reduction Equations for Each Ion Arranged in Decreasing Order of Potential:

Reduction equation Electrode Potentials using Zn as the standard, Ezn Accepted Electrode Potentials using Hydrogen as Standard, Eo Ezn - Eo
Ag+ + e- --> Ag 1.231 0.800 0.431
Cu2+ + 2e- --> Cu 0.801 0.340 0.461
Fe3+ + 3e- --> Fe 0.396 0.770 -0.374
Pb2+ + 2e- --> Pb 0.391 -0.013 0.404
Zn2+ +2e- --> Zn 0.000 -0.760 0.760
Mg2+ + 2e- --> Mg -0.625 -2.370 1.745














Measure Cell Potential:
Anode Cathode Equation for the cell reaction Predicted Potential from Experimental Data Measured Potential
Fe Mg Fe3+ + Mg --> Fe + Mg2+ 1.021 1.157
Ag Fe Fe3+ Ag --> Fe + Ag+ 1.627 0.879
Ag Pb Pb2+ + Ag --> Pb + Ag+ 1.622 0.918
Cu Mg Cu2+ Mg --> Cu + Mg2+ 1.426 1.588












Calculations:

In the first part, where we were determining the reduction potentials, we used the voltmeter to measure the potential difference between the two half cells. When the voltmeter reads a positive voltage, the black electrode, connected to the positive terminal is the cathode and is undergoing reduction, while the red electrode connected to the negative terminal, is the anode. Then we composed the reduction equations based on that. We assumed Zn|Zn2+ had the potential of 0.00 volts.

Example:

Cell Voltage Anode Cathode
Zn vs. Ag 1.231 Ag Zn


Then we recorded the standard potential using the hydrogren electrode as standard. We were then able to calculate the difference between the two values.
Example:

Reduction equation Electrode Potentials using Zn as the standard, Ezn Accepted Electrode Potentials using Hydrogen as Standard, Eo Ezn - Eo
Ag+ + e- --> Ag 1.231 0.800 0.431







The next part consisted of measuring cell potentials using at least 4 combinations of the various electrodes. We used the electrode potentials from the first table to predict the voltage and determine the reducing and oxidizing agents so both values can be compared.

Example:
Reduction: 2(Fe3+ + 3e-  Fe) E cell: .396 V
Oxidation: (Mg  Mg2+ + 2e-) E cell: .625 V
2Fe3+ + Mg  Fe + Mg2+ E cell: 1.021 V


Changing the concentration of the Ion:

2 drops of 1.0 M Cu(NO3)2 diluted in 18 drops of dionized water provided us with 0.10 M concentration. We did it two more times to give us the final concentration of 0.0010 M. So the concentration went from 1.0 M to 0.10 M, then 0.010 M, and finally 0.0010 M.
We had to write an ionic equation for the reaction occurring in the cell and then use the Nernst equation to calculate what the expected voltage should be, and compare to the measured value.

Reducing: Cu2++2 e-  Cu(s)
Oxidizing: Zn(s) Zn2++2 e-
Equation: Cu2++Zn(s) Zn2++ Cu(s)

Trial 1: Cu3+ + Zn  Cu + Zn3+ Voltage: .220
Trial 2: Cu3+ + Zn  Cu + Zn3+ Voltage: .224

Average: .220 + .224 = .232 V
2

Ecell = Eo cell – RT In Q
nF

= Eo cell – 0.0592 log Q
n


Finding Solubility Product of AgCl:

In this step, almost all of the silver ions combined with chloride ions to precipitate AgCl. Since there is a large excess of Cl-, it can be assumed that the concentration of Cl- is still 1.0 M. The concentration of the silver ions will be really small (reduced). After measuring the potential difference versus this half-cell and the zinc half-cell, we had to write a balanced equation for the reaction occurring in the well. Then we have to compare the two values with each other.

Reduction: 2 Ag2++2e- 2AgCu(s)
Oxidation: Zn(s) Zn2++ 2e-
Balanced: 2Ag2++Zn(s) Zn2++ 2Ag(s)

Next step involves using the Nernst Equation to calculate the concentration of the Ag+ ion. Therefore, we’d then calculate the solubility product of AgCl and compare the two values with each other.



2 Ag+ + Zn  2Ag + Zn2+ E cell: