Part 1: Concentration dependence of the electrochemical potential - confirming t
ID: 972781 • Letter: P
Question
Part 1: Concentration dependence of the electrochemical potential - confirming the Nerst equation for concentration cells In the first part of the exercise you will construct a series of five concentration cells with AgNO3 as the electrolyte. For each, a half-cell containing a 0.0100 M AgNO3 solution will be used as the reference potential. The other half-cells in the series will be made up of 10–5 M, 10–4 M, 10–3 M, 10–2 M, 10–1 M AgNO3 solutions respectively. The pH meters you have used in the laboratory also have a mode that will measure the electrochemical potential difference of a cell. Two silver electrodes will be connected to the pH meter for this purpose. Placing the electrodes in the solutions of each half-cell will give a reading in millivolts. One electrode will be marked to designate it as the "reference". This electrode should always be placed in the 0.01 M AgNO3 solution. When you have measured the voltages of the five concentration cells, you will enter your data in an Excel spreadsheet on the lab computer. Enter the values of the molarities of each AgNO3 solution contained in the NONreference half-cell, along with the corresponding measured potential differences in units of volts. Excel will produce a linear regression plot of voltage vs log (Ag+). The slope of the straight line will be close to the expected value of 2.303 RT / nF, called the Nerst constant.
1) What is the expected (calculated) value of the constant 2.303 RT / nF in units of volts ?
Use F = 96,500 C/mol
2) What is the value of 2.303 RT / nF , in volts, that is observed from the above plot of voltage vs. log [Zn2+] ?
Part 2 Determining the Solubility Product of an Unknown Silver AgX (X = Cl–, Br–, or I– )
AgX(s) ? Ag+(aq) + X– (aq)
- by measuring small concentrations of Ag+ in equilibrium with solid, AgX, electrochemically by utilizing the Nerst Equation
You must determine the value of the solubility product for this reaction,
Ksp = [Ag+] [X–]
Mixing aqueous solutions of AgNO3 and KX (potassium halide, where X = Cl–, Br–, or I– ) will cause a precipitate to form, if the solutions are concentrated enough. This precipitation reaction is the reverse of the above : Ag+(aq) + X– (aq) ? AgX(s)
For the potassium halide solution, knowing the molarity, the volume used, and the total volume of the mixture, the concentration of X– can be calculated.
The concentration of Ag+ that is in equilibrium with the solid precipitate can be found from the Nerst equation by constructing a concentration cell and measuring the potential difference between it and a reference cell. The reference cell that you will use in the laboratory will again contain a 0.0100 M solution of Ag+.
3) Below is the result of a student's efforts to determine the solubility product, Ksp, of zinc oxalate.
Note that the concentration of the reference cell is different than that of the one you will use.
The concentration of Zn2+ in the reference cell is 0.001 M. The concentration of oxalate in the second cell is 0.588 M. The voltage of this second cell relative to the reference is -128 mv.
The temperature is 25.0 C.
What is the concentration (M) of the Zn2+ in equilibrium with the precipitate ?
4) What is the solubility product, Ksp, of zinc oxalate ?
Part 3 - Determining Standard Reduction Potentials
You will prepare three half-cells of the form :
M (s) | M2+ (aq) where M = Zn, Pb, and Cu
For each, the counter ion is NO3– at 0.050 M.
From the half-cells you will construct three complete cells :
Zn/Cu Zn/Pb and Pb/Cu
By measuring the voltage of each cell you can determine the electrical potential of the half-cells relative to each other.
Since this involves three half-reactions and three cell voltages, there are three relationships and three unknowns. You will therefore be given one of the cell voltages as an initial starting point.
5) In a similar exercise a student constructed three cells of the form :
X | X2+ (0.1 M) || Y2+ (0.1 M) | Y
Their voltages of three combinations of the half-cell reactions are tabulated below:
The reduction potential for Sr2+/Sr is -2.90 V
NOTE : Ecell = Ecathode – Eanode
However, since the Ecell given here is negative, the reference lead of the meter used to measure the voltage was connected to the anode instead of the cathode. The actual voltage of the cell would be the positive value of that given above.
Also note that traditionally, the left hand half-cell is the anode.
Calculate the reduction potential (V) for Zn2+/Zn .
6) Calculate the standard reduction potential (V) for Cu2+/Cu ?
(Note that your answers to this Question and Question 6 can be combined to give the voltage of the remaining cell.)
Explanation / Answer
1) If we are talking bout the Zn concentratio cell then
2.303 RT / nF = 2.303 X 8.314 X 298 / n X 96500
For Zn , n=2
2.303 RT / nF = 0.05912 / 2 = 0.02956
2) The relation is
Ecell = E0cell - 2.303RT / nF log Q
E0cell = 0
Graph is between Ecell and log Q
so the slope = -2.303 RT / nF
slope = 0.0282
So value of 2.303RT / nF = 0.0282
3) Ecell = -0.0592 / 2 log [Zn+ as oxalate] / [Zn+]
-0.128 =- 0.0296 log[Zn+ as oxalate] / 0.001
21086.28 = [Zn+ as oxalate] / 0.001
[Zn+ as oxalate] = 21.08
Part 3)
Reduction potential of Sr+2/ Sr = -2.90 V
Ecell = Ecathode - Eanode
-2.138 = ESr - EZn = (EZn) -2.90
Ezn = +2.90 - 2.138 = 0.762 V
If we are takining all the given values as mentioned
However it will bd -0.762 V (actually) as the cell conventions used in wrong in the above case
6) Ecell = Ecathode - Eanode
Ecell = ECu - EZn = ECu - (0.762)
-1.104 = ECu - (0.762)
ECu = -0.342
However it will be +0.342 V actually