Assignment 3 - Solutions

Question 1 The concentration of chloride (Cl-) anion in solution can be measured with an electrode made with CuCl.  If an unknown solution of Cl- has a voltage of 0.320 V (vs. the SHE), what is the concentration of Cl- in that solution?  For Ksp values, see Appendix 1.  For Eo values, see Appendix 4.  If you get your values from a source other than the course text, be sure to note the source!
 
Solution This is a 2-step question. 
First, you must solve the Eo value for the electrode being used (CuCl), then use that electrode to find the [Cl-]:

Cu is the species undergoing a redox reaction,
 Cu+ (aq)  +  e-  --->  Cu (s)      (Eocell = 0.521 V vs. S.H.E.)
Therefore, under standard conditions, Eo for the new half-cell reaction will be:

Replacing [Cu+] with Ksp / [Cl-], where [Cl-] = 1 and Ksp = 1.9 x 10-7, gives the following:

Second, use this value to solve for the unknown Cl- concentration:


[Cl-] = 4.72 x 10-4 M
 
Question 2 In a stripping experiment, concentrations are generally between 10-4 and 10-9 M.  Show that it is possible to quantitatively separate (i.e., <10-11 M) a mixture of Mg2+ and Na+ in these concentration ranges.
 
Solution: The concentration range for each element is 10-4 to 10-11M.  Therefore, find the voltages corresponding to each concentration to find the voltage range for the two metals in the stripping experiment.  Note: Eo = -2.363 V (Mg2+) and -2.714 V (Na+)


Similarly,
EMg2+,10^-11 M = -2.69 V
ENa+,10^-4 M = -2.95 V
ENa+,10^-11 M = -3.37 V

Therefore, there is no overlap in the voltage ranges, as Mg2+ reduces from -2.48 to -2.69, while Na+ reduces in the range -2.95 to -3.37.
 
Question 3 An unknown solution contains Cd2+ and Cr3+, which can be quantitatively separated by coulometric methods.  Using 2.00 mL of the solution, it took 3.85 min to quantitatively remove the Cd2+ using a current of 12.0 mA.  Subsequently, at the same current, the Cr3+ was quantitatively deposited in 1.76 min.  What was the concentration (in mol/L) of the two metals in the original solution?
 
Solution:
similarly, using a 1:3 ratio of electrons: Cr3+, nCr3+ = 4.38 x 10-6.

dividing by the volume (0.00200 L) gives the concentrations:
[Cd2+] = 7.18 x 10-3 M ,  [Cr3+] = 2.19 x 10-3 M.
 
Question 4 The standard addition method was used to determine the concentration of aluminum in the water near a mine.  The following data were obtained:
 
volume of water sample (mL)
volume of 0.0800 M Al3+ (mL) 
potential (V)
25.0
0
 -1.623
25.0
5.0
-1.587

 Determine the concentration of Al3+ in the original solution.  Report the result as both molarity (mol/L) and parts per million (ppm = mg/L).  (For this question, we will not assume Ej = 0 and concentration = activity.)
 
Solution: for a cation:

Therefore, for sample #1,

and for sample #2,

Eq. #1 - Eq. #2:
log[Al3+] - log(0.833[Al3+] + 0.0133) = -1.824

[Al3+] - 0.0125[Al3+] = 2.00 x 10-4

Therefore, [Al3+] = 2.02 x 10-4 M.

Using the fact that ppm = mg/L, [Al3+] = (2.02 x 10-4 mol/L)(26.98 g/mol)(1000 mg / 1 g) = 5.45 ppm


 

Last updated, March 31, 2003
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