Which Ions Interfere with CrO4 2- Quantitation?

  • Thread starter Thread starter SebPT
  • Start date Start date
  • Tags Tags
    Cell Ions
Click For Summary
SUMMARY

The discussion focuses on the interference of specific ions with the quantitation of CrO4 2- in an electrochemical cell setup involving Ag2CrO4 and Ag electrodes. The ions identified as potential interferents include S2-, Br-, and Pb2+, while NO3- is deemed non-interfering due to its high solubility. The standard reduction potential for the reaction involving Ag2CrO4 is +0.446 V vs. NHE, and the Ksp values of the interfering ions are critical in determining their impact on the half-cell reaction. The Nernst equation for the measuring electrode is provided, emphasizing the importance of understanding the interactions at the electrode surface.

PREREQUISITES
  • Understanding of electrochemical cells and half-cell reactions
  • Familiarity with standard reduction potentials
  • Knowledge of solubility product constants (Ksp)
  • Proficiency in applying the Nernst equation
NEXT STEPS
  • Research the solubility product constants (Ksp) of S2-, Br-, and Pb2+ in relation to Ag+
  • Study the Nernst equation and its application in electrochemical measurements
  • Examine the standard reduction potentials of common cations to assess their interference
  • Explore methods to mitigate interference in electrochemical quantitation
USEFUL FOR

Chemists, electrochemists, and laboratory technicians involved in analytical chemistry and electrochemical measurements, particularly those working with chromate quantitation and interference analysis.

SebPT
Messages
3
Reaction score
0

Homework Statement



The electrochemical cell:

SCE // CrO4 2- (x M) / Ag2CrO4 (s) / Ag (s)

is used for the determination of CrO4 2- . The standard reduction potential for the reaction:

Ag2CrO4 (s) + e = 2 Ag + CrO42-

is +0.446 V vs. NHE.

(b) Which of the following ions would be expected to interfere with CrO4 2- quantitation if added to the analyte solution:

NO3 - , Br - , S 2- , K + , Pb 2+

3. To solve this I guess I need to understand how the half cell is working. So I think the electrons coming from the salt bridge react with Ag2CrO4 (s) at the surface of the Ag electrode to produce a deposit of Ag on the metal and CrO42- which stays in solution. However I am not sure where the interference will take place. I am not sure if adding K and Pb cations will just take in electrons and reduce current or if some of the anions interfere at the electrode surface. Any help would be greatly appreciated!
 
Physics news on Phys.org
Can you write Nernst equation for the measuring electrode?
 
S.C.E: 1/2Hg2Cl2 (s) + e-⇌Hg(l)+Cl-

Emeas=0.242V-0.05916*log[Cl-]

Note: the half cell rxn in the original post should be
Ag2CrO4 (s) + 2e- = 2 Ag(s) + CrO4 2- (aq)

I don't see how the reference electrode matters in this case, since the interfering ions are in the other half cell.
 
You are right, reference doesn't matter.

Hint: note that while you wrote reaction as the one between silver chromate and metallic silver, it is in fact Ag+ that gets reduced, just its concentration near the electrode is controlled by the Ksp of Ag2CrO4.
 
So, should I consider the Ksp values of the interfering ions with Ag+ and CrO42-

Then the nitrate ion won't interfere since it is highly soluble and will stay in solution. However, the Ksp of S2- and Br- with Ag+ is less than the Ksp of the silver chromate reaction. Does this mean that these ions will mostly form a solid product when interacting with Ag+ effectively interfering with the half-cell reaction?

When considering the cations, could I look at their standard reduction potentials and compare them to Ag+? I see that they are both lower and therefore, the reduction of Ag+ will be favoured, but lead chromate Ksp is also low, and will precipitate some CrO42- out of solution.

So the ions that interfere will be S2-, Br- and Pb2+?
 
SebPT said:
So the ions that interfere will be S2-, Br- and Pb2+?

That's what I would expect.

Sorry, no time for more detailed discussion, and I won't be able to read the forum for a week.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
10K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
9K
Replies
1
Views
2K
Replies
1
Views
2K
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K