December 25, 2024, 12:14:51 AM
Forum Rules: Read This Before Posting


Topic: Cadmium (II) Polarography  (Read 3680 times)

0 Members and 1 Guest are viewing this topic.

Offline tiger1987

  • New Member
  • **
  • Posts: 5
  • Mole Snacks: +0/-0
Cadmium (II) Polarography
« on: November 25, 2006, 02:11:29 AM »
I am doing a cadium (II) polarography experiment, and when I calculate the value of n for my polarograms, I get 1 e-. I should be getting 2 e- because the oxidation state of cadmium is +2. What could possibly be causing this problably? ( I have double and triple checked my values for n and it still comes out to 1 e-.

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27888
  • Mole Snacks: +1816/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Cadmium (II) Polarography
« Reply #1 on: November 25, 2006, 05:08:53 AM »
Not enough info IMHO. Post some details.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline chiralic

  • Full Member
  • ****
  • Posts: 234
  • Mole Snacks: +20/-3
  • Gender: Male
  • Test
Re: Cadmium (II) Polarography
« Reply #2 on: November 25, 2006, 07:41:25 PM »
Hi Tiger1987:

I recommend you read chapter 22 of book Problems and Experiments in Instrumental Analysis by Meloan-Kiser (I have it version in spanish), there you can find EXACTLY what are you looking for...

You must plot Log(I/Id-I) vs E, you'll get a "lineal" curve where the slop value is equal to n/0.0591.
Then you'll get a "n"

Inside of this book you can find an exercise from experimental data and they got n=1.75 for Cd+2

Sponsored Links