December 01, 2021, 12:43:42 AM
Forum Rules: Read This Before Posting

Topic: How does precipitation of ZnS by addition of Na2S effect cell potential?  (Read 351 times)

0 Members and 1 Guest are viewing this topic.

Offline JR19

  • Very New Member
  • *
  • Posts: 1
  • Mole Snacks: +1/-0
I have made voltaic cell as follows.

Zn(s) | ZnCl(aq) 0.1M  ||  CuCl(aq) 0,1M | Cu(s)

The EMS was measured with a voltmeter at 1,02V.

Then by addition of Na2S, to the cathode side I witnessed some precipitation. I'm reasoning that the precipitation is of ZnS(s)

According to the Nernst equation; since [Zn2+] < [Cu2+] now

Q<1 and therefore the EMS should rise. However, this is not what was observered, rather the measured EMS went down.

How can I find an explaination for this?
« Last Edit: March 19, 2021, 10:09:15 AM by JR19 »

Offline mjc123

  • Chemist
  • Sr. Member
  • *
  • Posts: 1975
  • Mole Snacks: +288/-12
I assume you mean ZnCl2 and CuCl2.
The Zn side is the anode.

Sponsored Links