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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: JR19 on March 19, 2021, 09:41:56 AM

Title: How does precipitation of ZnS by addition of Na2S effect cell potential?
Post by: JR19 on March 19, 2021, 09:41:56 AM

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?

Title: Re: How does precipitation of ZnS by addition of Na2S effect cell potential?
Post by: mjc123 on March 20, 2021, 06:58:49 PM

I assume you mean ZnCl₂ and CuCl₂.
The Zn side is the anode.