March 28, 2024, 09:04:11 PM
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Topic: How does precipitation of ZnS by addition of Na2S effect cell potential?  (Read 730 times)

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Offline JR19

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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

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

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