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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: apcurtiss on April 17, 2014, 07:42:03 PM

Title: Why would the voltage between the two leads of a simple lead-acid (Pb-H2SO4) bat
Post by: apcurtiss on April 17, 2014, 07:42:03 PM

As part of a lab I made a simple lead-acid battery with two solid lead strips and a sulfuric acid bath. The voltage was zero when the two strips were introduced into the bath, indicating they were clean of lead-dioxide. When I hooked the 9V battery up to the two lead (Pb) leads, the voltmeter shot up to 4.0 volts, then decreased over the course of a minute to 2.8 V at which point I unhooked the battery. Lead dioxide had formed at the positive electrode and the battery began discharging from about 2.0 V and it steadily decreased to 0.8 V after two minutes. I thought voltage didn't change much when a battery drained, only the current. And i certainly can't understand why the voltage would decrease while the battery was charging. Any help understanding this would be appreciated!

Title: Re: Why would the voltage between the two leads of a simple lead-acid (Pb-H2SO4) bat
Post by: Borek on April 18, 2014, 03:11:33 AM

1. Google for Nernst equation.

2. Observed potential depends not only on concentrations/activities of substances reacting, but also on internal resistance of battery (because of IR loses).