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Standard Electrode Potentials Lab Question, Need help!

Submitted by Anonymous (not verified) on Wed, 04/30/2008 - 03:44

In a cell Ag/AgCl/Cl^- (1.00 M) and Pb/PbI2/I^- (1.00) Is PbI2 formed or consumed and how do you know?  Also, how do you write the balanced equation for the overall reaction for this cell? 

You need to determine which is more active by referring to an activity chart.  If you do that you will find that Pb is more active than Ag, so the Pb metal will lose electrons while the Ag+ ions will gain them.
                  Pb  -->  Pb+2    +  2 e-
                  2 Ag+  +  2e-  -->  2 Ag

So, PbI2 is being formed as the Pb metal dissolves and forms the Pb+2 ions.

The overall reaction is:
                Pb  +  2 Ag+  --> Pb+2 *  +  2 Ag
* A secondary reaction then occurs in the Pb half cell where
                Pb+2  +  2 I-  -->  PbI2 (solid)

Submitted by spock on Wed, 04/30/2008 - 10:04 Permalink