. The reaction Pb(s) + 2HCl(aq)  H2(g) + PbCl2(s) occurs in a beaker and is found to be exothermic.

Calculate Eo

(So im assuming that CL does not change so i discluded it now im stuck with a reaction with a negative E which im pretty sure is wrong)

If you had shown your work I could find the error. You're right, the Cl doesn't change oxidation state; however, it enters the reaction (formation PbCl2) so it should be included in the equation.And I assume you meant to say USE Eo to calculate Ereaction.

To calculate Eo (standard cell potential) for the given reaction, we can use the standard reduction potentials of the half-reactions involved.

The half-reactions involved in this reaction are:

1. Pb2+(aq) + 2e- -> Pb(s) (Reduction half-reaction)
This half-reaction occurs at the cathode.

2. 2H+(aq) + 2e- -> H2(g) (Oxidation half-reaction)
This half-reaction occurs at the anode.

Now, let's find the standard reduction potentials for these half-reactions. The reduction potentials are typically given in tables.

From the table, we find:

Eo (Pb2+ -> Pb) = -0.13 V (reduction potential for reduction of Pb2+)
Eo (H+ -> H2) = 0.00 V (reduction potential for reduction of H+)

To calculate the standard cell potential (Eo) for the overall reaction, we subtract the reduction potential of the anode half-reaction from the reduction potential of the cathode half-reaction:

Eo = Eo (cathode) - Eo (anode)
= -0.13 V - 0.00 V
= -0.13 V

Note that the negative sign indicates that the reaction is exothermic and spontaneous in the direction written.