[xc630]

Hi, I would appreciate any help with the following question:

which of the following reactions should have the larger emf under standard conditions & why?

CuSO4 (aq) + Pb (s) goes to PbSO4 (s) + Cu (s)
Cu(NO3)2 (aq) + Pb(s) goes to Pb(NO3)2(aq) + Cu(s)

I believe emf is just standard potential so it would be the reduction potential of the cathode minus the anode.  however in this case I think the cathode and anode rxns would be the the same for both but how could that be? thanks

[Rabn]

start by writing out the redox reactions for each and then the overall reaction. the answer should be apparent after writing out the balanced reactions.

[LQ43]

CuSO4 (aq) + Pb (s) goes to PbSO4 (s) + Cu (s)
Cu(NO3)2 (aq) + Pb(s) goes to Pb(NO3)2(aq) + Cu(s)

The net ionic equations are different

consider the anode reactions

goes to PbSO4 (s)
vs
goes to Pb(NO3)2(aq)

[xc630]

what does the difference in being solid and aqueous mean?

[Arkcon]

what does the difference in being solid and aqueous mean?

The question you're being asked is:  What does an aqueous solution of a soluble salt mean?  And what does it mean, in terms of electrochemistry, when the aqueous salt plates out as free, solid, metal?

[LQ43]

What does an aqueous solution of a soluble salt mean?  And what does it mean, in terms of electrochemistry, when the aqueous salt plates out as free, solid, metal?

actually in this case, it is as the slightly soluble salt.

Firstly, the reduction is the same for both reactions 1) and 2)

1) CuSO4 (aq) + Pb (s) goes to PbSO4 (s) + Cu (s)
2) Cu(NO3)2 (aq) + Pb(s) goes to Pb(NO3)2(aq) + Cu(s)

but for the anode

1) Pb (s) goes to PbSO4 (s)
2) Pb(s) goes to Pb(NO3)2(aq)

look up the reduction potentials for PbSO4(s) and Pb(NO3)2(aq)

I believe emf is just standard potential so it would be the reduction potential of the cathode minus the anode. 

you should see the difference in the emf.