19. Given the following half-reactions, which are arranged in the same order as in your Chem 132 from top to bottom): Cu2* 2e Cu () Ered +0.337 V If you make a mixture of Pb+, Pb (s), Cu, and Cu (s) all in the same beaker, which way will Electrons flow spontaneously? a. from Cu (s) to Pb2+ b. from Cu to Pb (s) c. from Pb2 to Cu (s) d. from Pb (s) to Cu2 e. from Pb2+ to Cu\'+ Solution by defintion the one with most positive reduction potential undergoes reduction so Copper underoges reduction here Cu+2 + 2e- --> Cu (s) so the electrons are gained by Cu+2 and now lead undergoes oxidation Pb(s) ---> Pb+2 + 2e- so the electrons are lost from Pb(s) now the net reaction is Pb(s) + Cu+2 ---> Pb+2 + Cu(s) so the electrons flow spontaneously from Pb(s) to Cu+2 so the answer is d) from Pb(s) to Cu+2 .

1. 19. Given the following half-reactions, which are arranged in the same order as in your Chem
132 from top to bottom): Cu2* 2e Cu () Ered +0.337 V If you make a mixture of Pb+, Pb (s), Cu,
and Cu (s) all in the same beaker, which way will Electrons flow spontaneously? a. from Cu (s)
to Pb2+ b. from Cu to Pb (s) c. from Pb2 to Cu (s) d. from Pb (s) to Cu2 e. from Pb2+ to Cu'+
Solution
by defintion
the one with most positive reduction potential undergoes reduction
so
Copper underoges reduction here
Cu+2 + 2e- --> Cu (s)
so the electrons are gained by Cu+2
and now lead undergoes oxidation
Pb(s) ---> Pb+2 + 2e-
so the electrons are lost from Pb(s)
now
the net reaction is
Pb(s) + Cu+2 ---> Pb+2 + Cu(s)
so
the electrons flow spontaneously from Pb(s) to Cu+2
so

2. the answer is d) from Pb(s) to Cu+2