Untitled - Kelly Walsh High School

Aqueous Solutions 69
There will be no reaction because the substance on the left is not higher on the
activity series.
Li(s) H 2O(l) l
H OH
Li(s) l Li (aq) e
H 2(g) l 2 H (aq) 2 e
There will be a reaction, so we need to reverse the second equation.
We now have the following equations:
Ni(s) l Ni 2 (aq) 2 e
2 H (aq) 2 e l H 2(g)
Zn(s) l Zn 2 (aq) 2 e
Cu 2 (aq) 2 e l Cu(s)
Mg(s) l Mg 2 (aq) 2 e
2 H (aq) 2 e l H 2(g)
Al 2(SO 4) 3(aq) Fe(s) l NR
Li(s) l Li (aq) e
2 H (aq) 2 e l H 2(g)
We need to adjust these equations so the number of electrons in each pair of
reactions matches. All the reactions, except the lithium equation, have two electrons.
We need to multiply the lithium equation by two so its electrons will
match those in the hydrogen reaction. We then add each pair of reactions and
cancel the electrons:
Ni(s) 2 H (aq) 2 e l Ni 2 (aq) 2 e H 2(g)
Zn(s) Cu 2 (aq) 2 e l Zn 2 (aq) 2 e Cu(s)
Mg(s) 2 H (aq) 2 e l Mg 2 (aq) 2 e H 2(g)
Al 2(SO 4) 3(aq) Fe(s) l NR
2 Li(s) 2 H (aq) 2 e l 2 Li (aq) 2 e H 2(g)
Later, we will see that the first and second reactions are the net ionic equations
for these reactions.
To get to the molecular equation we must put the ions in the above reactions
into their original compounds.
Ni(s) 2 HCl(aq) l Ni 2 (aq) H 2(g)
Zn(s) CuBr 2(aq) l Zn 2 (aq) Cu(s)