CW half reactions 032812.doc
Half-reactions
Any redox reaction can be represented with two “half-reactions”. One
is an oxidation half-reaction, and one is a reduction half-reaction.
When the two half-reactions are combined we get the whole net ionic
equation.
Example 1: Consider the reaction of zinc metal in copper(II) chloride
solution. You can see in photograph the zinc metal above the level of
the solution, but the zinc in the solution has been “eaten away” and
formed zinc ions (Zn2+) in solution, and copper metal has been formed
in its place. The material adhering to what’s left of the zinc is copper
metal which has been transformed from copper ions (Cu2+).
The overall reaction is: Zn(s) + CuCl2(aq)  ZnCl2(aq) + Cu(s)
The net ionic equation is: Zn(s) + Cu2+  Zn2+ + Cu(s) …. Cl- is a spectator ion.
Make sure that you know why zinc is oxidized and copper ions are reduced. Look at the changes is the
oxidation states of zinc and copper. Zinc metal goes from an oxidation state of zero to +2 for the zinc
ions. An increase in oxidation state is oxidation. We can also look at in terms of the electrons lost by
zinc. OILRIG – Oxidation Is Loss, Reduction Is Gain. Copper ions are gaining two electrons to form
copper metal. The oxidation state of copper is decreased from +2 to zero. Both are indicators of
reduction.
We can write two half-reactions, one for the oxidation of zinc metal to zinc ions and one for the reduction
of copper ions to copper metal.
Zn(s)  Zn2+ + 2e- …. zinc loses two electrons. The two electrons lost are written on the product side
Cu2+ + 2e-  Cu(s) …. copper gains the two electrons lost by zinc, to form copper metal
(The loss and gain of electrons occurs when Cu2+ ions are in direct contact with the zinc metal.)
When the two half-reactions are added together, the number of electrons gained and lost cancel out and
we are left with the original net ionic equation.
Example 2: Aluminum metal (in the presence of chloride
ions, which remove the passivating Al2O3 layer) will react
with copper ions according to the following net ionic
equation.
2Al(s) + 3Cu2+  2Al3+ + 3Cu(s)
You can see that aluminum is oxidized to aluminum ions
and copper ions are reduced to copper metal. The copper
metal adheres to the surface of the aluminum foil. What
will be the oxidation and reduction half-reactions?
From the net ionic equation we can write two half-reactions.
2Al(s) + 3Cu2+  2Al3+ + 3Cu(s)
Al(s)  Al3+ + 3e- … oxidation half-reaction … electrons are lost … oxidation number increases
Cu2+ + 2e-  Cu(s) … reduction half-reaction … electrons are gained … oxidation number decreases
Suppose we are given two half reactions and we want to combine them into the net ionic equation. The
number of electrons lost and the number of electrons gained are not the same. We must multiple each of
the entire half reactions by an integer so as to make the number of electrons gained and lost the same.
2(Al(s)  Al3+ + 3e-) = 2Al(s)  2Al3+ + 6e3(Cu2+ + 2e-  Cu(s)) = 3Cu2+ + 6e-  3Cu(s)
2Al(s) + 3Cu2+  2Al3+ + 3Cu(s)
Now the number of electrons gained and lost
are the same. The electrons cancel out in our
finished net ionic equation.
Exercises – Do the following on notebook paper.
1. Write oxidation and reduction half-reactions for each balanced net ionic equation.
a. 2Ag+ + Cu(s)  2Ag(s) + Cu2+
b. Zn(s) + 2H+  Zn2+ + H2(g)
c. Fe(s) + Cd2+  Fe2+ + Cd(s)
d. F2(g) + 2Cl-  2F- + Cl2(g)
e. Cl2(g) + Sn2+  Sn4+ + 2Cl2. Add the following half-reactions together to get the balanced net ionic equation. Be sure to make the
number of electrons lost and gained the same.
a. 2H+ + 2e-  H2(g)
Zn(s)  Zn2+ + 2eb. Sn2+  Sn4+ + 2eAg+ + 1e-  Ag(s)
c. 2Cl-  Cl2(g) + 2eCr3+ + 3e-  Cr(s)
d. Mg(s)  Mg2+ + 2eFe3+ + 1e-  Fe2+
e. Fe3+ + 1e-  Fe2+
Sn2+  Sn4+ + 2e-
Half-reaction exercises – answers
1. Write oxidation and reduction half-reactions for each balanced net ionic equation.
a. Cu(s) + 2Ag+  Cu2+ + 2Ag(s)
Cu(s)  Cu2+ + 2e- ……. Oxidation
2Ag+ + 2e-  2Ag(s) …. Reduction
b. Zn(s) + 2H+  Zn2+ + H2(g)
Zn(s)  Zn2+ + 2e- ..…Oxidation
2H+ + 2e-  H2(g) ..….. Reduction
c. Fe(s) + Cd2+  Fe2+ + Cd(s)
Fe(s)  Fe2+ + 2e- …... Oxidation
Cd2+ + 2e-  Cd(s) … Reduction
Notice that each half-reaction and each
net ionic equation are both mass balanced
and charge balanced.
Mass balance: The same number of
atoms on each side of the arrow.
Charge balance: The same total charge
on each side.
d. F2(g) + 2Cl-  2F- + Cl2(g)
F2(g) + 2e-  2F- ……. Reduction
2Cl-  Cl2(g) + 2e- .….. Oxidation
e. Cl2(g) + Sn2+  Sn4+ + 2ClCl2(g) + 2e-  2Cl- …. Reduction
Sn2+  Sn4+ + 2e- . .…... Oxidation
2. Add the following half-reactions together to get the balanced net ionic equation. Be sure to make the number
of electrons lost and gained the same.
a. 2H+ + 2e-  H2(g)
Zn(s)  Zn2+ + 2eZn(s) + 2H+  Zn2+ + H2(g)
b. Sn2+  Sn4+ + 2e2(Ag+ + 1e-  Ag(s))
Sn2+ + 2Ag+  Sn4+ + 2Ag(s)
c. 3(2Cl-  Cl2(g) + 2e-)
2(Cr3+ + 3e-  Cr(s))
2Cr3+ + 6Cl-  2Cr(s) + 3Cl2(g)
d. Mg(s)  Mg2+ + 2e2(Fe3+ + 1e-  Fe2+ )
Mg(s) + 2Fe3+  Mg2+ + 2Fe2+
e. 2(Fe3+ + 1e-  Fe2+)
Sn2+  Sn4+ + 2e2Fe3+ + Sn2+  2Fe2+ + Sn4+
Notice that in some cases, one or both
half-reactions must be multiplied by an
integer so that the number of electrons
gained and lost are the same.