Video Transcript
A student conducts two electrolysis
experiments using copper(II) sulfate as an electrolyte. He uses graphite electrodes in the
first experiment and copper electrodes in the second experiment. Which line in the table correctly
identifies what will happen at the electrodes in the two experiments?
Before we begin discussing this
problem, let’s remove this table so we can free up some screen space. Electrolysis experiments are
performed using an electrolytic cell. Electrolytic cells are made of two
electrodes in an electrolyte solution, which is a solution that contains ions, which
are connected by wires to a power supply. The power supply causes a redox
reaction to occur in the cell. Negatively charged ions in the
solution are attracted to the positive electrode, which is called the anode.
At the anode, the negatively
charged ion is oxidized. This produces electrons, which
travel through the wire to the negative electrode, which is called the cathode. Positive ions in the solution are
attracted to the cathode. There, they accept the electrons
that traveled through the wire and are reduced. Let’s apply this understanding to
the experiments in this question.
The first experiment uses graphite
electrodes and copper(II) sulfate as the electrolyte. So what will happen at the
electrodes in this experiment? We know that copper two plus ions
will be attracted to the cathode and sulfate ions will be attracted to the
anode. But there are other chemical
species that might be oxidized or reduced. You see, molecules of water are
also present in the cell. The electricity flowing in the
electrolytic cell causes some of the water molecules to be split up into hydrogen
and hydroxide ions. These ions are also attracted to
the electrodes. So how do we know which chemical
species will be oxidized at the anode, hydroxide ions or sulfate ions?
This shows us the general rule for
the tendency of different chemical species to be oxidized at the anode. Ions of the halogens, the chloride
ion, bromide ion, and iodide ion, have a greater tendency to be oxidized, as well as
most electrodes made of reactive materials. Next is the hydroxide ion. Other ions like the sulfate ion and
the nitrate ion are the least likely to be oxidized. Graphite is inert, so it won’t be
oxidized at the anode. The hydroxide ion will. This will produce water and oxygen
gas. So we’ve determined that oxygen gas
will be produced at the anode.
Now let’s determine what will
happen at the cathode. The most stable element is the one
that reacts at the cathode; or we might think of this as the less reactive
element. Between copper and hydrogen, copper
is more stable. So copper will be the species
that’s reduced at the cathode. Copper ions in the solution are
reduced to form solid copper. So copper metal is formed at the
cathode. In experiment two, the only change
is that the electrodes are made of copper. Graphite was inert, but copper is
reactive. This means that copper will be
oxidized at the anode instead of the hydroxide ions. The copper metal that makes up the
electrode will gradually disappear as it dissolves to form copper two plus ions. Copper two plus ions will still be
reduced to form copper metal.
Now we can finish up this
problem. We know that in experiment one,
oxygen is produced at the anode and copper is formed at the cathode. In experiment two, the anode
dissolves and copper is formed at the cathode. So the line in the table that
correctly identifies what will happen at the electrodes in the two experiments is
line three.
