Balancing Redox Reactions in Acidic and Basic Conditions

Balancing Redox Reactions in Acidic and Basic Conditions

Hey it’s Professor Dave, let’s learn how
to balance complex redox reactions. We know that redox reactions are reactions
in which electron transfer occurs. One substance is oxidized, which means it
loses electrons, and another substance is reduced, meaning it gains electrons, which
will occur when these electrons are transferred from one substance to another. Sometimes redox reactions are very simple,
where a neutral metal atom is oxidized to become a metal cation, when it transfers electrons
to some other metal cation to form a different neutral metal. These reactions are easy to balance, because
we simply need to make sure that the charges are balanced, by adding electrons and coefficients
if necessary. But redox reactions can become more complicated
if molecules of solvent are involved in the chemical reaction, as water molecules can
be if the reaction is carried out in aqueous solution. Hydrogen atoms or oxygen atoms from water
molecules can be incorporated into these substances, which will make it a little trickier to balance
the corresponding equations. Let’s learn how to balance these kinds of
redox reactions now, considering both acidic and basic conditions, as the approach will
differ slightly for these scenarios. For example, in aqueous solution, the dichromate
ion will react with iron two to yield chromium three and iron three. How do we balance this? The first step is to split this up into half
reactions. For the oxidation half reaction, iron two
becomes iron three. This is the oxidation half-reaction because
the ion has lost another electron. For the reduction half-reaction, dichromate
becomes chromium three. We can see by looking at the oxidation numbers
for chromium that it has been reduced. Now we must balance each half-reaction, starting
with elements other than hydrogen or oxygen. We can see that for the reduction, we need
to put a two here to get two chromium atoms on each side. Next, we balance oxygen by adding water molecules. We have seven oxygen atoms on the left, so
we need to add seven water molecules to the right, since there is one oxygen per molecule. Now we can balance hydrogen atoms by adding
hydrogen ions. Since there are fourteen hydrogen atoms on
the right, from the seven water molecules, we will need fourteen hydrogen ions on the left. This means all of the elements are balanced. Now we can balance the electrical charge for
each half-reaction by adding electrons. In the oxidation, in order for iron two to
become iron three, it must lose an electron, so there must be one electron as a product here. For the reduction, it’s a little trickier. We have 2- on the dichromate and 14 H plusses,
for a total of 12+ on the left. On the right we have two chromium three ions
for a total of 6+. In order to balance this, we will need six
electrons on the left to bring the charge down to 6+ and balance the charge. Now we will need to combine these in such
a way so as to cancel out the electrons. That means we will have to multiply this half-reaction
by six in order to get six electrons to match the other half-reaction. So six of each iron ion and the six electrons
we need. Now we can finally combine these half reactions. Once they are combined, we can cancel out
any species that appear on both sides, which in this case is just the electrons. So we can now see the exact number of ions
that must react in order to produce this particular redox reaction. Let’s make sure to realize that this is
the algorithm we can use under acidic conditions, which is why hydrogen ions are present. If this is occurring under basic conditions,
the algorithm will be just a little bit different, but it will start out by balancing as though
we are under acidic conditions, so let’s do that first, just as we have learned. Let’s say we have solid silver reacting
with zinc two plus to become silver oxide and solid zinc. Again, we will start by separating this into
half reactions. Silver has been oxidized, going from zero
to plus one, and zinc has been reduced, going from plus two to zero. Then, just as before, we balance elements
other than hydrogen and oxygen. That will involve placing a two here before
solid silver. Next, we balance oxygen with water molecules,
that means one on the left here. Then we balance hydrogens with protons, which
means two on the right here. Then we balance charge by including electrons. That means two on the right for this one,
and two on the left for this one. Because the number of electrons in each half
reaction is equal, we can go ahead and combine them, cancel out the electrons, and we get this. If conditions were acidic, we would be finished,
but since they are basic, there is a little more work to do, because it doesn’t make
sense to say that protons exist in basic solution. First, we note how many protons there are,
and we add that number of hydroxides to both sides, so that’s two hydroxides on each
side. Then, anywhere we see protons and hydroxides,
we can combine these to form water molecules, so that makes two waters on the right. Then, we cancel any like terms that remain. In this case, we can eliminate one water molecule
from both sides. And this will be the balanced redox reaction
under basic conditions, with hydroxides listed instead of protons. We can see that in either case, we are involving
solvent molecules in order to supply the hydrogens and oxygens necessary for the reaction, and
we simply follow a series of steps in order to balance all the components involved. Here is that list of steps summarized one
more time, in case you need these instructions for reference, with instructions for acidic
conditions listed first, and then the continuation if conditions are basic. Let’s check comprehension.

22 thoughts on “Balancing Redox Reactions in Acidic and Basic Conditions

  1. This was good but the step where you add electrons could use a bit more clarification. I get it, but only after rewriting the half reactions as one long equation. Maybe annotate that part of the video to eliminate ambiguity as to what left and right actually mean. Again, I got it, but not at first until I stared at it for a little.

  2. I'm from Philippines, it really help a lot Professor Dave, please make new ones to help a lot of people, Thank you so much

  3. x KMnO4 + y HCl = c KCl + b MnCl2 + d H2O + a Cl2

    Find y and b ?
    Please provide step by step solution by ionic method😥
    Thank you in advance

Leave a Reply

Your email address will not be published. Required fields are marked *