UNDERSTANDING BALANCING OF REDOX EQUATIONS SIMPLY

In this lesson, we shall be discussing the step by step process on how to balance redox equation in the basic medium as taught in our schools. By the end of our discussion, you should be able compute questions on the topic above with no difficulty at all. We have dealt with balancing redox equations in acidic medium in our previous lesson and we shall apply the same technique involving basic medium. Balancing redox equations in basic medium means to have a standard basic substance in the equation. The generally accepted base indicator is the OH-.
Let’s take time to analyze the unbalanced equation below and attempt to balance it in the basic medium
MnO4 - + Cr3+ ----------- MnO2- + Cr2O72- (unbalanced equation)
We shall begin our discussions in balancing the above equation in basic medium by first realizing that the equation above consists of two sets of look-alike compounds. We have compounds, some containing Manganese (Mn) and others of chromium (Cr) elements and thus we split the lookalike compounds into what we term as half equations. These half equations are sometimes given names such as reduction and oxidation depending on what exactly is happening in each half equation.

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UNDERSTANDING BALANCING OF REDOX EQUATIONS IN
BASIC MEDIUM
“Nothing Like You Were Taught in School”

In this lesson, we shall be discussing the step by step process on how to balance redox
equation in the basic medium as taught in our schools. By the end of our discussion,
you should be able compute questions on the topic above with no difficulty at all. We
have dealt with balancing redox equations in acidic medium in our previous lesson and
we shall apply the same technique involving basic medium. Balancing redox equations
in basic medium means to have a standard basic substance in the equation. The
generally accepted base indicator is the OH-.
Let’s take time to analyze the unbalanced equation below and attempt to balance it in
the basic medium
MnO -
4 + Cr3+ ----------- MnO2 + Cr2O7 (unbalanced equation)
We shall begin our discussions in balancing the above equation in basic medium by
first realizing that the equation above consists of two sets of look-alike compounds.
We have compounds, some containing Manganese (Mn) and others of chromium (Cr)
elements and thus we split the lookalike compounds into what we term as half
equations. These half equations are sometimes given names such as reduction and
oxidation depending on what exactly is happening in each half equation.
Let’s split the unbalanced equation first into the two half equations, before naming
them as oxidation or reduction.

MnO -
4 ------------------ MnO2
Cr3+ -------------------- Cr
2-
2O7
The two equations above show the two half equations. We shall name them either
reduction or oxidation half, depending on whether as each equation proceeds from left
to right, there is a decrease or increase in the number of oxygen atoms or whether
there is an increase or decrease in oxidation number of the central atoms in the
equation. When there is an increase in the number of oxygen atoms as the reaction
proceeds from left to right, then we are dealing with an oxidation equation, otherwise
we are looking at a reduction equation. Also, when the oxidation number of the central
atom in the equation is increasing from left to right, then we are dealing with an
oxidation equation, but if decreasing, then we are looking at a reduction equation.
However, remember that knowing which half equation reduction or oxidation is not
necessary in successfully balancing the overall equation.
For an unbalanced redox equation, when one of the half equations are known to be
reduction, then the other is oxidation and vice versa. Thus, let’s proceed to name them
accordingly;
MnO -
4 ----------------- MnO2 (reduction half equation)
Cr3+ -------------------- Cr
2O7 (oxidation half equation)
We will learn to balance each of the half equations separately. Let’s begin by balancing
the reduction half equation in basic medium;
MnO -
4 ----------------- MnO2
Before we can balance this equation very well, we will have to first ensure that the
amounts of the central atom on both sides of the equation are the same. As we can
see from the reduction half equation above, the amounts of the central atom which is
manganese (Mn) on both sides of the equations are the same. This means to ask
oneself, how many manganese atoms are on the left- and right-hand sides. If the
amounts of the central atoms on both sides of the equation are not the same, we will
have to ensure that they are the same before we can effectively proceed to continue
the balancing process. Since the amounts of the central atom are the same, we proceed
to the next step, which is to ensure that the oxygen atoms are balanced. Oxygen atoms
are usually found in the water molecule (H2O) and so we will ensure to balance the
oxygen atoms on both sides of the equation with equal number of water molecules.
From the unbalanced reduction equation MnO -
4 ----------------- MnO2 we can see
that the right- and left-hand side of the equation have four and two oxygen atoms
respectively, thus the right hand side is short of two atoms of oxygen atoms. Thus, we
would have need two water molecules (oxygen atoms) on the right-hand side in order
to ensure a balance. However, in balancing redox equation in basic medium, the
number of oxygen atoms needed to balance the oxygen atoms on both sides of the
equation, will be placed on the side of the equation with the greater number of oxygen
atoms instead of the usual side with the least number of oxygen atoms. With this
understanding, we can proceed to place two water molecules on the right hand side
of the equation as shown below;
MnO -
4 + 2H2O ----------------- MnO2 (balancing oxygen atoms with water
molecules)
After balancing the oxygen atoms with water molecules, we will proceed to balance the
other side of the equation with hydroxide ions, whose number must be twice the