Balancing Redox Equation — Tips

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This is my first article on chemistry. I would like to share the technique used in balancing redox equations.

There are two methods suggested in textbooks. They are

  1. half equation(reaction) method
  2. oxidation number method

Here, I illustrate the pros and cons of the method(s).

The advantage of the first method is that it does not require calculation of oxidation number and reduces the chance of mistake. For example, the oxidation number of Cr in dichromate ion, if you mis-calculate the result as +9, the whole redox equation will be wrong. However, you need to separate the unbalanced equation into oxidation part and reduction part. It will cost you more steps and more time for solving the question.

For the second method, the situation is as opposite to the first method. You don’t need to separate the unbalanced equation into two parts. That means fewer steps and faster time(if you are familiar with). However, you have to calculate the change in oxidation number correctly. In general case it is easy to calculate (make sure you have no mistake) but in situation such as oxidation of ethanol, it is not so easy to identify the oxidation number of the carbon involved in oxidation. In this case, using method 1 will be better.

If you insist on using the second method, you should draw the structure of ethanol. Then, consider the electronegativity of each atom in an ethanol molecule.

  • The two carbon atoms have different oxidation number here. You still can use this method although it may be ambiguous when you consider the oxidation number in average (i.e. (-1 + -3 )/2 = -2

Example

Here I use a redox reaction to demonstrate the steps of balancing redox equations. HKDSE Chemistry 2017 Paper 1A Q15. It was a multiple choice question. I found it interesting to have a discussion on the methods for balancing redox equation because it is not so familiar. If we follow the steps tightly we still can get the answer — no matter how difficult the question is.

The original equation is written in the form like this:

Since medium does not support subscript or LaTeX, I have to type separately.

Half reaction method in step-by-step illustration:

oxidation number method in step-by-step illustration:

Method 2 requires a calculation on oxidation number where method 1 does not. The tricky one is the last step — addition of six chloride ions on both sides since we want to “construct” back HCl by combining the spectator ions.

In fact, we can add back spectator ions to get the full equation from ionic equation. e.g.

By adding nitrate ions on both sides, we can construct a full equation.

Balancing redox equation in Acidic medium and Alkaline medium

In the steps of balancing redox reactions, no matter which method you use — half equation method or oxidation number method, we use hydrogen ion and water to balance the number of hydrogen atoms and number of oxygen atoms.

But why? for redox reactions involving aqueous solution, water is present.

It is a property of water — self ionization to give out equal amount of hydrogen ion and hydroxide ion.

In acidic medium, concentration of hydrogen ion is predominate and in alkaline medium concentration of hydroxide ion is predominate. Hence, In acidic or neutral medium, we generally use this set to balance number of oxygen atoms and hydrogen atoms in redox equations.

In alkaline medium, we generally use this set to balance number of oxygen atoms and hydrogen atoms in redox equations.

Let’s walk the talk.

Given an unbalanced redox equation of a reaction between acidified potassium permanganate solution and sodium sulphite solution:

Follow the usual way to balance the redox equation (in acidic medium), here I use half reaction method. For oxidation number please refer to another article of my channel.

Let’s say the condition changes from acidic to alkaline. Assume there is no change in the product (very often it does — so I just use it as an example to demonstrate the way to balance equations in alkaline condition)

The steps are almost the same except the way of balancing H atoms in the redox equation. We have to add both sides a water molecule and a hydroxide ion (as shown in yellow region)

The steps are very similar.

In textbook or on some teaching website, it is suggested to balance the equation in the way of acidic medium, and then add sufficient number of hydroxide ion on both sides to get rid of hydrogen ion. It is actually a better method. Using the above example, we have

How about this?

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