Overview of Redox Equations

I assume that if you chose to download this program that you know what redox equations are, but to make this explanation completely clear I have chosen to include a brief overview of redox equations. You may skip this section if you are confident in your knowledge of redox equations.

Redox equations are a special class of chemical equations that are more difficult to balance than ordinary chemical equations. What makes redox equations so difficult is the concept of charge. In redox equations not only must elements be balanced but charge must be balanced as well. In a redox reaction electrons are transferred, raising the oxidation state of one element while lowering the oxidation state of another. This process of transferring electrons is often promoted by the presence of oxygen, which is the only element with a constant oxidation state. However, this is not always the case, for the presence of oxygen is not required for electron transfer to occur. But when oxygen is present, it is most always coupled with its partner, hydrogen. This is because the majority of redox reactions occur in aqueous solutions, and as we will see, it is important to know the type of solution in which the reaction takes place (acidic or basic). All redox reactions generally look the same. There are rarely more than four elements in a redox equation, and if there are four elements, two of these are always hydrogen and oxygen. The other two are the oxidizing agent (the substance that is gaining electrons) and the reducing agent (the substance that is losing electrons). There may be fewer than four elements because, as I said before, oxygen need not be present for oxidation (electron transfer) to occur. There must be at least two elements, however, an oxidizer and a reducer. A typical skeletal redox equation, that is the equation before it is balanced may look like this:

H₃PO₄(aq)+Cu(s)→CuO(aq)+P(s).

In this reaction copper is being oxidized (it is the reducing agent) while phosphorous is being reduced (it is the oxidizing agent). However, the reducing agent and oxidizing agent need not be known to balance the equation. All that must be known are the subscripts of hydrogen, oxygen, and the elements in every species (i.e. compound) in the reaction. I am not going to explain how to balance a redox equation. There are several methods, but the most widely taught and probably the easiest is the half-reactions method, which can be found in nearly every chemistry textbook on the planet. The algorithm of my redox equation-balancing program is, however, not based on this method. It uses a more systematic method. But who cares about methods anyway. You downloaded this program and your method is to use this program.

How to use Redox Equation Balancer

There is one thing you must know to use my redox equation-balancing program, that is the general form of a skeletal redox equation. It is given below.

H_aE_1bO_c^charge1+H_dE_2eO_f^charge2→H_gE_1hO_i^charge3+H_jE_2kO_l^charge4.

In the equation above H and O are hydrogen and oxygen and E₁ and E₂ are the elements being oxidized or reduced in the first and second reactants, respectively. As I said above, you need not know which element is the oxidizer and which is the reducer. What is important is that you keep them in order when you enter their subscripts into the calculator. That means that the first reactant carries the same element as the first product. If you do not enter them in the order given in the generalized equation above, the calculator will not balance the equation properly. You must keep the right order, even if the equation given to you by your textbook or teacher does not have this order. Let’s look at an example to see how we would enter the subscripts into the calculator. Suppose we were given the skeletal equation

H₃PO₄(aq)+Cu(s)→CuO(aq)+P(s)

and told that the reaction occurs in acidic solution. Ah ha! What do we first notice about this equation? That’s right. It is not in the correct order. If we entered the subscripts from left to right, the calculator would return an obsequious solution. We must change the order. There are two ways we could change it. We could change it to:

H₃PO₄(aq)+Cu(s)→P(s)+CuO(aq)

or we could change it to:

Cu(s)+ H₃PO₄(aq)→CuO(aq)+P(s).

Either of these will work, but we will use the first in this example. Again, the first reactant must carry the same element as the first product. The same applies to the second reactant and the second product.

When we run the redox program, we find that it first asks us for the solution type. It displays a menu at the bottom of the screen with “acid” and “base” as choices. Because we were told that the reaction takes place in acidic solution, we press the button under acid to select the acidic solution type. We now find that calculator screen clears and the calculator now displays “reactant 1 subscripts”, “hydrogen=”. The calculator is prompting us for the subscripts in our skeletal equation. We now look at eh hydrogen subscript in our first reactant (H₃PO₄). We see that it is 3 so we enter 3 at the first prompt. The calculator now says “oxygen=”. It is now asking for the oxygen subscript in our first reactant. We see that it is 4, so we enter 4 at the second prompt. The calculator now says “element=”. The element in our first reactant is phosphorous and its subscript is 1, so we enter 1 at the third prompt. The calculator now says “charge=”. It wants to know the charge on our first reactant. As we can see, our reactant has no net charge, so we enter 0 at the fourth prompt. The calculator’s screen will now clear and display “reactant 2 subscripts”, and as before, it prompts us for the hydrogen subscript. In our second reactant there is no hydrogen, so we enter 0 at the “hydrogen=” prompt. Our second reactant also has no oxygen, so we must enter 0 at the “oxygen=” prompt. The element in our second reactant is clearly copper (Cu). Its subscript is 1, so we enter 1 at the “element=” prompt. Like our first reactant, our second reactant has no net charge, so we enter 0 at the “charge=” prompt once again. The method for entering subscripts is the same for products as it is for reactants. I think that you should understand this procedure by now.

Once you have entered the subscripts and charges for all species in the redox equation, the calculator will process the data you entered to balance the equation. Don’t be surprised if it takes the calculator a few seconds to balance the equation. This redox program utilizes one processing-hungry function called the simult function, which is pre-programmed into the calculator, and a processing-hungry routine that simplifies the data returned by the simult function. If you are familiar with the simult function, you know that it is used to solve systems of linear algebraic equations. The method of balancing redox equations on which this program’s algorithm is based, is balancing redox equation by solving linear systems. This method is impractical to use when balancing equations by hand, but it is much easier to program than any other method I know of and can be generalized quite easily.

When the calculator is finished balancing the equation it will first tell you on which side of the equation water molecules appear, if any appear, and which side hydrogen ions (H+ ions) appear if the reactions occurs in acidic solution, or which side hydroxide (OH-) ions appear if the reaction occurs in basic solution. For our equation the calculator displays “H+ on left”, “H20 on right”. It then displays the coefficients of H+ and H₂O. It says “H+=0” and “H20=3”. Note that the calculator is telling us that the H+ should appear on the left side of the equation with a coefficient of 0. This is the same as leaving it out of the balanced the equation entirely. The calculator now pauses so we may read this information. When we press enter it displays the coefficients of the reactants and products in our skeletal equation. The display reads:

coefficients

reactant 1=2

reactant 2=5

product 1=2

product 2=5

Again order is vital to writing the balanced equation correctly. Recall that reactant 1 is H₃PO₄, reactant 2 is Cu, product 1 is P, and product 2 is CuO. We can now write the balanced equation.

2H₃PO₄(aq)+5Cu(s)→2P(s)+5CuO(aq)+3H₂O(l)

And that’s how it works.

Now you know how to use my redox equation-balancing program. I apologize if this explanation is confusing or too long. I just thought it necessary to fully explain this program so that you use it correctly. Thanks again for downloading it. I hope that it helps you and that you enjoy it.

Jerry Vigil

TI Programmer