Question

$K_{2}Cr{O}_7+KOH\to H_{2}O+K_{2}Cr{O}_4$
Is the above equation correct? How to balance a chemical equation?

Answer 1

This equation is correct and the balanced chemical equation is given by:
$K_{2}C{r}_2{O}_7+2KOH=2K_{2}Cr{O}_4+H_{2}O$
Balancing Of Chemical Equation Based On Reactants And Products Balanced chemical equation
The representation of a chemical reaction in the form of symbols (substances) is known as chemical equation. A chemical equation consists of reactants, products and an arrow showing the direction of reaction. The equation in which number of atoms of all the molecules is equal on both sides of the equation is known as balanced chemical equation.

Law of conservation of mass governs the balancing of a chemical equation. According to this law, mass can neither be created nor be destroyed in a chemical reaction and obeying this law total mass of the elements or molecules present on the reactant side should be equal to the total mass of elements or molecules present on the product side. If the number of atoms of the elements/molecules on the reactant side are not equal to the number of atoms of the elements/molecules on the product side, then it is said to be an unbalanced chemical equation. For example:
$Fe+H_{2}O\to F{e}_3{O}_4+H_2$
In the above example, law of conservation of mass is not applied. The number of iron atoms on the left side is not equal to the number of iron atoms on the right side; there is one iron atom on the left side whereas there are three iron atoms on the right side of the equation.
Unbalanced equation can be represented in given table below:
$\begin{array}{ccc}Element& Noofatomsonreactantside& Noofatomsonproductside\\ Fe\left(Iron\right)& 1& 3\\ H\left(Hydrogen\right)& 2& 2\\ O\left(Oxygen\right)& 1& 4\end{array}$
Now, we need to balance the above chemical reaction.
We will first take the maximum number of atoms present on either side of the reaction, that we can find on product side. (Oxygen = 4).
Then we multiply the number of oxygen atoms on reactant side by 4, such that the number of oxygen atoms on both sides of the reaction is balanced.
The equation becomes $Fe+4H_{2}O\to F{e}_3{O}_4+H_2$
Now the number of hydrogen atoms is 8 on the reactant side which is more than that of the product side, hence we need to balance hydrogen atoms, and hence we multiply $H_2$ on the product side by 4.
The equation becomes $Fe+4H_{2}O\to F{e}_3{O}_4+4H_2$
Now, we can see that number of hydrogen and oxygen atoms on both sides of the equation are same. Now, we need to balance the number of iron atoms on both side of the equation.
We need to multiply the number of iron atoms on the reactant side by 3, so the reaction now becomes,
$3Fe+4H_{2}O\to F{e}_3{O}_4+4H_2$
Now the number of atoms of each element on both sides of the equation is represented in the table given below:
$\begin{array}{ccc}Element& Noofatomsonreactantside& Noofatomsonproductside\\ Fe\left(Iron\right)& 3& 3\\ H\left(Hydrogen\right)& 8& 8\\ O\left(Oxygen\right)& 4& 4\end{array}$
$3Fe+4H_{2}O\to F{e}_3{O}_4+4H_2$
Therefore, we can say that is the above reaction is a balanced chemical equation.
Balancing Redox Reactions
In balancing redox reaction Reduction is defined as the gain of electrons and oxidation is defined as loss of electrons. In other words, oxidation is loss whereas reduction is the gain. The reaction which consists of both oxidation and reduction is termed as a redox reaction. Now we need to learn how to balance these redox reactions.
There are two ways of balancing a redox reaction. One method is by using the change in oxidation number of oxidizing agent and the reducing agent and the other method is based on dividing the redox reaction into two half reactions-one of reduction and other of oxidation.
Balancing by oxidation number
This method can be best explained with the help of the following steps:
1. Correct formula should be written for both reactant and product.
2. Assign the oxidation number to all the elements present in the reaction and hence identify the atoms that will undergo a change in oxidation number.
3. Increase or decrease in oxidation number should be calculated. Make the total ionic charges of reactant and product equal by adding $H^{+}orO{H}^{–}$ when the reaction is taking place in water.
4. Hydrogen atoms in the reaction are made equal by adding $H_{2}O$ molecule either to reactant side or product side.
Balancing by half-reaction method
In this method, the redox reaction is split into two half reactions-one of reduction and other of oxidation and then combined back to give a balanced reaction.
This method can be best explained with the help of the following steps:
1. Put the unbalanced equation in the ionic form for the given reaction.
2. Divide the equations into half reactions: Oxidation half and reduction half
3. For reactions occurring in acidic medium, add $H_{2}O$ to balance O atoms and $H^{+}$ to balance H atoms.
4. Balancing of charges should be done by adding electrons
5. Add the two half-reactions when the number of electrons exchanged is equal to form a redox reaction.
Let’s explain these steps with the help of an example.
1. $Fe(OH{)}_2+H_2{O}_2\to Fe(OH{)}_3+H_{2}O$
2. We balance the half reactions:
$\left(Fe\right(OH{)}_2+O{H}^{-}\to Fe(OH{)}_3+e^{-})x2$
$H_2{O}_2+2e^{-}\to 2O{H}^{-}$
After balancing we can find that, there is no presence of $H_{2}O$. The balanced equation is: $2Fe(OH{)}_2+H-2O_2\to 2Fe(OH{)}_3$