How would you balance the following chemical equation. \[Fe+C{{l}_{2}}+{{H}_{2}}O\to { }{{\left[ Fe{{\left( {{H}_{2}}O \right)}_{6}} \right]}^{3+}}+C{{l}^{-}}\]

Hint: The balancing of an equation is a process in which the number of atoms of each element present in every molecule is balanced on both the sides along with the charge transfers which took place. The charges are balanced by adding the electrons wherever required, and then making sure it gets canceled out when we write the whole balanced equation.

Complete step-by-step answer:
In order to balance the equation we will follow a stepwise method. At first we will identify the two half reactions which are taking place in the following reaction, to get a better idea.
Here we can see that the chlorine is getting reduced and the iron is getting oxidised. The two half reactions would be
\[Fe\to {{[Fe{{({{H}_{2}}O)}_{6}}]}^{3+}}\]
\[C{{l}_{2}}\to C{{l}^{-}}\]
Here as we can see the oxidation state of chlorine decreases hence it is getting reduced and the oxidation state of iron is increasing hence it is getting oxidised.

Now in the next step we will balance the atoms other than hydrogen and oxygen. Since, the number of atoms of iron is one on both sides, we would not write any coefficient for it now. As per the chlorine, it is clear that on the left side of the equation it is two atoms and on the right side there is only one. So we will add a coefficient $2$.
$Fe\to {{[Fe{{({{H}_{2}}O)}_{6}}]}^{3+}}$
\[C{{l}_{2}}\to 2C{{l}^{-}}\]

In the next step we will balance the number of oxygen atoms on both the sides by adding molecules of water on the other side. We get,
$Fe+6{{H}_{2}}O\to {{[Fe{{({{H}_{2}}O)}_{6}}]}^{3+}}$
\[C{{l}_{2}}\to 2C{{l}^{-}}\]

Now in the next step we will check if the number of hydrogen is balanced too. And as we can see it is balanced in both the sides. Now we will balance the charges on both the sides by adding electrons on the required sides. The charge on the right side of the equation containing iron is $+3$ so we will add three electrons on the same side. In case of the reduction half, there is $-2$ charge on the right hand side, so we add two electrons on the opposite side to balance it out.
$Fe+6{{H}_{2}}O\to {{[Fe{{({{H}_{2}}O)}_{6}}]}^{3+}}+3{{e}^{-}}$
$C{{l}_{2}}+2{{e}^{-}}\to 2C{{l}^{-}}$


Now we will equalize the electrons on both the half reactions.
\[2\times \left\{ Fe+6{{H}_{2}}O\to {{[Fe{{({{H}_{2}}O)}_{6}}]}^{3+}}+3{{e}^{-}} \right\}\]
$3\times \left\{ C{{l}_{2}}+2{{e}^{-}}\to 2C{{l}^{-}} \right\}$

Now the electron on both the reactions becomes six, so it will cancel out each other when we add both these equations. We get,
\[2Fe+3C{{l}_{2}}+12{{H}_{2}}O\to {{[Fe{{({{H}_{2}}O)}_{6}}]}^{3+}}+6C{{l}^{-}}\]
Now finally we will check the mass and charge balance on both sides which is the same on both sides of the equation.

Note:
i) Balancing a chemical equation is done by following a number of steps. At first the identification of two half reactions, where one is oxidation and one is reduction. Then balancing the atoms without the oxygen and hydrogen atoms, eventually we balance the number of oxygen atoms by adding water molecules.
ii) Then we balance the charge on both sides by adding the required amount of electrons and then equalize those numbers of electrons on both sides and write the whole equation by adding both the half equations.