Question

Balance the equation $C + F{e_2}{O_3} \to Fe + CO$

Answer 1

For balancing a chemical equation, the atoms on each side, i.e. the reactant side and product side should be equal. The no. of atoms on the reactant side should be equal to the no. of atoms on the product side. To make the numbers on both sides equal, we need to multiply the no. of atoms of each element on both sides, until they become equal.

Complete answer:
The given reaction is a redox reaction where Iron (III) Oxide is reduced to Iron (0) and carbon is being oxidised from C(0) to C (+2). We will balance the given redox reaction according to balancing the redox reaction in the acidic medium.
First consider the Reduction Half Reaction: $F{e_2}{O_3} \to Fe$
Balancing the no. of Fe atoms on both sides and adding water molecules to balance the 3 oxygen atoms on the reactant side.: $F{e_2}{O_3} \to 2Fe + 3{H_2}O$
Now balancing the no. of Hydrogen on the reactant side by adding ${H^ + }$ ions: $F{e_2}{O_3} + 6{H^ + } \to 2Fe + 3{H_2}O$
Balancing the charges: $F{e_2}{O_3} + 6{H^ + } + 6{e^ - } \to 2Fe + 3{H_2}O$ -- (1)
Similarly considering the Oxidation half reaction: $C \to CO$
Balancing no. of oxygen and adding corresponding no. of ${H^ + }$ ions on the opposite side: $C + {H_2}O \to CO + 2{H^ + }$
Balancing the charges, we get the final reaction as: $C + {H_2}O \to CO + 2{H^ + } + 2{e^ - }$ -- (2)
Multiplying (2) by 3 and adding to (1) we get the final redox reaction as:
$F{e_2}{O_3} + 3C \to 2Fe + 3CO$
This is the required answer.

Note:
When redox reactions are given, always start balancing according to the acidic medium, if it is not mentioned. If the medium is basic then an equal number of $O{H^ - }$ is added, as that of ${H^ + }$ ions. This results in balancing redox reactions in Basic medium.