Question

Balance the redox reaction by half reaction method.
$F{e^{2 + }}(aq)\, + C{r_2}{O_7}^{2 - }(aq)\xrightarrow{{(acid\,medium)}}F{e^{3 + }}(aq) + C{r^{3 + }}(aq)$

Answer 1

Hint: The half reduction method is based on the principle that the electrons lost during oxidation half reaction in a particular redox reaction is equal to the electrons gained in the reduction half reaction. The method is called the half reduction method.

Complete step-by-step answer:The balancing by the half reduction method is done by the following steps:
1.Write the oxidation number of all the atoms above their symbols that appear in the chemical equation.
Now, let us write the oxidation number of all the atoms in our reaction.
$F{e^{2 + }} + {(C{r^{ + 6}}_2{O^{2 - }}_7)^{2 - }}\, \to \,F{e^{3 + }} + C{r^{3 + }}$
2.We will then find out the species which are getting oxidized and also which are getting reduced.
We can easily figure out that $F{e^{2 + }}$ is getting oxidized while $C{r_2}{O_7}^{2 - }$ is getting reduced.
Then we will split the whole equation into two half reactions- oxidation half reaction and reduction half reaction. While balancing each half reaction, we will add electrons to whichever side, in order to balance the oxidation number of the species involved in the half reactions.
Oxidation half reaction: $F{e^{2 + }}(aq)\xrightarrow{{}}F{e^{3 + }}(aq) + {e^ - }$
Reduction half reaction: $C{r_2}{O_7}^{2 - } + 6{e^ - } \to 2C{r^{3 + }}$
3.In the acidic medium, we will balance the oxygen atoms by adding the required number of ${H_2}O$ molecules to the side which is deficient in oxygen atoms. Then we will balance the hydrogen atoms by adding ${H^ + }$ to the side deficient in H atoms.
We can see that the oxidation half reaction is already balanced in our case.
In order to balance O atoms in reduction half reaction, we will add $7$${H_2}O$ molecules to the product side. Then, we need to add $14{H^ + }$ to the reactant side in order to balance H atoms.
The balanced half reactions are as follows:
Oxidation half reaction: $F{e^{2 + }}(aq)\xrightarrow{{}}F{e^{3 + }}(aq) + {e^ - }$
Reduction half reaction: $C{r_2}{O_7}^{2 - }(aq) + 6{e^ - } + 14{H^ + }\xrightarrow[{}]{}2C{r^{3 + }} + 7{H_2}O$
4.Last step is to add the two half reactions.
In order to equate the electrons, we will multiply the oxidation half reaction by $6$ and then the two reactions.
The final reaction will be as follows:
$6F{e^{2 + }} + C{r_2}{O_7}^{2 - } + 14{H^ + } \to 2C{r^{3 + }} + 6F{e^{3 + }} + 7{H_2}O$

Note:We need to remember the difference while balancing a reaction in acidic and basic medium.In the basic medium, we will balance the charges by adding the required number of $O{H^ - }$ ions to the site which is deficient in oxygen atoms. Then we will balance the hydrogen atoms by adding the required number of ${H_2}O$ molecules to the side which is deficient in H atoms.