Question

Given a reaction: $C{r_2}O_7^{2 - } + F{e^{2 + }} + {H^ \oplus }\, \to \,C{r^{3 + }} + F{e^{3 + }} + \,{H_2}O$ . The balanced equation is $6F{e^{2 + }} + 14{H^ \oplus } + C{r_2}O_7^{2 - }\, \to \,6F{e^{3 + }} + 2C{r^{3 + }} + 7{H_2}O$
If true enter 1, else enter 0.
A. 1
B. 0

Answer 1

Hint: While balancing the redox reactions, first and foremost find out which element is oxidized and which element is undergoing reduction, after that split it into two half reactions i.e. one for reduction and the other for oxidation.

Complete step-by-step answer:
A redox reaction can be termed as a chemical reaction where electrons are transferred between two reactants participating in it. This transfer of electrons can be recognized by observing the changes in the oxidation states of the reacting species.
Now let us start balancing the reaction: $C{r_2}O_7^{2 - } + F{e^{2 + }} + {H^ \oplus }\, \to \,C{r^{3 + }} + F{e^{3 + }} + \,{H_2}O$
So the very first step as discussed in the Hint section is to find which element is undergoing oxidation and also which element is undergoing reduction.
Since, Chromium’s oxidation state changes from +6 to +3 ($Cr_2^{ + 6}O_7^{2 - }\xrightarrow{{\operatorname{Re} duction}}C{r^{3 + }}$ ), it undergoes reduction. Similarly Iron changes its oxidation state from +2 to +3 ($F{e^{2 + }} \to F{e^{3 + }}$ ) and undergoes oxidation.
Now splitting the reactions on the basis of which element underwent oxidation and reduction simultaneously.
Let us balance the number of atoms except oxygen and hydrogen (for now):
$C{r_2}O_7^{2 - } \to 2C{r^{3 + }}$
$F{e^{2 + }} \to F{e^{3 + }}$
The next step is to balance the number of oxygen atoms. For each extra oxygen atom needed, we will add a water molecule (${H_2}O$).
$C{r_2}O_7^{2 - } \to 2C{r^{3 + }} + 7{H_2}O$ (As $C{r_2}O_7^{2 - }$has 7 oxygen atoms, therefore we add 7 water molecules at the products side)
$F{e^{2 + }} \to F{e^{3 + }}$(No oxygen atom)
After balancing oxygen, the next step is to balance the hydrogen atoms by adding ${H^ \oplus }$ ions (as the reaction takes place in an acidic medium).
$C{r_2}O_7^{2 - } + 14{H^ \oplus } \to 2C{r^{3 + }} + 7{H_2}O$
$F{e^{2 + }} \to F{e^{3 + }}$ (still remains same)
Now the final step is to balance out the charges on both sides by adding electrons to the comparatively positive side and then simplify the half reactions to combine them into a complete balanced reaction.
$C{r_2}O_7^{2 - } + 14{H^ \oplus } + 6{e^ - } \to 2C{r^{3 + }} + 7{H_2}O$ -(a) (Since the net charge on the reactants side is +12 whereas it is +6 on the products side)
$F{e^{2 + }} \to F{e^{3 + }} + {e^ - }$-(b) (Only one electron needed on the products side to balance out the charge)
Now multiplying equation-(b) by 6 to balance the number of reactants in both the half equations
$6F{e^{2 + }} \to 6F{e^{3 + }} + 6{e^ - }$-(c)
Now combining the reactions (a) and (c) and on simplification we get:
$6F{e^{2 + }} + 14{H^ \oplus } + C{r_2}O_7^{2 - }\, \to \,6F{e^{3 + }} + 2C{r^{3 + }} + 7{H_2}O$
Therefore option A is correct i.e. the balanced redox reaction is true.

Note: Keep in mind that while balancing a redox reaction, the number of electrons of one half reaction must cancel out the number of electrons in the other half reaction. This technique will always be handy and save your time.