Question

Balance the equation reaction oxidation number method.
$F{e^{2 + }}_{(aq)} + C{r_2}{O_7}{^{ - 2}_{(aq)}} \to F{e^{3 + }}_{(aq)} + C{r^{3 + }}_{(aq)}$ (acidic medium)

Answer 1

Hint:One approach to adjust redox responses is by monitoring the electron move utilizing the oxidation quantities of every one of the iotas/atoms. For the oxidation-number-change technique, start with the unbalanced skeleton condition.

Complete step by step answer:
Let us start with calculating the oxidation number of Cr:
$C{r_2}{O_7}^{2 - }$ oxidation number of Cr = $2(x) + ( - 2)7$= $ - 2$
$ \Rightarrow 2x - 14 = - 2$
Let us know calculate the value of x from the above equation:
$ \Rightarrow 2x = - 2 + 14$
$ \Rightarrow 2x = 12$
Let us now shift the all the values we know to the right side and the value we need to find on the left side that is; the value x:
$ \Rightarrow x = 6$ or $x = + 6$
By, careful observation of the given equation we come to know that:
Iron is oxidized and Cr is reduced.
Those reactions are:
$F{e^{2 + }} \to F{e^{3 + }} + 1{e^ - }$ ….(1)
$2C{r^{6 + }} + 6{e^ - } \to 2C{r^{3 + }}$ …..(2)
Let us balance the above reactions. In the above reactions we need to balance the electrons.
So, let us add 6 electrons to the first reactions and balance it as following:
$F{e^{2 + }} \to F{e^{3 + }} + 6{e^ - }$
By further balancing we get;
$6F{e^{2 + }} \to 6F{e^{3 + }} + 6{e^ - }$
This equation is now balanced with the second equation.
Now, when we check the second equation, we notice that it is balanced.
Let us calculate the charge on both the sides;
Charge on LHS is $ + 12 - 2 = + 10$
Charge on RHS is $ + 18 + 6 = + 24$.
Finally, let us balance the above given equation:
$6F{e^{ + 2}} + C{r_2}{O_7}^{2 - } + 14{H^ + } \to 6F{e^{3 + }} + 2C{r^{3 + }}$
Let us now balance the Hydrogen atoms on both the sides of the equation:
$6F{e^{ + 2}} + C{r_2}{O_7}^{2 - } + 14{H^ + } \to 6F{e^{3 + }} + 2C{r^{3 + }} + 7{H_2}$
So, the balanced equation is $6F{e^{ + 2}} + C{r_2}{O_7}^{2 - } + 14{H^ + } \to 6F{e^{3 + }} + 2C{r^{3 + }} + 7{H_2}$

Additional Information:
Rules for balancing redox reactions using oxidation number rule-
1.Write the condition, show the oxidation number of every component and distinguish the component which is going through change in oxidation number. Identify the oxidizing just as a reducing agent.
2.Multiply the equation of the oxidizing and reducing agent by suitable integers to adjust the absolute increment or diminishes in oxidation number. Balance all atoms other than $H$ and $O$, at long last balance $H$ and $O$ by adding water atoms.

Note:
The oxidation number technique, monitors electrons picked up when a substance is diminished and the electrons lost when a substance is oxidized. Dole out oxidation numbers to every one of the particles in the condition and compose the numbers over the iota/atom.