Question

In the redox: $ C{r_2}{O_7}^{2 - } + F{e^{2 + }} \to C{r^{3 + }} + F{e^{3 + }} $ . What would the balanced reaction look like, and would $ Cr{O_7}^{2 - } $ be the Reducing Agent, and $ F{e^{2 + }} $ be the Oxidizing Agent?

Answer 1

Hint: There are a number of methods for balancing the oxidation-reduction reaction, the two methods are very much important. These are:
-Oxidation number method
-Ion-electron method or half-equation method.

Complete answer:
We will solve this question in following steps:
Step-1: First of all, divide the equation into two half reactions. The oxidation numbers of various atoms are shown below:
 $ C{r_2}{O_7}^{2 - } + F{e^{2 + }} \to C{r^{3 + }} + F{e^{3 + }} $
In this case, chromium undergoes reduction, the oxidation number of chromium decreases from $ + 6\left( {in\;C{r_2}{O_7}^{2 - }} \right) $ to $ + 3\left( {in\;C{r^{3 + }}} \right) $
And the oxidation number of iron increases from $ + 2\left( {in\;F{e^{2 + }}} \right) $ to $ + 3\left( {in\;F{e^{3 + }}} \right) $ , thus it undergoes oxidation. The species undergoing oxidation and reduction are:
Oxidation: $ F{e^{2 + }} \to F{e^{3 + }} $
Reduction: $ C{r_2}{O_7}^{2 - } \to C{r^{3 + }} $.

Step-2: Now balance the each half separately, like:
 $ F{e^{2 + }} \to F{e^{3 + }} $
Balance all the atoms separately. Here it is already balanced. The oxidation number on the left side is $ + 2 $ and on the right side is $ + 3 $ . To remove the difference, we will add the electron to the right as:
 $ F{e^{2 + }} \to F{e^{3 + }}\; + \;{e^ - }\;\;\;\;\;\;\;\;\;\;\{ Eq.1\} $
Here we can see that $ F{e^{2 + }} $ will be a reducing agent as it reduces itself. Charge is already balanced and no need to add $ H $ or $ O $ .
Now considering the second half equation:
 $ C{r_2}{O_7}^{2 - } \to C{r^{3 + }} $
Balance the atoms on the both side:
 $ C{r_2}{O_7}^{2 - } \to 2C{r^{3 + }} $
The oxidation number of chromium on the left side is $ + 6 $ whereas $ + 3 $ on the right side. Each chromium atom must have three electrons. Since there are two $ Cr $ atoms, add $ 6{e^ - } $ on the left side.
 $ C{r_2}{O_7}^{2 - } + \,6{e^ - } \to 2C{r^{3 + }} $
Here we can see that $ C{r_2}{O_7}^{2 - } $ will be an oxidizing agent as it oxidises itself.
Since the reaction takes place in an acidic medium, add $ 14{H^ + } $ on the left side to equate the net charge on both sides.
 $ C{r_2}{O_7}^{2 - } + \,6{e^ - } + 14{H^ + } \to 2C{r^{3 + }} $
To balance $ H $ atoms, add $ 7{H_2}O $ molecules on the right.
 $ C{r_2}{O_7}^{2 - } + \,6{e^ - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O\;\;\;\;\;\;\;\;\;\;\{ Eq.2\} $
This is the balanced half equation.

Step-3: Now add up the two half equations. Multiply the $ Eq.1 $ by $ 6 $ so that electrons are balanced.
 $ \left[ {F{e^{2 + }} \to F{e^{3 + }} + {e^ - }} \right] \times 6 $
 $ C{r_2}{O_7}^{2 - } + 6{e^ - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O $
 $ \Rightarrow 6F{e^{2 + }} + C{r_2}{O_7}^{2 - } + 14{H^ + } \to 6F{e^{3 + }} + 2C{r^{3 + }} + 7{H_2}O $
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}O $

Note:
There are two important characteristics which are:
-Reduction-oxidation reactions are coupled which means in any oxidation reaction a reciprocal reduction occurs.
-They involve a net chemical change which means an atom or electron goes from one unit of matter to another.