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Question: In the redox: \( C{r_2}{O_7}^{2 - } + F{e^{2 + }} \to C{r^{3 + }} + F{e^{3 + }} \) . What would the ...

In the redox: Cr2O72−+Fe2+→Cr3++Fe3+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 CrO72−Cr{O_7}^{2 - } be the Reducing Agent, and Fe2+F{e^{2 + }} be the Oxidizing Agent?

Explanation

Solution

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:
Cr2O72−+Fe2+→Cr3++Fe3+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(in  Cr2O72−)+ 6\left( {in\;C{r_2}{O_7}^{2 - }} \right) to +3(in  Cr3+)+ 3\left( {in\;C{r^{3 + }}} \right)
And the oxidation number of iron increases from +2(in  Fe2+)+ 2\left( {in\;F{e^{2 + }}} \right) to +3(in  Fe3+)+ 3\left( {in\;F{e^{3 + }}} \right) , thus it undergoes oxidation. The species undergoing oxidation and reduction are:
Oxidation: Fe2+→Fe3+F{e^{2 + }} \to F{e^{3 + }}
Reduction: Cr2O72−→Cr3+C{r_2}{O_7}^{2 - } \to C{r^{3 + }}.

Step-2: Now balance the each half separately, like:
Fe2+→Fe3+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+ 2 and on the right side is +3+ 3 . To remove the difference, we will add the electron to the right as:
Fe2+→Fe3+  +  e−                    Eq.1F{e^{2 + }} \to F{e^{3 + }}\; + \;{e^ - }\;\;\;\;\;\;\;\;\;\;\\{ Eq.1\\}
Here we can see that Fe2+F{e^{2 + }} will be a reducing agent as it reduces itself. Charge is already balanced and no need to add HH or OO .
Now considering the second half equation:
Cr2O72−→Cr3+C{r_2}{O_7}^{2 - } \to C{r^{3 + }}
Balance the atoms on the both side:
Cr2O72−→2Cr3+C{r_2}{O_7}^{2 - } \to 2C{r^{3 + }}
The oxidation number of chromium on the left side is +6+ 6 whereas +3+ 3 on the right side. Each chromium atom must have three electrons. Since there are two CrCr atoms, add 6e−6{e^ - } on the left side.
Cr2O72−+ 6e−→2Cr3+C{r_2}{O_7}^{2 - } + \,6{e^ - } \to 2C{r^{3 + }}
Here we can see that Cr2O72−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 14H+14{H^ + } on the left side to equate the net charge on both sides.
Cr2O72−+ 6e−+14H+→2Cr3+C{r_2}{O_7}^{2 - } + \,6{e^ - } + 14{H^ + } \to 2C{r^{3 + }}
To balance HH atoms, add 7H2O7{H_2}O molecules on the right.
Cr2O72−+ 6e−+14H+→2Cr3++7H2O                    Eq.2C{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.1Eq.1 by 66 so that electrons are balanced.
[Fe2+→Fe3++e−]×6\left[ {F{e^{2 + }} \to F{e^{3 + }} + {e^ - }} \right] \times 6
Cr2O72−+6e−+14H+→2Cr3++7H2OC{r_2}{O_7}^{2 - } + 6{e^ - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O
⇒6Fe2++Cr2O72−+14H+→6Fe3++2Cr3++7H2O\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:
6Fe2++Cr2O72−+14H+→6Fe3++2Cr3++7H2O6F{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.