What is the colour change when aqueous iron $ \left( {II} \right) $ ions are treated with an acidified solution of dichromate $ \left( {VI} \right) $ ?
Answer
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Hint: Oxidation-reduction reaction which is mainly known as redox reaction, is the chemical reaction in which the oxidation number of the reactants changes. There are many common redox reactions. Some of them are rusting, fire and dissolution of metals, the browning of fruits or vegetables, respiration etc.
Complete answer:
Basically, there are two types of redox reaction, one of which is the Oxidation number method and the other is Ion-electron method or half-reaction method. Most of the redox reaction involves the transfer of oxygen atoms, hydrogen atoms or electrons.
We have to write the balanced chemical equation which is in following steps:
Step-1: Firstly, we have to write the equation by using reactant which is:
$ C{r_2}{O_7}^{2 - } + F{e^{2 + }} \to C{r^{3 + }} + F{e^{3 + }} $
The Chromium atom is reduced, as the oxidation number of chromium decreases from $ + 6 $ to $ + 3 $ and the Oxidation number of iron has been increased from $ + 2 $ to $ + 3 $ , thus it undergoes oxidation.
Oxidation reaction: $ F{e^{2 + }} \to F{e^{3 + }} $
Reduction reaction: $ C{r_2}{O_7}^{2 - } \to C{r^{3 + }} $
Step-2: We will balance each half separately after that we will add both the equation:
Oxidation reaction: $ F{e^{2 + }} \to F{e^{3 + }} + {e^ - } $
Reduction reaction: $ C{r_2}{O_7}^{2 - } + 6{e^ - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O $
$ \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 $
Iron $ \left( {II} \right) $ solution have hexaaquairon $ \left( {II} \right) $ ion $ {\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }} $ which is a pale blue or green colour. Now after the chemical reaction this solution of iron $ \left( {II} \right) $ changes into a solution of iron $ \left( {III} \right) $ in the aqueous conditions. The solution of iron $ \left( {III} \right) $ contains the hexaaquairon $ \left( {III} \right) $ ion $ {\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }} $ which is of yellow or brown colour.
Note:
The colour of pure iron $ \left( {III} \right) $ ions are pale violet. However, these colours will be mixed while the reaction is carried out. Potassium chromate $ \left( {VI} \right) $ is orange in colour. This is converted into $ C{r^{3 + }} $ ions which are of green colour. After that we get the mixture of these ions and iron $ \left( {III} \right) $ .
Complete answer:
Basically, there are two types of redox reaction, one of which is the Oxidation number method and the other is Ion-electron method or half-reaction method. Most of the redox reaction involves the transfer of oxygen atoms, hydrogen atoms or electrons.
We have to write the balanced chemical equation which is in following steps:
Step-1: Firstly, we have to write the equation by using reactant which is:
$ C{r_2}{O_7}^{2 - } + F{e^{2 + }} \to C{r^{3 + }} + F{e^{3 + }} $
The Chromium atom is reduced, as the oxidation number of chromium decreases from $ + 6 $ to $ + 3 $ and the Oxidation number of iron has been increased from $ + 2 $ to $ + 3 $ , thus it undergoes oxidation.
Oxidation reaction: $ F{e^{2 + }} \to F{e^{3 + }} $
Reduction reaction: $ C{r_2}{O_7}^{2 - } \to C{r^{3 + }} $
Step-2: We will balance each half separately after that we will add both the equation:
Oxidation reaction: $ F{e^{2 + }} \to F{e^{3 + }} + {e^ - } $
Reduction reaction: $ C{r_2}{O_7}^{2 - } + 6{e^ - } + 14{H^ + } \to 2C{r^{3 + }} + 7{H_2}O $
$ \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 $
Iron $ \left( {II} \right) $ solution have hexaaquairon $ \left( {II} \right) $ ion $ {\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }} $ which is a pale blue or green colour. Now after the chemical reaction this solution of iron $ \left( {II} \right) $ changes into a solution of iron $ \left( {III} \right) $ in the aqueous conditions. The solution of iron $ \left( {III} \right) $ contains the hexaaquairon $ \left( {III} \right) $ ion $ {\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{3 + }} $ which is of yellow or brown colour.
Note:
The colour of pure iron $ \left( {III} \right) $ ions are pale violet. However, these colours will be mixed while the reaction is carried out. Potassium chromate $ \left( {VI} \right) $ is orange in colour. This is converted into $ C{r^{3 + }} $ ions which are of green colour. After that we get the mixture of these ions and iron $ \left( {III} \right) $ .
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