Question

Balance the following equation by oxidation number method.
 $ {K_2}C{r_2}{O_7}\; + \;14\;HCl\; \to \;2\;CrC{l_3}\; + \;2\;KCl\; + \;3\;C{l_2}\; + \;7\;{H_2}O $

Answer 1

Hint: The acquisition of electrons is known as reduction, and the loss of electrons is known as oxidation. Redox reaction/redox process refers to the combination of reduction and oxidation reactions. As previously stated, understanding "balancing redox reactions" is critical. In a redox process, there are typically two ways for balancing redox reactions (chemical equations). The Oxidation Number Method and the Half-Reaction Method are the two techniques.

Complete Step By Step Answer:
Recognize the oxidation and reduction parts of a redox equation before attempting to balance it. Then you strike a balance by ensuring that the electron loss and gain are equal. You determine the oxidation numbers of all atoms using the oxidation number technique. The atoms that have altered are then multiplied by tiny whole numbers. You're making the entire electron loss equal to the total electron gain. The remaining atoms are then balanced. Determine the oxidation numbers and write two half-reactions using the half-reaction technique. Then you multiply them by tiny whole numbers to equalise the electron loss and gain. Then you combine the two partial reactions and balance the remaining atoms. You must know what the oxidation numbers are and what they become during the reaction in both techniques. Occasionally, one approach is more convenient than the other.
Writing the oxidation numbers of all the atoms.
 $ {{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7} + {\text{HCl}} \to {\text{KCl}} + {\text{CrC}}{{\text{l}}_3} + {{\text{H}}_2}{\text{O}} + {\text{C}}{{\text{l}}_2} $
The Oxidation number of Cr has decreased while that of chlorine has increased. $ {{\text{K}}_2}\mathop {{\text{C}}{{\text{r}}_2}}\limits^{ + 6} {{\text{O}}_7} \to 2\mathop {{\text{Cr}}}\limits^{ + 3} {\text{C}}{{\text{l}}_3} $ --(i)
 $ {\text{H}}\mathop {{\text{Cl}}}\limits^{ - 1} \to \mathop {{\text{Cl}}}\limits^0 $ -- (ii)
Decrease in Oxidation number of Cr = 6 units per molecule $ {{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7} $
 Increase in Oxidation number of Cl =1 unit per molecule $ {\text{HCl}} $
Eq. (ii) is multiplied by 6 .
 $ {{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7} + 6{\text{HCl}} \to 2{\text{CrC}}{{\text{l}}_3} + 3{\text{C}}{{\text{l}}_2} $
To balance chlorine and potassium, 14 molecules of HCl are required.
 $ {{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7} + 14{\text{HCl}} \to 2{\text{CrC}}{{\text{l}}_3} + 3{\text{C}}{{\text{l}}_2} + 2{\text{KCl}} $
To balance hydrogen and oxygen, $ 7{{\text{H}}_2}{\text{O}} $ are added to RHS. Hence, the balanced equation is,
 $ {{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7} + 14{\text{HCl}} \to 2{\text{KCl}} + 2{\text{CrC}}{{\text{l}}_3} + 3{\text{C}}{{\text{l}}_2} + 7{{\text{H}}_2}{\text{O}} $ .

Note:
The oxidation number technique is based on the difference between the oxidising and reducing agents' oxidation numbers. The half-reaction technique is based on dividing redox processes into oxidation and reduction halves. It is up to the person to decide which strategy to utilise. It is critical, like with any other reaction, to write the right components and formulae. When writing oxidation-reduction reactions, it's critical to remember to write the compositions and formulae of the substances and products involved in the chemical reaction accurately.