Writing the oxidation number of all the atoms.
`overset(+1)(K_(2))overset(+6)(Cr_(2))overset(-2)(O_(7))+overset(+1)(K)overset(-1)(Cl)+overset(+3)(Cr)overset(-1)(Cl_(3))+overset(+1 -2)(H_(2)O)+overset(0)(Cl_(2))`
The Ox. No. of Cr has decreased while has sdecreased while that of chlorine has increased.
`K_(2)overset(+6)(Cr_(2))O_(7) to 2overset(+3)(CrCl_(3)...(i)`
`overset(-1)(HCl) to overset(0)(Cl)..(ii)`
Decrease in Ox. no. of Cr=6 units per molecule `K_(2)Cr_(2)O_(7)`
Increase in Ox. no. of Cl=1 unit per moleule HCl Eq. (ii) is multiplied by 6.
`K_(2)Cr_(2)O_(7) +6HCl to 2CrCl_(3)+3Cl_(2)`
To balance chlorine and potassium, 14 molecules of HCl of required.
`K_(2)Cr_(2)O_(7)+14HCl to 2CrCl_(3)+3CrCl_(3)+3Cl_(2)+2KCl`
To balance hydrogen and oxygen. `7H_(2)O` are added to RHS Hence, the balanced equation is
`K_(2)Cr_(2)O_(7)+14HCl to 2KCl+2CrCl_(3)+3Cl_(2)+7H_(2)O`
