Balance the following equation by oxidation number method.
$K_{2} C r_{2} O_{7} + H C l \to K C l + C r C l_{3} + H_{2} O + C l_{2}$ .

Open in App

Solution

Writing the oxidation numbers of all the atoms.
$_{2} + 6 C r_{2}_{7} + + 1 H - 1 C l \to + 1 K - 1 C l + + 3 C r - 1 C l_{3} +_{2} - 2 O + 0 C l_{2}$
The Ox. no. of Cr has decreased while that of chlorine has increased.
$K_{2} + 6 C r_{2} O_{7} \to + 3 2 C r C l_{3}$ .........(i)
$H - 1 C l \to 0 C l$ ...........(ii)
Decrease in Ox. no. of $C r = 6$ units per molecule $K_{2} C r_{2} O_{7}$
Increase in Ox. no. of $C l = 1$ unit per molecule $H C l$
Eq. (ii) is multiplied by $6$ .
$K_{2} C r_{2} O_{7} + 6 H C l \to 2 C r C l_{3} + 3 C l_{2}$
To balance chlorine and potassium, $14$ molecules of $H C l$ are required.
$K_{2} C r_{2} O_{7} + 14 H C l \to 2 C r C l_{3} + 3 C l_{2} + 2 K C l$
To balance hydrogen and oxygen, $7 H_{2} O$ are added to RHS.
Hence, the balanced equation is,
$K_{2} C r_{2} O_{7} + 14 H C l \to 2 K C l + 2 C r C l_{3} + 3 C l_{2} + 7 H_{2} O$ .

Suggest Corrections

Similar questions

_{2}

K_{2} C r_{2} O_{7} + H C l

\to K C l + C r C l_{3} + H_{2} O + C l_{2}

. . . K_{2} C r_{2} O_{7} + . . . H C l \to . . . K C l + . . . C r C l_{3} + . . . H_{2} O + . . . C l_{2}

K_{2} C r_{2} O_{7} + H C l \to K C l + C r C l_{3} + H_{2} O + C l_{2}

K_{2} C r_{2} O_{7} + F e S O_{4} + H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + H_{2} O

K_{2} {C r}_{2} O_{7} + H C l \to C r {C l}_{3} + H_{2} O + {C l}_{2} + K C l

Join BYJU'S Learning Program