Balance the following equation by oxidation number method.
$A l + K M n O_{4} + H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + K_{2} S O_{4} + M n S O_{4} + H_{2} O$ .

Open in App

Solution

Writing oxidation numbers of all atoms,
$0 A l + + 1 K + 7 M n_{4} +_{2} + 6 S_{4} \to + 3 A l_{2} (+ 6 S_{4})_{3} +_{2} + 6 S_{4} + + 2 M n + 6 S_{4} +_{2} - 2 O$
The oxidation numbers of $A l$ and $M n$ have changed
$0 A l \to + 3 A l_{2} (S O_{4})_{3}$ .....(i)
$K + 7 M n O_{4} \to + 2 M n S O_{4}$ ........(ii)
Increase in Ox. no. of $A l$ $= 3$ units per $A l$ atom
$= 6$ units per $A l_{2} (S O_{4})_{3}$ molecule
Decrease in Ox. no. of $M n$ $= 5$ units per $K M n O_{4}$ molecule
Multiplying eq. (i) by $10$ and eq. (ii) by $6$ as to make increase and decrease equal.
$10 A l + 6 K M n O_{4} \to 5 A l_{2} (S O_{4})_{3} + 6 M n S O_{4} + 3 K_{2} S O_{4}$
To balance $S O_{4}^{2 -}$ ions, $24 H_{2} S O_{4}$ molecules be added on LHS.
$10 A l + 6 K M n O_{4} + 24 H_{2} S O_{4} \to 5 A l_{2} (S O_{4})_{3} + 6 M n S O_{4} + 3 K_{2} S O_{4}$
To balance hydrogen and oxygen, $24 H_{2} O$ molecules be added on RHS. Hence, the balanced equation is
$10 A l + 6 K M n O_{4} + 24 H_{2} S O_{4} \to 5 A l_{2} (S O_{4})_{3} + 6 M n S O_{4} + 3 K_{2} S O_{4} + 24 H_{2} O$ .

Suggest Corrections

Similar questions

K M n O_{4} + C_{2} O_{4} H_{2} + H_{2} S O_{4} \to K_{2} S O_{4} + M n S O_{4} + C O_{2} + H_{2} O

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 M n O_{4} + H_{2} S O_{4} + K_{2} C_{3} O_{4} \to M n S O_{4} + C O_{2} + K_{2} S O_{4}

K M n O_{4} + H 2 S O_{4} + F e S O 4 \to K 2 S O_{4} + M n S O_{4} + F e_{2} (S O_{4})_{3} + H_{2} O

F e C_{2} O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + C O_{2} + M n S O_{4} + K_{2} S O_{4} + H_{2} O

Join BYJU'S Learning Program