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Oxidation Number Method

Balance the f...

Question

Balance the following equation by oxidation number method.
$K_{2} C r_{2} O_{7} + F e S O_{4} + H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + H_{2} O$ .

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Solution

Writing oxidation numbers of all the atoms.
$_{2} + 6 C r_{2}_{7} + + 2 F e + 6 S_{4} +_{2} + 6 S_{4} = + 3 C r_{2} (+ 6 S_{4})_{3} + + 3 F e_{2} (+ 6 S_{4})_{3} +_{2} + 6 S_{4} +_{2} - 2 O$
Change in Ox. no. has occurred in chromium and iron.
$K_{2} + 6 C r_{2} O_{7} \to + 3 C r_{2} (S O_{4})_{3}$ ...........(i)
$+ 2 F e S O_{4} \to + 3 F e_{2} (S O_{4})_{3}$
Decrease in Ox. no. of $C r$ per molecule $= (2 \times 6 - 2 \times 3) = 6$ units
Increase in Ox. no. of $F e$ per molecule $= 1$ unit
Hence, eq. (ii) should be multiplied by $6$ ,
$K_{2} C r_{2} O_{7} + 6 F e S O_{4} \to C r_{2} (S O_{4})_{3} + 3 F e_{2} (S O_{4})_{3}$
To balance sulphate ions and potassium ions, $7$ molecules of $H_{2} S O_{4}$ are needed.
$K_{2} C r_{2} O_{7} + 6 F e S O_{4} + 7 H_{2} S O_{4} = C r_{2} (S O_{4})_{3} + 3 F e_{2} (S O_{4})_{3} + K_{2} S O_{4}$
To balance hydrogen and oxygen, $7 H_{2} O$ should be added on RHS. Hence, balanced equation is,
$K_{2} C r_{2} O_{7} + 6 F e S O_{4} + 7 H_{2} S O_{4} = C r_{2} (S O_{4})_{3} + 3 F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + 7 H_{2} O$ .

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