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Ion Electron Method

Balance the f...

Question

Balance the following equation by ion electron method.
$F e (C N)_{6}^{4 -} + H^{+} + M n O_{4}^{-} \to F e^{3 +} + C O_{2} + N O_{3}^{-} + M n^{2 +}$ .

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Solution

Write the reaction in two half reactions,oxidation half and reduction half as:
Oxidation: $F e (C N)_{6}^{4 -} \to F e^{3 +} + C O_{2} + N O_{3}^{-}$
Reduction: ${M n O}_{4}^{-} \to M n^{2 +}$
Balance all other atoms except $O$ and $H$ .
$F e (C N)_{6}^{4 -} \to F e^{3 +} + 6 C O_{2} + 6 N O_{3}^{-}$
${M n O}_{4}^{-} \to M n^{2 +}$
Now balance the oxygen atoms by adding $H_{2} O$ molecules.
$F e (C N)_{6}^{4 -} + 30 H_{2} O \to F e^{3 +} + 6 C O_{2} + 6 N O_{3}^{-}$
${M n O}_{4}^{-} \to M n^{2 +} + 4 H_{2} O$
Now balance hydrogen atoms by adding $H^{+}$ ions.
$F e (C N)_{6}^{4 -} + 30 H_{2} O \to F e^{3 +} + 6 C O_{2} + 6 N O_{3}^{-} + 60 H^{+}$
${M n O}_{4}^{-} + 8 H^{+} \to M n^{2 +} + 4 H_{2} O$
To balance the charge,add electrons to more positive side to equal the less positive side of the half reaction.
$F e (C N)_{6}^{4 -} + 30 H_{2} O \to F e^{3 +} + 6 C O_{2} + 6 N O_{3}^{-} + 60 H^{+} + 61 e^{-}$
${M n O}_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$
Now multiply oxidation half reaction by 5 and reduction half reaction by 61 and add both the reactions we will get
$5 F e (C N)_{6}^{4 -} + 61 {M n O}_{4}^{-} + 188 H^{+} \to 5 F e^{3 +} + 30 C O_{2} + 30 N O_{3}^{-} + 94 H_{2} O + 61 M n^{2 +}$

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