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Ion Electron Method

Balance the f...

Question

Balance the following equation by ion electron method.
$K_{3} F e (C N)_{6} + C r_{2} O_{3} + K O H \to K_{4} F e (C N)_{6} + K_{2} C r O_{4} + H_{2} O$ (basic medium).

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Solution

Write the reaction in two half reactions,oxidation half and reduction half as:
Oxidation: $F e (C N)_{6}^{3 -} \to F e (C N)_{6}^{4 -}$
Reduction: $C r_{2} O_{3} \to C r O_{4}^{2 -}$
Balance all atoms other than $O$ and $H$ .
$F e (C N)_{6}^{3 -} \to F e (C N)_{6}^{4 -}$
$C r_{2} O_{3} \to 2 C r O_{4}^{2 -}$
Now balance the oxygen atoms by adding $H_{2} O$ molecules.
$F e (C N)_{6}^{3 -} \to F e (C N)_{6}^{4 -}$
$C r_{2} O_{3} + 5 H_{2} O \to 2 C r O_{4}^{2 -}$
Now balance hydrogen atoms by adding $H^{+}$ ions.
$F e (C N)_{6}^{3 -} \to F e (C N)_{6}^{4 -}$
$C r_{2} O_{3} + 5 H_{2} O \to 2 C r O_{4}^{2 -} + 10 H^{+}$
As reaction takes place in basic medium, add one $O H^{-}$ each side for every $H^{+}$ ion present in the reaction.Combine $O H^{-}$ and $H^{+}$ to form $H_{2} O$ .
$F e (C N)_{6}^{3 -} \to F e (C N)_{6}^{4 -}$
$C r_{2} O_{3} + 10 O H^{-} \to 2 C r O_{4}^{2 -} + 5 H_{2} O$
To balance the charge,add electrons to more positive side to equal the less positive side of the half reaction.
$F e (C N)_{6}^{3 -} + e^{-} \to F e (C N)_{6}^{4 -}$
$C r_{2} O_{3} + 10 O H^{-} \to 2 C r O_{4}^{2 -} + 6 e^{-} + 5 H_{2} O$
Now multiply oxidation half reaction by 6 and add both the reactions we will get
$C r_{2} O_{3} + 6 F e (C N)_{6}^{3 -} + 10 O H^{-} \to 2 C r O_{4}^{2 -} + 6 F e (C N)_{6}^{4 -} + 5 H_{2} O$
Or
$C r_{2} O_{3} + 6 K_{3} [F e (C N)_{6}] + 10 K O H \to 2 K_{2} C r O_{4} + 6 K_{4} [F e (C N)_{6}] + 5 H_{2} O$

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