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Ion Electron Method

Balance the f...

Question

Balance the following equation by ion electron method.
$Z n + B r O_{4}^{-} + O H^{-} + H_{2} O \to [Z n (O H)_{4}]^{2 -} + B r^{-}$ .

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Solution

Write the reaction in two half reactions,oxidation half and reduction half as:
Oxidation: $Z n \to Z n (O H)_{4}^{- 2}$
Reduction: ${B r O}_{4}^{-} \to B r$
Now balance the oxygen atoms by adding $H_{2} O$ molecules.
$Z n + 4 H_{2} O \to Z n (O H)_{4}^{- 2}$
${B r O}_{4}^{-} \to {B r}^{-} + 4 H_{2} O$
Now balance hydrogen atoms by adding $H^{+}$ ions.
$Z n + 4 H_{2} O \to Z n (O H)_{4}^{- 2} + 4 H^{+} {B r O}_{4}^{-} + 8 H^{+} \to {B r}^{-} + 4 H_{2} O$
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 $H_{2} O$ .
$Z n + 4 {O H}^{-} \to Z n (O H)_{4}^{- 2} {B r O}_{4}^{-} + 4 H_{2} O \to {B r}^{-} + 8 {O H}^{-}$
To balance the charge,add electrons to more positive side to equal the less positive side of the half reaction.
$Z n + 4 {O H}^{-} \to Z n (O H)_{4}^{- 2} + 2 e^{-} {B r O}_{4}^{-} + 4 H_{2} O + 8 e^{-} \to {B r}^{-} + 8 {O H}^{-}$
Now multiply oxidation half reaction by 4 and add both the reactions we will get
$4 Z n + {B r O}_{4}^{-} + 4 H_{2} O + 8 {O H}^{-} \to 4 Z n (O H)_{4}^{- 2} + {B r}^{-}$

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