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Ion Electron Method

Write half-re...

Question

Write half-reactions using electrons.

$3 {C l}_{2} + 6 N a O H ⟶ 5 N a C l + N a C l O_{3} + 3 H_{2} O$

Oxidation

: \frac{1}{2} {C l}_{2} + 6 {O H}^{-} ⟶ C l O_{3}^{-} + 3 H_{2} O + 5 e^{-}

; Reduction

: \frac{1}{2} {C l}_{2} + e^{-} ⟶ C l^{-}

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Oxidation

: \frac{1}{2} {C l}_{2} + 3 {O H}^{-} ⟶ C l O_{3}^{-} + 2 H_{2} O + 5 e^{-}

; Reduction

: \frac{1}{2} {C l}_{2} + e^{-} ⟶ C l^{-}

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Oxidation

: \frac{1}{2} {C l}_{2} + 6 {O H}^{-} ⟶ C l O_{3}^{-} + 2 H_{2} O + 5 e^{-}

; Reduction

: \frac{1}{2} {C l}_{2} + e^{-} ⟶ C l^{-}

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Oxidation

: \frac{1}{2} {C l}_{2} + 3 {O H}^{-} ⟶ C l O_{3}^{-} + 3 H_{2} O + 5 e^{-}

; Reduction

: \frac{1}{2} {C l}_{2} + e^{-} ⟶ C l^{-}

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Solution

The correct option is A Oxidation $: \frac{1}{2} {C l}_{2} + 6 {O H}^{-} ⟶ C l O_{3}^{-} + 3 H_{2} O + 5 e^{-}$ ; Reduction $: \frac{1}{2} {C l}_{2} + e^{-} ⟶ C l^{-}$
The balanced chemical equation for the reaction of chlorine with NaOH is as follows:
$3 {C l}_{2} + 6 N a O H ⟶ 5 N a C l + N a C l O_{3} + 3 H_{2} O$
The oxidation half reaction is $: \frac{1}{2} {C l}_{2} + 6 {O H}^{-} ⟶ C l O_{3}^{-} + 3 H_{2} O + 5 e^{-}$ .
It represents the oxidation of chlorine to chlorate ions.
The reduction half reaction is $: \frac{1}{2} {C l}_{2} + e^{-} ⟶ C l^{-}$ .
It represents reduction of chlorine to chloride ion.

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Similar questions

What is the standard reduction potential of $C l O_{3}^{-} \to \frac{1}{2} C l_{2}$ in acidic medium?

Q. The

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Write half-reactions using electrons.3Cl2+6NaOH⟶5NaCl+NaClO3+3H2O

Write half-reactions using electrons.

$3 {C l}_{2} + 6 N a O H ⟶ 5 N a C l + N a C l O_{3} + 3 H_{2} O$