Find the values of x and y in the following redox reaction,
$x C l_{2} + 6 O H^{-} ⟶ C l O_{3}^{-} + y C l^{-} + 3 H_{2} O$

x = 3, y = 5

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x = 5, y = 3

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x = 4, y = 2

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x = 2, y = 4

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Solution

The correct option is A x = 3, y = 5
$C l_{2} + 6 O H^{-} ⟶ C l O_{3}^{-} + C l^{-} + 3 H_{2} O$

Step-1: Splitting into two half cell reaction,
$C l_{2} + 6 O H^{-} \to C l O_{3}^{-} + 3 H_{2} O$
$C l_{2} \to C l^{-}$

Step-2: Balancing half cell reaction,
$C l_{2} + 12 O H^{-} \to 2 C l O_{3}^{-} + 6 H_{2} O$
$C l_{2} \to 2 C l^{-}$

Step -3 Adding electron to sides deficient in electrons,
$C l_{2} + 12 O H^{-} \to 2 C l O_{3}^{-} + 6 H_{2} O + 10 e^{-}$
$C l_{2} + 2 e^{-} \to 2 C l^{-}$

Step-4 Balancing electron in both half reactions,
$C l_{2} + 12 O H^{-} \to 2 C l O_{3}^{-} + 6 H_{2} O + 10 e^{-}$
$5 [C l_{2} + 2 e^{-} \to 2 C l^{-}]$

Step-5 Adding both half reaction
$6 C l_{2} + 12 O H^{-} \to 2 C l O_{3}^{-} + 10 C l^{-} + 6 H_{2} O$

Dividing by 2,
$3 C l_{2} + 6 O H^{-} ⟶ C l O_{3}^{-} + 5 C l^{-} + 3 H_{2} O$

Thus, the redox reaction is as follows:
$3 C l_{2} + 6 O H^{-} ⟶ C l O_{3}^{-} + 5 C l^{-} + 3 H_{2} O$

Therefore, x and y are 3 and 5 respectively.

Hence, option (c) is correct.

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