Identify the oxidant and the reductant in the following reaction respectively:
$(i) F e C l_{2} + H_{2} O_{2} + H C l \to F e C l_{3} + H_{2} O$

F e C l_{2}

&

H_{2} O_{2}

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H_{2} O_{2}

&

F e C l_{2}

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H C l

&

F e C l_{2}

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F e C l_{2}

&

H C l

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Solution

The correct option is B $H_{2} O_{2}$ $&$ $F e C l_{2}$
$(+ 2) F e C l_{2} + (- 1) H_{2} O_{2} + H C l \to (+ 3) F e C l_{3} + (- 2) H_{2} O .$
$(+ 2) F e C l_{2} (reductant) \to (+ 3) F e C l_{3} (oxidation half) .$
$(- 1) H_{2} O_{2} (oxidant) \to (- 2) H_{2} O (reduction half) .$

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