Identify the oxidant and the reductant in the following reaction respectively:
$2 K M n O_{4} + 10 K C l + 8 H_{2} S O_{4} \to 2 M n S O_{4} + 6 K_{2} S O_{4} + 8 H_{2} O + 5 C l_{2}$

K M n O_{4}

&

K C l

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K M n O_{4}

&

H_{2} S O_{4}

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M n S O_{4}

&

K M n O_{4}

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K C l

&

K M n O_{4}

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Solution

The correct option is A $K M n O_{4}$ $&$ $K C l$
The given redox reaction is
$(+ 7) 2 K M n O_{4} + (- 1) 10 K C l + 8 H_{2} S O_{4} \to (+ 2) 2 M n S O_{4} + 6 K_{2} S O_{4} + 8 H_{2} O + (0) C l_{2} .$
$Reduction half reaction$
A species which gains electron in a redox reaction is called as oxidising agent/oxidant.
$(+ 7) 2 K M n O_{4} (oxidant) \to (+ 2) 2 M n S O_{4}$ .
Here $K M n O_{4}$ is gaining electron so it is an oxidant.

$Oxidation half reaction$
A species which looses electron in a redox reaction is called as reducing agent/reductant.
$(- 1) 10 K C l (reductant) \to (0) 5 C l_{2}$ .
Here KCl is losing reagent so it is a reductant.

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Similar questions

2 K M n O_{4} + 10 K C l + 8 H_{2} S O_{4} \to 2 M n S O_{4} + 6 K_{2} S O_{4} + 8 H_{2} O + 5 C l_{2}

K M n O_{4} + H_{2} S O_{4} + H C l \to K_{2} S O_{4} + M n S O_{4} + C l_{2} + H_{2} O

2 M n S O_{4} + 2 K M n O_{4} + 8 H_{2} S O_{4} \to 2 M n S O_{4} + 5 F e_{2} (S O_{4})_{3} + K_{2} S O_{4} + 8 H_{2} O

2 K M n O_{4} + 16 H C l \to 5 C l_{2} + 2 M n C l_{2} + 2 K C l + 8 H_{2} O

F e S O_{4} + K M n O_{4} + H_{2} S O_{4} \to F e_{2} (S O_{4})_{3} + M n S O_{4} + H_{2} O + K_{2} S O_{4}

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