In the following reaction, identify the species oxidised, the species reduced, the oxidising agent and the reducing agent.
$4 H C l + M n O_{2} \to M n C l_{2} + 2 H_{2} O + C l_{2}$ .

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Solution

$M n O_{2} + 4 H C l = M n C l_{2} + C l_{2} + 2 H_{2} O$
Writing the oxidation numbers on various atoms,
$+ 4 M n_{2} + + 1 4 H - 1 C l = + 2 M n - 1 C l_{2} + 0 C l_{2} + + 1 2 H_{2} - 2 O$
Ox.no. of $M n$ changes from $+ 4$ to $+ 2$ , i.e., decrease in oxidation number. $M n O_{2}$ is thus reduced. It acts as an oxidising agent. Ox. no. of $C l$ changes from $- 1$ to $0$ , i.e., increase in oxidation number. $H C l$ is thus oxidised. It acts as a reducing agent.

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In the following reaction, identify the species oxidised, the species reduced, the oxidising agent and the reducing agent.4HCl+MnO2→MnCl2+2H2O+Cl2.

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