1.

In the reaction 1. H_(2)O_(2)+O_(3) to H_(2)O+2O_(2) 2. PbS+4O_(3) to PbSO_(4)+4O_(2)

Answer»

`O_2` is REDUCED both in (a) and (b)
`O_(3)` is enodined both in (a) and (b)
`O_(3)` is OXIDIZED in (a) and reduced in (b)
`O_(3)`, is reduced in (a) and oxidized in (b)

Solution :Reaction (a) occurs as FOLLOWS:
`O_3 to O_(2)+O`
`H_(2)O_(2)+O to H_2O+O_(2)`
=`H_(2)O_(2)+O_(3) to H_(2)O+2O_(2)`
In this reaction, `O_(3)` is also reduced to `O_(2)`, Reaction
(b) occurs as follows:
`4O_(3) to 4O_(2)+4O`
`PbS+4O to PbSO_(4)+4O_(2)`
In this reaction, `O_(3)` is also reduced to `O_(2)`


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