Question

the value of n factor for the phenol in NH4 twice cr2 O7​

Answer 1

Given that,

The phenol in $(NH_{4})_{2}Cr_{2}O_{7}$

We need to calculate the value of n factor

Using given reaction

This reaction show the oxidation of phenol.

We know that,

The number of factor = difference of total oxidation number

In L.H.S,

The sum of oxidation of the total 6 carbon is 4.

In R.H.S,

Here, 4 electrons added.

The oxidation state of the 6 carbon is zero.

According to figure,

$6x=-4\Rightarrow C_{6}H_{4}O_{2}+4e^{-}$

$6x=-4\Rightarrow 6x=0$

Here, the total difference is 4 electron.

Hence, The value of n factor will be 4.