Question

In the following reaction :
${4P+3KOH+3H}_{2}O⟶{3KH}_2{PO}_2+{PH}_3$

• A. P is oxidised only
• B. P Is reduced only
• C. P is reduced only is oxidised as well as reduced
• D. None of above

Answer 1

The correct option is C P is reduced only is oxidised as well as reduced

$4P+3KOH+3H_{2}O\to 3K{H}_{2}P{O}_2+P{H}_3$

On the reactant side, oxidation number for $4P$ is zero.

Oxidation number of K in $KOH$ is $1$ and hydrogen is $1$.

In the product side,

oxidation number of $P$in $K{H}_{2}P{O}_2$

Let $P\to x$

$1+2\times 1+x+2(-2)=1$

oxidation number in $P{H}_3$

$x+3\times 1=-3$

so, the oxidation number of $P$ is increase from zero to one and decrease from zero to $-3$.

When oxidation number increases it means substance oxidized while the oxidation number decreases it means substance reduced.

Hence, option (c) is the correct answer.