Observe the following reaction $\begin{aligned} & \text { a } \mathrm{P}_4(\mathrm{~s})+\mathrm{b}^{-}…
Observe the following reaction
$\begin{aligned} & \text { a } \mathrm{P}_4(\mathrm{~s})+\mathrm{b}^{-} \mathrm{OH}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \\ & \qquad \mathrm{cPH}_3(\mathrm{~g})+\mathrm{dH}_2 \mathrm{PO}_2^{-}(\mathrm{aq})\end{aligned}$
$\mathrm{a}, \mathrm{b}, \mathrm{c}$ and $\mathrm{d}$ are respectively
- 1,3,3,1
- $1,3,2,3$
- $3,1,3,1$
- $1,3,1,3$
Solution
The balanced equation is :-
$\mathrm{P}_4(\mathrm{~s})+3 \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{PH}_3(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{PO}_2^{-}$(aq.)
Thus, $a=1, b=3, c=1$ and $d=3$
Asked in: AP EAMCET 2023 (17 May Shift 2)
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