Question

Balance the following equation in basic medium by ion electron method and oxidation number methods and identify the oxidising agent and the reducing agent: $P_{4}(s)+OH^{-}(aq)\longrightarrow PH_{3}(g)+HPO_{2}^{-}(aq)$

Answer 1

Reaction skeleton

${ P }_{ 4 }\quad +\quad { OH }^{ - }\quad +\quad { H }_{ 2 }O\quad \rightarrow \quad { PH }_{ 3 }\quad +\quad { H }_{ 2 }{ PO }_{ 2 }^{ - }$

Assign oxidation number

${ P }_{ 4 }^{ 0 }\quad +\quad { O }^{ -2 }{ H }^{ +1- }\quad +\quad { H }_{ 2 }^{ +1 }{ O }^{ -2 }\quad \rightarrow \quad { P }^{ -3 }{ H }_{ 3 }^{ +1 }\quad +\quad { H }_{ 2 }^{ +1 }{ P }^{ +1 }{ O }_{ 2 }^{ -2- }$

Identify redox couple in the reaction

$O:\quad { P }_{ 4 }^{ 0 }\quad \rightarrow \quad { H }_{ 2 }^{ +1 }{ P }^{ +1 }{ O }_{ 2 }^{ -2- }$

$R:\quad { P }_{ 4 }^{ 0 }\quad \rightarrow \quad { P }^{ -3 }{ H }_{ 3 }^{ +1 }$

Combine redox reactions

$O:\quad { P }_{ 4 }^{ 0 }\quad \rightarrow \quad { H }_{ 2 }^{ +1 }{ P }^{ +1 }{ O }_{ 2 }^{ -2- }$

$R:\quad { P }_{ 4 }^{ 0 }\quad \rightarrow \quad { P }^{ -3 }{ H }_{ 3 }^{ +1 }$

Balance all atoms except hydrogen and oxygen

$O:\quad { P }_{ 4 }\quad \rightarrow \quad 4{ H }_{ 2 }{ PO }_{ 2 }^{ - }$

$R:\quad { P }_{ 4 }\quad \rightarrow \quad 4{ PH }_{ 3 }$

Balance oxygen atoms

$O:\quad { P }_{ 4 }\quad +\quad 8{ H }_{ 2 }O\quad \rightarrow \quad 4{ H }_{ 2 }{ PO }_{ 2 }^{ - }$

$R:\quad { P }_{ 4 }\quad \rightarrow \quad 4{ PH }_{ 3 }$

Balance hydrogen atoms

$O:\quad { P }_{ 4 }\quad +\quad 8{ H }_{ 2 }O\quad \rightarrow \quad 4{ H }_{ 2 }{ PO }_{ 2 }^{ - }\quad +\quad 8{ H }^{ + }$

$R:\quad { P }_{ 4 }\quad +\quad { 12H }^{ + }\quad \rightarrow \quad 4{ PH }_{ 3 }$

For basic medium reaction add OH- on either side

$O:\quad { P }_{ 4 }\quad +\quad 8{ H }_{ 2 }O\quad +\quad 8{ OH }^{ - }\quad \rightarrow \quad 4{ H }_{ 2 }{ PO }_{ 2 }^{ - }\quad +\quad 8{ H }^{ + }$

$R:\quad { P }_{ 4 }\quad +\quad 12{ H }_{ 2 }O\quad \rightarrow \quad 4{ PH }_{ 3 }\quad +\quad 12{ OH }^{ - }$

Balance the charge on both sides

$O:\quad { P }_{ 4 }\quad +\quad 8{ H }_{ 2 }O\quad +\quad 8{ OH }^{ - }\quad \rightarrow \quad 4{ H }_{ 2 }{ PO }_{ 2 }^{ - }\quad +\quad 8{ H }_{ 2 }O\quad +\quad 4{ e }^{ - }$

$R:\quad { P }_{ 4 }\quad +\quad 12{ H }_{ 2 }O\quad +\quad 12{ e }^{ - }\quad \rightarrow \quad 4{ PH }_{ 3 }\quad +\quad 12{ OH }^{ - }$

Make electron gain and electron lost

$O:\quad { P }_{ 4 }\quad +\quad 8{ H }_{ 2 }O\quad +\quad 8{ OH }^{ - }\quad \rightarrow \quad 4{ H }_{ 2 }{ PO }_{ 2 }^{ - }\quad +\quad 8{ H }_{ 2 }O\quad +\quad 4{ e }^{ - }$

$R:\quad { P }_{ 4 }\quad +\quad 12{ H }_{ 2 }O\quad +\quad 12{ e }^{ - }\quad \rightarrow \quad 4{ PH }_{ 3 }\quad +\quad 12{ OH }^{ - }$

$O:\quad 3{ P }_{ 4 }\quad +\quad 24{ H }_{ 2 }O\quad +\quad 24{ OH }^{ - }\quad \rightarrow \quad 12{ H }_{ 2 }{ PO }_{ 2 }^{ - }\quad +\quad 24{ H }_{ 2 }O\quad +\quad 12{ e }^{ - }$

$R:\quad { P }_{ 4 }\quad +\quad 12{ H }_{ 2 }O\quad +\quad 12{ e }^{ - }\quad \rightarrow \quad 4{ PH }_{ 3 }\quad +\quad 12{ OH }^{ - }$

Add half reactions together

$4{ P }_{ 4 }\quad +\quad 36{ H }_{ 2 }O\quad +\quad 24{ OH }^{ - }\quad +\quad 12{ e }^{ - }\quad \rightarrow \quad 12{ H }_{ 2 }{ PO }_{ 2 }^{ - }\quad +\quad 4{ PH }_{ 3 }\quad +\quad 24{ H }_{ 2 }O\quad +\quad 12{ OH }^{ - }\quad +\quad 12{ e }^{ - }$

Simplifying the equation, we get

${ P }_{ 4 }\quad +\quad 3{ H }_{ 2 }O\quad +\quad 3{ OH }^{ - }\quad \rightarrow \quad 3{ H }_{ 2 }{ PO }_{ 2 }^{ - }\quad +\quad { PH }_{ 3 }$

Answer 2

Answer:

tha y chor ek cr9uir the x7b rc 6re

Explanation:

h82so3j2s9uxu9d3x3ei020iuv3sx1su3d8h3dijkne1