Question

Balance this equation in basic medium :


P4 + OH- =======> PH3 + H2PO2^-​

Answer 1

$<b> <u> <i>Heya ! </b> </u> </i>$

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$<Font color= "red"> <u>★Balancing Redox Equations : Alkaline Medium ★ </u> </font>$
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$<b> Given Equation → P4 + OH- -----------> PH3 + H2PO2- </b>$
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$<u>Step.1 ]→ Split the equation into two Half Reactions .</u>$ One for the oxidation and one for the reduction reaction .

Oxidation Reaction - P4 ----------> H2PO2-

Reduction Reaction - P4 -----------> PH3

[ This can be noted by indicating the oxidation number of each species. An increase in oxidation number means oxidation and decrease means reduction ]

$<u>Step. 2]→ . Balance the atoms in each of the equation except oxygen and hydrogen. </u>$

Oxidation Reaction - P4 ----------> 4H2PO2-

Reduction Reaction - P4 -----------> 4PH3

$<u> Step. 3]→ . Balance the Oxygen atoms next by adding water.</u>$

Oxidation Reaction - P4 + 8H2O ----------> 4H2PO2-

Reduction Reaction - P4 -----------> 4PH3

$<u>Step. 4]→ balance the hydrogen atoms next by adding positive H atoms (H+)</u>$

Oxidation Reaction - P4 + 8H2O ---------->( 4H2PO2- ) + (8H+)

Reduction Reaction - P4 + 12 H+ -----------> 4PH3

⚫$<i>Now since we've gotta balance the equation in alkaline medium , so there can't be H+ atoms! Therefore eliminate them by substituting OH- atoms in place of each H+ on the opposite side . </i>$

$<u>Step. 4]→ Add OH- atoms to balance in Alkaline Medium .</u>$

Oxidation Reaction - P4 + 8H2O + 8OH- ---------->4H2PO2- + 8H2O

Reduction Reaction - P4 + 12 H2O -----------> 4PH3 + 12 OH-

$<u>Step. 5 ]→ balance the overall charge by adding electrons :</u>$

Oxidation Reaction - P4 + 8H2O + 8OH- ---------->4H2PO2- + 8H2O + 4e-

Reduction Reaction - P4 + 12 H2O + 12e- -----------> 4PH3 + 12 OH-

after we're done with balancing the charge , we can notice that the number of electrons in both the equations isn't the same. We have 12 electrons in (2) equation and only 4 in (1) equation .So in order to make it same , we multiply equation (1) with 3

$<u>Step. 6]→ Multiply Eqaution (1) By 3</u>$ :-

Oxidation Reaction - 3P4 + 24H2O + 24OH- ---------->12H2PO2- + 24H2O + 12e-

Reduction Reaction - P4 + 12 H2O + 12e- -----------> 4PH3 + 12 OH-

$<u>Step. 7 ]→ : After we are done , we can combine the two equations . So add the two Half equations and cancel out the common terms on both sides .</u>$

→ 4P4 + 12H2O + 12 OH- ----------> 4PH3 + 12H2PO2-

$<u>Step. 8 This is the last basic step. Take 4 as a common factor . </u>$

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$<u> <b>Final Equation → P4 + 3H2O + 3OH- ----------> 3H2PO2- + PH3 </b> </u>$

Thanks !!
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Answer 2

P4 + OH- =======> PH3 + H2PO2^-

$P {}^{4} + {3}OH( - ) + 3H {}^{2} O = > 3H{2} \: PO2( - ) +PH3$