Question

Refer the attached picture. ​

Answer 1

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

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 ]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$<u> Step. 7 ]→</u>$ : 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 .

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

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

$<b> <i>P4 + 3H2O + 3OH- ----------> 3H2PO2- + PH3 </b> </i>$

$<b> <u> Thanks !! </u> </b>$

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Answer 2

Balanced Equation :-

P4 + 3H2O + 3OH- ----------> 3H2PO2- + PH3