Question

balance the following redox reaction in basic medium by oxidation number method p4(s)+oh-(aq) gives ph3(g) + h2po21-( aq)​

Answer 1

P4 + 3H20 + 3OH- ==> Ph3 + 3H2PO2

Answer 2

$P_{4}$ (s) + 3 $OH^{-}$ (aq) + 3 $H_{2}O$ (l) → $PH_{3}$ (g) + 3 $H_{2} PO^{-} _{2}$ (aq)

Explanation:

Oxidation number method:

• The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence, $P_{4}$  is oxidizing as well as reducing agent.

• During reduction, the total decrease in the oxidation number for 4 P atoms is 12.

• During oxidation, the total increase in the oxidation number for 4 P atoms is 4.

• The increase in the oxidation number is balanced with $H_{2} PO^{-} _{2}$ decrease in the oxidation number by multiplying H  with 3.

         $P_{4}$ (s) +  $OH^{-}$ (aq)  → $PH_{3}$ (g) + 3 $H_{2} PO^{-} _{2}$ (aq)

• To balance O atoms, multiply $OH^{-}$  ions by 6

         $P_{4}$ (s) + 6 $OH^{-}$ (aq)  → $PH_{3}$ (g) + 3 $H_{2} PO^{-} _{2}$ (aq)

• To balance H atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.

         $P_{4}$ (s) + 6 $OH^{-}$ (aq) + 3 $H_{2}O$ (l) → $PH_{3}$ (g) + 3 $H_{2} PO^{-} _{2}$ (aq) + 3 $OH^{-}$ (aq)

• Subtract 3 hydroxide ions from both sides.

        $P_{4}$ (s) + 3 $OH^{-}$ (aq) +  3 $H_{2}O$ (l) → $PH_{3}$ (g)  + 3 $H_{2} PO^{-} _{2}$ (aq)