Question

Balance the following equation in basics medium by oxidation number method and identify the oxidizing
and reducing agent P4 (s)+OH- (aq) → PH3(g)+H2PO2​

Answer 1

Answer:

Oxidation number method:

The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence, P4 is oxidizing as well as reducing agent.

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

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

The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying H2PO2− with 3.

P 4 (s)+OH−(aq) → PH 3 (g)+3H2PO2− (aq)

To balance O atoms, multiply OH − ions by 6.

P4 (s)+6OH − (aq)→PH 3(g)+3H 2 PO2− (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)+6OH −(aq)+3H 2O(l)→PH 3(g)+3H2 PO 2-(aq)+3OH − (aq)

Subtract 3 hydroxide ions from both sides.

P 4 (s)+3OH − (aq)+3H2O(l)→PH3 (g)+3H2PO2−(aq)

Answer 2

There four oxygen is present is carbon dioxide are true solution