balance the following equation in basic medium by ion electron method and oxidation number method and identify the oxidising agent and reducing agent .
Answers
Answer:
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, 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 H
2
PO
2
−
with 3.
P
4
(s)+OH
−
(aq)→PH
3
(g)+3H
2
PO
2
−
(aq)
To balance O atoms, multiply OH
−
ions by 6.
P
4
(s)+6OH
−
(aq)→PH
3
(g)+3H
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)+6OH
−
(aq)+3H
2
O(l)→PH
3
(g)+3H
2
PO
2
−
(aq)+3OH
−
(aq)
Subtract 3 hydroxide ions from both sides.
P
4
(s)+3OH
−
(aq)+3H
2
O(l)→PH
3
(g)+3H
2
PO
2
−
(aq)
Ion electron method:
The oxidation half reaction is P
4
(s)→H
2
PO
2
−
(aq).
The P atom is balanced.
P
4
(s)→4H
2
PO
2
−
(aq)
The oxidation number is balanced by adding 4 electrons on RHS.
P
4
(s)→4H
2
PO
2
−
(aq)+4e
−
The charge is balanced by adding 8 hydroxide ions on LHS.
P
4
(s)+8OH
−
(aq)→4H
2
PO
2
−
(aq)
The O and H atoms are balanced.
The reduction half reaction is P
4
(s)→PH
3
(g).
The oxidation number is balanced by adding 12 eelctrons on LHS.
P
4
(s)+12e
−
→PH
3
(g)
The charge is balanced by adding 12 hydroxide ions on RHS.
P
4
(s)+12e
−
→PH
3
(g)+12OH
−
The oxidation half reaction is multiplied by 3 and the reduction half reaction is multiplied by 2.
The half reactions are then added to obtain balanced chemical equation.
P
4
(s)+3OH
−
(aq)+3H
2
O(l)→PH
3
(g)+3H
2
PO
2
−
(aq)
Explanation: