P⁴ + OH- -> pH³ + H₂Po²
Answers
Answer:
ANSWER
P
0
4
+
O
−2
H
−
+1
⇌
P
−3
H
+1
3
+
H
+1
2
P
+1
O
−2
2
−
⇒ P
4
0
⟶4
H
+1
2
P
+1
O
−2
2
+4e
−
(oxidation)
⇒
P
0
4
+12e
−
⟶4
P
−3
H
+1
3
(Reduction).
Balance the oxygen atoms in oxidation reaction.
P
0
4
+8OH
−
⟶4
H
+1
2
P
+1
O
2
2
+4e
−
3
P
0
4
+24OH
−
⟶12
H
+1
2
P
+1
O
+2
2
+12e
−
(O)
P
0
4
+12e
−
+12H
2
O⟶4
P
−3
H
+1
3
+12OH
−
(R)
4
P
0
4
+24OH
−
+12e
−
+12H
2
O⟶12
H
+1
2
P
+1
O
−2
2
+4
P
−3
H
+1
3
+12e
−
+12OH
−
Simplity the above equation.
P
0
4
+3OH
−
+3H
2
O⟶3
H
+1
2
P
+1
O
−2
2
+
P
−3
H
+1
3
P
4
+3OH
−
+3H
2
O⇌PH
3
+3H
2
PO
2
Answer:
4 P4 + 12 H2O + 12 OH- => 4 PH3 + 12 H2PO2 -
Explanation:
It is a disproportionation reaction of P4, i.e., P4 is oxidised as well as reduced in the reaction. P4 is the oxidising as well as the reducing agent. P4 is oxidised to H2PO2 - and reduced to PH3. So the(unbalanced) half-equations are as follows.
P4 => PH3, (1) and
P4 => H2PO2- (2)
In eqs (1) and (2) , the number of P atoms are unequal on both sides, so make them equal by multiplying the RHS by 4. The equations are now
P4 => 4 PH3 ( 1 a) and
P4 => 4 H2PO2 - (2a)
Equalise the number of Oxygen atoms by adding water molecules to the side deficient in Oxygen atoms . There is no Oxygen in (1a) , but the LHS of (2a) is short by 8 O atoms. So add 8 H2O molecules to the lhs of (2a) .
P4 + 8 H2O => 4 H2PO2 - (2b)
Next we equalise the number of H atoms in the half- equations. For this we add H+ ions ( if the solution is acidic) to the side deficient in H atoms. In case of an alkaline or a neutral solution , add H2O molecules to the side deficient in H atoms and am equal number of OH- ions to the opposite side. Eq(1a) requires 12 H atoms on lhs so we add 12 H2O on the lhs and an equal number of OH- on the rhs.
P4 + 12 H2O => 4 PH3 + 12 OH- ( 1b)
Similarly there is an excess of 8 H atoms on the lhs of (2b), so add 8 H2O on the rhs and 8 OH- on the lhs of (2b).
P4 + 8 H2O + 8 OH- => 4 H2PO2- + 8 H2O (2c)
After the atoms of different elements have been equalised on both sides balance charges by adding electrons (e-). (1b) needs 12 e- on the lhs and (2c) needs 4 e- on the rhs.
P4 + 12 H2O + 12 e- => 4 PH3 + 12 OH- (1c)
P4 + 8H2O + 8 OH- => 4 H2PO2- + 8 H2O + 4e- (2d)
Since (2d) has 8 H2O on both sides we can cancel them out.
P4 + 8 OH- => 4 H2PO2- +4 e- (2e)
So the balanced half-equations are
P4 + 12 H2O + 12 e- => 4 PH3 + 12 OH- (1c)
P4 + 8 OH- => 4 H2PO2 - + 4 e- (2e)
Eq(1c) is the reduction half-equation and Eq(2e) is the oxidation half equation.
Multiplying Eq (2e) by 3 so that the two half equations have equal number of electrons,
we get
3 P4 + 24 OH- => 12 H2PO2 - + 12 e- (2f)
Adding Eq (1c) and Eq(2f) and cancelling the common terms , we get the balanced ionic equation
4 P4 + 12 H2O + 12 OH- => 4 PH3 + 12 H2PO2 -