Question

n factor for p4h2 in
p2h4=ph3+p4h2

Answer 1

Answer:

P

2

H

4

⟶PH

3

+P

4

H

2

Oxidation state of P in P

2

H

4

⇒2x+4=0

⇒x=−2

Oxidation state of P in P

4

H

2

⇒4x+2=0 (Oxidation)

⇒x=−

2

1

Oxidation state of P in PH

3

⇒−3 (Reduction)

This is a disproportionation reaction.

P

2

H

4

n

factor

in oxidation =2∣(−2+

2

1

)∣=3

P

2

H

4

n

factor

in reduction =2∣(−2+3)∣=2

Net n

factor

=

n

red

+n

oxid

n

red

×n

oxid

=

3+2

3×2

=

5

6

E

equ

=n

factor

×m

=

6

31×5

=25.833