Question

how long should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66g of diborane?

Answer 1

The combustion formula for diborane is given by :

B2H6 + 3O2  —> B2O3 + 6H2O

We need 3 moles of Oxygen for every mole of diborane.

In Electrolysis of water Oxygen gas is deposited at the anode.

The electrolysis formula is given as :

4OH-  —> O2 + 2H2O + 4e-

1 mole of oxygen requires = 4 × 96500 = 386000 Coulombs

How about 3 moles :

3 × 386000 = 1158000 C

From the formula :

Charge = Current × time

1158000 = 100t

t = 1158000/100

t = 11580 Seconds

11580/3600 = 3 hrs 13 minutes

Answer 2

Hey mate,

● Answer-

t = 11580 s

◆ Explaination-

Burning of diborane occurs according to reaction-

B2H6 + 3O2 ---> B2O3 + 6H20

Moles of diborane burnt = 27.66/27.66 = 1 mol

Moles of oxygen = 3 mol

Electrolysis of water happens as-

4OH- ---> O2 + 2H2O + 4e-

For 3 mol of oxygen 12 e- are liberated.

Faraday's law of mass deposition states that-

W = QIt

Here,

100 × t / 96500 = 12

t = 11580 s

Water should be electrolyzed to at least 11580 s.

Hope this helps...