Answer to Question #167375 in General Chemistry for Tjamilon

Question

 A sodium sulfate solution was electrolyzed using inert Pt electrodes. The cathode reaction was: 2 H2 0  +  2 e -  →  H2  + 2 OH-. If a current of 3.0 amp was used for 30 min, what weight of H2 gas would be produced? 

 

Answer 1

2 H2O + 2 e - → H2 + 2 OH-

For every 2 moles of e (electron) transfer there produce 1 mole of H2 gas.

1 amp current = 1 coulomb is pussing through solution in every second.

current of 3.0 amp was used for 30 min

Means, (3 × 30 × 60) coulomb will pusses through the solution in this time period.

= 5400 Coulomb

So, number of moles of H2 gas produced = (5400 Coulomb)÷(charge need to puss through the solution for each moles of H2 gas to production)

= 5400 coulomb ÷ ( charge of 2 moles of electron)

= 5400 coulomb ÷ ( charge of 1 mole electron× 2)

= [5400 ÷ (96500×2)] mol

Mass of H2 gas produced

= (number of moles of H2 gas produced× mass of 1 mole H2 gas)

= [5400 ÷ (96500×2)] mol × 2 g/mol

= 0.056 g

Hence, 0.056 g of H2 gas will produce.