Question

on the electrolysis of acidified water the mass of oxygen liberated at the anode by passing 9650 Faraday electricity is​

Answer 1

we have to find the mass of oxygen of acidified water at anode by passing 9650 Faraday.

solution : electrolysis of water is given by,

       2H₂O   ⇄ 2H₂ + O₂

at anode,

     4OH⁻ ⇆ 2H₂O + O₂ + 4e⁻

here you can see that one mole of oxygen gas is liberated when 4 moles of electrons are released.

   we know, one mole of electrons liberates 96500c charge.

so, 9650 C charge is equivalent to 0.1 mol of electrons.

now, no of moles of oxygen liberated at anode = 0.1/4 = 0.025 mol

mass of oxygen = moles × molecular mass of oxygen gas

                           = 0.025 × 32 = 0.8 g

Therefore mass of liberated oxygen at anode is 0.8g