1.

When water is electrolysed, hydrogen and oxygen gas are produced. If 1.008 g of `H_(2)` is liberated at the cathode. What mass of `O_(2)` is formed at the anode?

Answer» `because(w_(1))/(w_(2))=(E_(1))/(E_(2)),` where `E_(1)` and `E_(2)` are eq.wt. of `H_(2)` and `O_(2)` respectively
`therefore(1.008)/(w_(2))=(1.008)/(8)rArrw_(2)=8g`


Discussion

No Comment Found

Related InterviewSolutions