Question

When a current of 0.5 amperes was passed for 32 minutes and 10 seconds through fused chloride of metal P 0.44 of P was deposited. Determine the charge of the ion of metal P. (IF = 96,500C, P =88)    

Answer 1

Q=IF
  =0.5×((32×60)+10)
  =0.5×1930
  =965c
It 965c→0.44g
      ?   →88
965×88=193,000c
   0.44
If IF=96,500c
    ?=193,000c
193,000 =2F
 96,500
Thus the charge =2+