Please fast. Explain properly
dear student,
mass of copper deposited in 1F charge = 63.5/2g = 31.75g
charge needed for deposition of 3 g copper= (1/31.75)*3 = 0.094 F
in coulumbs = 0.094*96500 =9118 C
charge supplied =4*1.5*60*60 =21600 C
efficiency% = (9118/21600)*100 = 42%
regards
mass of copper deposited in 1F charge = 63.5/2g = 31.75g
charge needed for deposition of 3 g copper= (1/31.75)*3 = 0.094 F
in coulumbs = 0.094*96500 =9118 C
charge supplied =4*1.5*60*60 =21600 C
efficiency% = (9118/21600)*100 = 42%
regards