Question

Suppose that you have to silver plate two spoons. The surface area to be plated with silver is 2.0 cm² for each spoon. The thickness of the silver should be 4.5 mm. Knowing the density of silver as 10.5 g/cm³ and that the plating machine is only available for 3 hours 30 minutes, determine the electric current you have to use to achieve your two spoon silver plating for this period of time.

Answer 1

According to Faraday's 1st law, we have wa Q [when E constant ] Where W = cut of deposited & quando passed E = Chemical equivalent aut. Again, according to Faraday's End Saw, we have C WRE [When a constant 2 combination of 1 law and and love of Faraday When I and a are changing 1 - Proportionality I W = F F = Faraday (quantity of current rent op E Constant Q=c.t where c= t= Current time in see - Spoon Now, Surface area of to be plated = 2 cm2 for each Thickness of silver = 4.5 mm = 0.45cm. Volume = (2 cm 2 (0.45cm) = 0.9 cm3 Weight of silver = 0.9 cm3 x 1015 g = 9.45 g for each Spoon. for two Spoons total weight of silver 2 2X 9.459

According Egh 0 We E.c.t W = 1899 Ef Ag = 187.868 g t = 3 hour 30 min = (3600x3 + 30 x 60 sec a = 12600 see F = 96500 conlomb o ng cz 2 (107.968)()( 12600) 96500 - ca 18.9 x 96500 Conlomb 18.9= (107-868) (12600). a = 1.3419 conlomb. - . 1.3419 coulomb of electric current have to use achieve two Spoons silver plating for this period of time