A layer of 0.25 mm thick chromium metal is to be electroplated on the rims of autovehicle with a total area of 32 m2 from a solution of CrO42 - . What amount of electric current is required for electroplating the rims in 60 sec? The density of chromium metal is 7.2 g/cc.

Question

A layer of 0.25 mm thick chromium metal is to be electroplated on the rims of autovehicle with a total area of 32 m2 from a solution of CrO42 - . What amount of electric current is required for electroplating the rims in 60 sec? The density of chromium metal is 7.2 g/cc. (A) 4.9× 103 A (B) 1.78× 103 A (C) 5.3× 104 A (D) 10.69× 106 A

Tardigrade Admin · Accepted Answer

Volume of coated lay = Area × thickness =32× 104× 0.025 =8000 cm3 Mass of chromium deposited = volume × density =8000× 7.2 =57600 g W=(I t E/96500) 57600=(I × 60 × (. 52 / 6 .)/96500) I=10.69× 106 amp.

Q. A layer of 0.25 mm thick chromium metal is to be electroplated on the rims of autovehicle with a total area of 32 $m^{2}$ from a solution of $C r O_{4}^{2 -}$ . What amount of electric current is required for electroplating the rims in 60 sec? The density of chromium metal is 7.2 g/cc.

$4.9 \times 1 0^{3}$ A

$1.78 \times 1 0^{3}$ A

$5.3 \times 1 0^{4}$ A

$10.69 \times 1 0^{6}$ A

Volume of coated lay = Area $\times$ thickness
$= 32 \times 1 0^{4} \times 0.025$
$= 8000 c m^{3}$
Mass of chromium deposited = volume $\times$ density
$= 8000 \times 7.2$
$= 57600$ g
$W = \frac{I tE}{96500}$
$57600 = \frac{I \times 60 \times ( 52/6 )}{96500}$
$I = 10.69 \times 1 0^{6}$ amp.

Q. A layer of 0.25 mm thick chromium metal is to be electroplated on the rims of autovehicle with a total area of 32 m2 from a solution of CrO42−​ . What amount of electric current is required for electroplating the rims in 60 sec? The density of chromium metal is 7.2 g/cc.

4.9×103 A

1.78×103 A

5.3×104 A

10.69×106 A