How long it will take to deposit 1.0 g of chromium when a current of 1.25 A flows through a solution of chromium (III) sulphate ? (Molar mass of Cr =52 )

Step-by-step solution

Cr ³⁺+3 e⁻ Cr (s) 3 mol or 3 96500 C of electricity are needed to deposit 1 mol or 52 g of Cr . 52 g of Cr require current =3 96500 C 1 g of Cr will require current aligned & = 3 96500 52 C & =5567.3 C aligned Now, number of coulombs aligned & = current (ampere) t (seconds) & Time (s) required = no. of coulombs current (ampere) & = 5567.3 C 1.25 ~A & =4453.8 ~s & or & =1.24 ~h & aligned