In an electrolysis experiment current was passed for 5 hours through two cells connected in series. The first cell contains a solution of gold at the oxidation state of + 3 and second contain CuSO 4 solution. During electrolysis in the first cell 9.85 gm of gold is deposited, determine the average gm of copper deposited per ampere of electricity passed in the nearest possible integers.
Cu – 63.5 ; Au – 197
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
0006
Sol. Amount of Au deposited = 
= 0.05 mol
∴ Amount of electrons passed
= 0.05 × 3 = 0.15 mol
∴ Quantity of electricity passed
= 0.15F = 96500 × 0.15 C
∴ Quantity of current passed
= 
= 0.8042 amp
Mass of Cu deposited
= 
= 4.7625 gm
∴ 
= 
= 5.92 ~ 6 
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