1.5 A current is reduced to 10 minutes in a solution of CuSO4,
How many mass of copper is deposited on the cathode?
A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?
Given :
Time (t) = 10 minute = 10 x 60 = 600 s
Current = 1.5 A
Formula used :
Charge = Current × time
=>1.5 A × 600 s = 900 C
Reaction involved :
Cu2+(aq) + 2e– = Cu(s)
Using concept of Faraday :
For 2e– one need 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu.
Thus for 900 C, the mass of Cu deposited := (63 g mol–1 × 900 C)/(2 × 96487 C mol–1) = 0.2938 g.
Given :
Time (t) = 10 minute = 10 x 60 = 600 s
Current = 1.5 A
Formula used :
Charge = Current × time
=>1.5 A × 600 s = 900 C
Reaction involved :
Cu2+(aq) + 2e– = Cu(s)
Using concept of Faraday :
For 2e– one need 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu.
Thus for 900 C, the mass of Cu deposited := (63 g mol–1 × 900 C)/(2 × 96487 C mol–1) = 0.2938 g.