Question

If 50 amperes current is passed for 965 seconds to a copper sulphate solution, then the mass of copper deposited at cathode is (Atomic weight of copper 63.5)

31.8 g

15.87 g

63.5 g

7.93 g​

Answer 1

Answer:

option a 31.8 g

this is the correct ans

Answer 2

Answer:

The mass of the copper ( Cu ) deposited at the cathode, m, calculated is $15.87g$.

Therefore, option b) $15.87g$ is correct.

Explanation:

Given data,

The amount of the current passing through the copper sulphate solution, I = $50A$

The time for the passage of $50A$ current through the copper sulphate solution, t = $965s$

The mass of the copper ( Cu ) deposited at the cathode, m =?

As we know,

• Charge ( Q ) = $It$
• Q = $50\times 965$ = $48250C$

The dissociation reaction of copper sulphate follows:

• $CuSO_{4} \rightarrow Cu^{2+} + SO_{4}^{2-}$

At anode:

• $SO_{4}^{2-} \rightarrow SO_{2} + O_{2} + 2e^{-}$

At cathode:

• $Cu^{2+} + 2e^{-}\rightarrow Cu$
•             2F         1mol

In the cathodic reaction, by flowing two faradays of charge, one mole of copper is formed.

As we know, the molar mass of copper = $63.5g/mol$

Therefore, 1 mole of copper = $63.5g$

Also 1F = $96500C$

Thus,

• $(2\times 96500)C$ = $63.5g$
• 1C = $\frac{63.5}{2\times 96500}$
• $48250C$ = $\frac{63.5}{2\times 96500}\times 48250$ = $15.87g$

Hence, the mass of the copper ( Cu ) deposited at the cathode, m = $15.87g$.