The time required to deposit 4.5g of copper from CuSO$_4$ solution by passing a current of 2.5 Amperes is (Cu = 64g ; 1F = 96500C/mol)

Analysis

To solve this problem, we use Faraday’s First Law of Electrolysis, which relates the mass of a substance deposited to the quantity of electricity (charge) passed through the electrolyte.

Concept

The formula is:

Where:

• $m$ = mass of the substance deposited (4.5g)
• $M$ = Molar mass of the substance (64g/mol)
• $I$ = Current in Amperes (2.5A)
• $t$ = Time in seconds (what we need to find)
• $n$ = Valency or number of electrons transferred (For Copper in $CuSO_4$, $Cu^{2+} + 2e^- \rightarrow Cu$, so $n = 2$)
• $F$ = Faraday’s constant (96500 C/mol)

Step-by-Step Solution

Step 1: Rearrange the formula to solve for $t$:

Step 2: Substitute the given values:

• $m = 4.5$
• $n = 2$
• $F = 96500$
• $M = 64$
• $I = 2.5$

Step 3: Perform the calculation:

• Numerator: $4.5 \times 2 = 9$
• $9 \times 96500 = 868500$
• Denominator: $64 \times 2.5 = 160$

Rounding to the nearest whole number, we get 5428 seconds.

Conclusion

Comparing our result with the options provided, it matches option C.

Exam Tip

Always check the valency ($n$) of the metal in the salt. For Copper ($Cu^{2+}$), $n=2$; for Silver ($Ag^+$), $n=1$; for Aluminum ($Al^{3+}$), $n=3$. A common mistake is forgetting to multiply by the valency, which would lead you to the wrong answer!