Question

A solution of copper sulphate is electrolysed for \(10\) minutes with a current of \(1.5\) ampere. The mass of copper deposited at the cathode is: \[ \text{Given: Molar mass of Cu}=63.5\text{ g mol}^{-1} \] \[ F=96487\text{ C mol}^{-1} \]

• A. \(0.296\text{ g}\)
• B. \(1.7018\text{ g}\)
• C. \(2.4036\text{ g}\)
• D. \(0.5876\text{ g}\)

Hint:

For electrolysis, use \(m=\frac{ItM}{nF}\), where \(n\) is the number of electrons involved.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks for the quantitative mass of copper metal that will be plated out of a copper sulfate ($CuSO_4$) solution onto a cathode during an electrolytic process. We are provided with the duration of the electrolysis, the amount of current passed, the molar mass of copper, and Faraday's constant. This is a direct application of the stoichiometry of electrolysis.
Step 2: Key Formula or Approach:
We utilize Faraday’s First Law of Electrolysis, which states that the mass ($w$) of a substance deposited is proportional to the quantity of electricity ($Q$) passed through the electrolyte.
$Q = I \times t$ (where $I$ is current in Amperes and $t$ is time in seconds).
$w = \frac{E \cdot I \cdot t}{F}$
Equivalent Weight (E): $E = \frac{\text{Molar Mass}}{\text{Valency factor (n)}}$.
For $CuSO_4$, copper is in the $+2$ oxidation state. The reduction half-reaction is $Cu^{2+} + 2e^- \rightarrow Cu(s)$, so $n = 2$.
Step 3: Detailed Explanation:

Time conversion: The current is applied for $10$ minutes. To use standard units, we convert this to seconds: $t = 10 \times 60 = 600$ seconds.
Calculating Total Charge: The total charge passed is $Q = 1.5 \text{ A} \times 600 \text{ s} = 900 \text{ Coulombs}$.
Determining Equivalent Weight: Using the molar mass ($63$ g/mol) and the valency ($2$), we get $E = \frac{63}{2} = 31.5$ g/eq.
Applying the mass formula: \[ w = \frac{31.5 \times 900}{96487} \]
Final calculation: Multiply $31.5$ by $900$ to get $28350$. Dividing $28350$ by $96487$ results in approximately $0.293823$ g.
Rounding the result to four decimal places gives us $0.2938$ g, which matches option (B).
Step 4: Final Answer:
The calculated mass of copper deposited at the cathode is $0.2938$ g.