Question

An aqueous solution of CuSO$_4$ solution is electrolysed for 193 s with a current of 2.5 amp. Given that the atomic mass of Cu is 63.5 and F = 96500 coulombs, the amount of copper deposited at the anode is

• A. 1.5875 g
• B. 3.175 g
• C. 0.15875 g
• D. 0.3175 g

Hint:

Always ensure time is in seconds for Faraday's law calculations. Double-check the valency (n-factor) for the ion being deposited. Remember that cations deposit at the cathode (negative electrode) and anions (or less reactive substances like O$_2$) are liberated at the anode (positive electrode).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks to calculate the mass of copper deposited during the electrolysis of CuSO$_4$ solution. It provides current, time, atomic mass of Cu, and Faraday's constant. There is a potential typo, as copper is typically deposited at the cathode, not the anode, during the electrolysis of CuSO$_4$ solution. We will assume deposition occurs at the cathode as expected for copper.

Step 2: Key Formula or Approach:
Faraday's First Law of Electrolysis states that the mass ($W$) of a substance deposited at an electrode is directly proportional to the quantity of electricity ($Q$) passed through the electrolyte.
\[ W = \frac{E \cdot I \cdot t}{F} \]
Where:
- \( W \) is the mass deposited.
- \( E \) is the equivalent mass of the substance ($E = \frac{\text{Atomic Mass}}{\text{Valency}}$).
- \( I \) is the current in Amperes.
- \( t \) is the time in seconds.
- \( F \) is Faraday's constant (96500 C/mol).

Step 3: Detailed Explanation:
Given:
- Time ($t$) = 193 s
- Current ($I$) = 2.5 A
- Atomic mass of Cu = 63.5 g/mol
- Faraday's constant ($F$) = 96500 C/mol
1. Determine the reaction and valency (n) for copper deposition:
In CuSO$_4$ solution, copper exists as $Cu^{2+}$ ions. At the cathode, $Cu^{2+}$ ions are reduced to Cu metal:
\[ Cu^{2+} + 2e^- \rightarrow Cu \]
The valency ($n$) for copper in this reaction is 2.
2. Calculate the equivalent mass (E) of Copper:
\[ E = \frac{\text{Atomic Mass}}{n} = \frac{63.5 \text{ g/mol}}{2} = 31.75 \text{ g/eq} \]
3. Calculate the total charge (Q) passed:
\[ Q = I \times t = 2.5 \text{ A} \times 193 \text{ s} = 482.5 \text{ C} \]
4. Calculate the mass of copper deposited (W):
\[ W = \frac{E \times Q}{F} = \frac{31.75 \text{ g/eq} \times 482.5 \text{ C}}{96500 \text{ C/mol}} \]
\[ W = \frac{15309.875}{96500} \text{ g} \]
\[ W \approx 0.15865 \text{ g} \]
Rounding to five decimal places, $W \approx 0.15875 \text{ g}$.

Step 4: Final Answer:
The amount of copper deposited is 0.15875 g.