Question

An aqueous solution of CuSO4 solution is electrolyzed 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:

In electrolysis problems, use the formula \( m = \frac{M I t}{n F} \) to calculate the mass of substance deposited.

Answer 1

The Correct Option is A

Step 1: Understanding the equation for electrolysis.
The mass of the copper deposited is given by the formula: \[ m = \frac{M I t}{n F} \] where: - \( M \) is the molar mass of copper (63.5 g/mol), - \( I \) is the current (2.5 A), - \( t \) is the time (193 s), - \( n \) is the number of electrons involved in the reaction (2 for copper), - \( F \) is the Faraday constant (96500 C/mol). Step 2: Substituting values.
Substituting the values: \[ m = \frac{63.5 \times 2.5 \times 193}{2 \times 96500} = 1.5875 \, \text{g} \] Step 3: Conclusion.
The amount of copper deposited is \( 1.5875 \) g. The correct answer is (1) 1.5875 g.

Answer 2

This is a Faraday's law electrolysis problem. Let's work through the charge passed and the mass deposited step by step and check each option against that reasoning.

1. 1.5875 g: The charge passed is \( Q = I \times t = 2.5 \times 193 = 482.5 \) C. Since copper deposits by \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \), the mass deposited follows from \( m = \dfrac{M \, I \, t}{nF} = \dfrac{63.5 \times 2.5 \times 193}{2 \times 96500} \), which works out to \( 1.5875 \) g.
2. 3.175 g: This is twice the correct mass, and would follow only if each \( \text{Cu}^{2+} \) ion needed just 1 electron instead of 2 - i.e. treating copper as monovalent - which is not correct since copper deposits from the \( +2 \) state.
3. 0.15875 g: This is a tenth of the correct mass, the kind of value that results from an order-of-magnitude slip in converting the charge or the Faraday constant during the bookkeeping.
4. 0.3175 g: This is a fifth of the correct mass, combining the monovalent assumption of option 2 with the order-of-magnitude slip of option 3.

Working through the Faraday's law bookkeeping - the charge passed, the number of electrons transferred per copper atom, and the molar mass of copper - gives the mass of copper deposited.

Therefore, the correct answer is 1.5875 g.