Question

In an electrochemical cell, Ag\(^+\) ions in AgNO\(_3\) are reduced to Ag metal at the cathode and Cu is oxidized to Cu\(^{2+}\) at the anode. A current of 0.7 A is passed through the cell for 10 min. The mass (in grams) of silver deposited and copper dissolved, respectively, are:

• A. 0.469 and 0.138
• B. 0.235 and 0.138
• C. 0.469 and 0.069
• D. 0.235 and 0.069

Hint:

To calculate the mass of a substance deposited or dissolved in an electrochemical reaction, use the formula \( m = \frac{I \cdot t \cdot M}{n \cdot F} \), where \( I \) is the current, \( t \) is the time, \( M \) is the molar mass, \( n \) is the number of electrons involved, and \( F \) is the Faraday constant.

Answer 1

The Correct Option is A

We are given the following data:
- Faraday constant \( F = 96,485 \, \text{C/mol} \)
- Atomic weight of Ag = 107.9
- Atomic weight of Cu = 63.55
- Current \( I = 0.7 \, \text{A} \)
- Time \( t = 10 \, \text{min} = 600 \, \text{s} \)

Step 1: Calculate the amount of silver deposited.
The reaction at the cathode is: \[ \text{Ag}^+ + e^- \rightarrow \text{Ag}. \] The amount of silver deposited is given by: \[ m_{\text{Ag}} = \frac{I \cdot t \cdot M_{\text{Ag}}}{n \cdot F}, \] where \( n = 1 \) (since 1 mole of Ag requires 1 mole of electrons), \( M_{\text{Ag}} = 107.9 \, \text{g/mol} \), and \( F = 96,485 \, \text{C/mol} \). Substituting the values: \[ m_{\text{Ag}} = \frac{0.7 \cdot 600 \cdot 107.9}{1 \cdot 96,485} = 0.469 \, \text{g}. \]

Step 2: Calculate the amount of copper dissolved.
The reaction at the anode is: \[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-. \] The amount of copper dissolved is given by: \[ m_{\text{Cu}} = \frac{I \cdot t \cdot M_{\text{Cu}}}{2 \cdot F}, \] where \( n = 2 \) (since 1 mole of Cu requires 2 moles of electrons), \( M_{\text{Cu}} = 63.55 \, \text{g/mol} \), and \( F = 96,485 \, \text{C/mol} \). Substituting the values: \[ m_{\text{Cu}} = \frac{0.7 \cdot 600 \cdot 63.55}{2 \cdot 96,485} = 0.138 \, \text{g}. \]

Step 3: Conclusion.
The mass of silver deposited is 0.469 g, and the mass of copper dissolved is 0.138 g.

Final Answer: \[ 0.469 \, \text{g of Ag} \text{ and } 0.138 \, \text{g of Cu}. \]