Question

On passing current through two cells, connected in series, containing solutions of \( {AgNO}_3 \) and \( {CuSO}_4 \), 0.18 g of Ag is deposited. The amount of Cu deposited is:

• A. \( 0.529 \) g
• B. \( 10.623 \) g
• C. \( 0.0529 \) g
• D. \( 1.2708 \) g

Hint:

For electrolysis problems, use Faraday’s Law: \[ \frac{m_1}{E_1} = \frac{m_2}{E_2} \] when the same current is passed through different cells in series.

Answer 1

The Correct Option is C

Step 1: Using Faraday’s Law of Electrolysis.
The mass of the substance deposited is given by: \[ m = \frac{E \times I \times t}{96500} \] Since the same current flows through both cells, \[ \frac{m_{{Ag}}}{E_{{Ag}}} = \frac{m_{{Cu}}}{E_{{Cu}}} \] where: - \( m_{{Ag}} = 0.18 \) g, \( E_{{Ag}} = 108 \), - \( E_{{Cu}} = 63.5 \), - \( m_{{Cu}} \) is unknown.
Step 2: Solving for \( m_{{Cu}} \).
\[ \frac{0.18}{108} = \frac{m_{{Cu}}}{63.5} \] \[ m_{{Cu}} = \frac{0.18 \times 63.5}{108} = 0.0529 { g} \]