Question

Three electrolytic cells A, B and C containing electrolytes of zinc sulphate, silver nitrate and copper sulphate respectively were connected in series. A steady current of 1.5 A was passed through them until 1.45 g of silver were deposited at the cathode of cell B. (i) How long did the current flow? (ii) What weight of copper and zinc get deposited?

Hint:

In electrolytic cells connected in series, the same charge flows through each cell, so masses deposited are proportional to their equivalent weights.

Answer 1

Step 1: According to Faraday’s first law of electrolysis: \[ m = \frac{E}{F} \, Q \] where \(m\) = mass deposited, \(E\) = equivalent weight, \(F = 96500\ \text{C mol}^{-1}\), \(Q = It\).
Step 2: For silver: \[ \text{Equivalent weight of Ag} = \frac{108}{1} = 108 \]
Step 3: Substitute values: \[ 1.45 = \frac{108}{96500} \times (1.5 \times t) \]
Step 4: Solve for time \(t\): \[ t = \frac{1.45 \times 96500}{108 \times 1.5} \] \[ t \approx 930\ \text{s} \]
Step 5: Since the cells are connected in series, the same quantity of electricity passes through all cells.
Step 6: Calculate mass of copper deposited: \[ E_{\text{Cu}} = \frac{63.5}{2} = 31.75 \] \[ m_{\text{Cu}} = \frac{31.75}{108} \times 1.45 = 0.44\ \text{g} \]
Step 7: Calculate mass of zinc deposited: \[ E_{\text{Zn}} = \frac{65}{2} = 32.5 \] \[ m_{\text{Zn}} = \frac{32.5}{108} \times 1.45 = 0.22\ \text{g} \]