Question 15 of 29advanced🔧 ApplyNumerical5 marks

Three electrolytic cells A,B,C containing solutions of ZnSO₄, AgNO₃ and CuSO₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

Correct Answer

(copper_mass) 0.426 g

(time) 863.73 seconds (or approximately 864 seconds)

(zinc_mass) 0.439 g

Exercise: EXERCISES | Q: 2.16 | (Chapter: Page 30)
For More Understanding

Explanation

This question applies Faraday's laws of electrolysis. Since the cells are connected in series, the same quantity of electricity (charge) passes through all three cells. First, we calculate the time using the silver deposition data, then use the same charge to find the masses of copper and zinc deposited.

Solution Steps

  1. Step 1: Calculate time using silver deposition For cell B (AgNO₃): Ag⁺ + e⁻ → Ag Atomic mass of Ag = 108 g/mol, n = 1 Using Faraday's law: m = (M × I × t)/(n × F) 1.45 = (108×1.5108 \times 1.5 × t)/(1×965001 \times 96500) t = (1.45×965001.45 \times 96500)/(108×1.5108 \times 1.5) = 139925/162 = 863.73 seconds

  2. Step 2: Calculate total charge passed Charge (Q) = Current × Time = 1.5×863.731.5 \times 863.73 = 1295.6 C

  3. Step 3: Calculate mass of copper deposited in cell C For CuSO₄: Cu²⁺ + 2e⁻ → Cu Atomic mass of Cu = 63.5 g/mol, n = 2 m(Cu) = (M × Q)/(n × F) = (63.5×1295.663.5 \times 1295.6)/(2×965002 \times 96500) = 0.426 g

  4. Step 4: Calculate mass of zinc deposited in cell A For ZnSO₄: Zn²⁺ + 2e⁻ → Zn Atomic mass of Zn = 65.4 g/mol, n = 2 m(Zn) = (M × Q)/(n × F) = (65.4×1295.665.4 \times 1295.6)/(2×965002 \times 96500) = 0.439 g

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