Question

250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO₃ and 0.1 M AuCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electrodeposited will be ______.

(\[\ce{E^{\circ}_{Ag^+/Ag}}\] = 0.80 V, \[\ce{E^{\circ}_{Au^{2+}/Au}}\] = 1.69 V)

Options

• silver and gold in proportion to their atomic weights

• silver and gold in equal mass proportion

• only silver

• only gold

Answer 1

250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M AgNO₃ and 0.1 M AuCl. The solution was electrolyzed at 2 V by passing a current of 1 A for 15 minutes. The metal/metals electrodeposited will be only gold.

Explanation:

\[\ce{Charge (Q) = \frac{I \times t}{96500} F}\]

= \[\ce{\frac{1 \times 15 \times 60}{96500}}\]

= \[\ce{\frac{900}{96500}}\]

= \[\ce{\frac{9}{965} F}\]

= 0.0093 F

Species with larger E° values will be deposited first at the cathode. The number of moles of Au⁺ and Ag⁺ is 0.025 each. Since only 0.0093 mol of electrons are available, only 0.0093 mol of Au will get deposited.

\[\ce{Au^+_{ (aq)} + e^- -> Au_{(s)}}\]

So only gold (Au) will get deposited.