JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Redox Reactions and Electrochemistry

250 mL of a waste solution obtained from the workshop of a goldsmith contains 0.1 M

AgNO_{3}

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:

(E_{Ag^{+} / Ag}^{o} = 0.8 V, E_{Au^{+} / Au}^{o} = 1.69 V)

Answer: (C)

Solution

The charge passed is

q = \frac{i t}{F} = \frac{1 \times 15 \times 60}{96500} \approx 9.3 \times 1 0^{- 3} F

, while the solution holds

0.025 mol

each of

A g^{+}

and

A u^{+}

. During electrolysis the ion with the higher reduction potential deposits first, and since

E_{A u^{+} /Au}^{\circ} = 1.69 V

exceeds

E_{A g^{+} /Ag}^{\circ} = 0.8 V

, gold is reduced preferentially. The available charge can deposit only about

9.3 \times 1 0^{- 3} mol

of metal, well below the

0.025 mol

A u^{+}

present, so the gold is not exhausted and only gold is deposited.

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option