JEE Main 2020 · 4 September · evening · Single correct · +4 / −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 2V by passing a current of 1A for 15 minutes. The metal/metals electrodeposited will be
[ E_(Ag + /Ag)⁰ = 0.80 V, E_(Au + /Au)⁰ = 1.69 V ]
- (a) Silver and gold in equal mass proportion
- (b) Silver and gold in proportion to their atomic weights
- (c) Only gold
- (d) Only silver
Answer and solution
Answer (c) Only gold
Solution
Chemistry ma’am · 4 steps
The ion easiest to reduce plates out first, so check whether the charge runs out before gold does.
- Who deposits first. At the cathode, the ion with the higher reduction potential is reduced first. E⁰_(Au⁺/Au) = 1.69 V is much higher than E⁰_(Ag⁺/Ag) = 0.80 V, so Au⁺ takes electrons before Ag⁺ does.
- Charge passed. Q = It = 1 A × 900 s = 900 C. That is 900/96500 = 9.33 × 10⁻³ mol of electrons.
- Gold available. Moles of Au⁺ = 0.25 L × 0.1 M = 0.025 mol. Each Au⁺ needs one electron, so removing all the gold needs 0.025 mol of electrons.
9.33 × 10⁻³ mol e⁻ < 0.025 mol Au⁺
- So. The charge runs out while plenty of Au⁺ is still left, about 1.84 g of gold deposits, and silver never gets its turn. Only gold is deposited, option (c).
The idea to keep. In a mixture, ions are reduced in order of their reduction potential. A later metal deposits only if enough charge passes to use up all the earlier one first.
Answer from the NTA final answer key. Solution by PrepWiser.