JEE Main 2024 Chemistry question
The emf of cell Tl|Tl⁺(0.001 M)||Cu²⁺(0.01 M)|Cu is 0.83 V at 298 K. It could be increased by:
Answer: (D)
Solution
2Tl + Cu²⁺ → 2Tl⁺ + Cu, so E = E° − (0.059/2)log([Tl⁺]²/[Cu²⁺]); E rises when [Cu²⁺] increases or [Tl⁺] decreases.