JEE Main 2024 · 8 April · evening · Single correct · +4 / −1
The emf of cell Tl|Tl⁺| Cu²⁺ | Cu is 0.83 V at 298 K. It could be increased by :
- (a) increasing concentration of Tl⁺ ions
- (b) increasing concentration of Cu²⁺ ions
- (c) increasing concentration of both Tl⁺ and Cu²⁺ ions
- (d) decreasing concentration of both Tl⁺ and Cu²⁺ ions
Answer and solution
Answer (b) increasing concentration of Cu²⁺ ions
Solution
Chemistry ma’am · 3 steps
Write the cell reaction first; then ask which concentration change makes Q smaller.
- Cell reaction. Tl on the left is the anode and is oxidised. Cu²⁺ on the right is reduced. Balancing electrons gives 2Tl + Cu²⁺ → 2Tl⁺ + Cu, with n = 2.
E_(cell) = E°_(cell) - 0.0591/2 log[Tl⁺]²/[Cu²⁺]
- Read the equation. The emf rises when the log term gets smaller, that is, when Q falls. Cu²⁺ is a reactant and sits in the denominator, so raising it lowers Q. Tl⁺ is a product and sits in the numerator, so raising it raises Q.
- So. Increasing [Cu²⁺] lowers Q and increases the emf, so the answer is (b).
The idea to keep. In the Nernst equation, adding reactant ions or removing product ions lowers Q and raises the emf. Always write the balanced cell reaction first so you know which ion goes on top.
Answer from the NTA final answer key. Solution by PrepWiser.