MixedMediumQuestion 710017
Question
The EMF of the cell: Zn(s)|Zn²⁺(0.01 M)||Cu²⁺(1 M)|Cu(s) compared to E°cell (+1.10 V) at 298 K is:
- AGreater than 1.10 VCorrect
- BEqual to 1.10 V
- CLess than 1.10 V
- DZero
Correct answer
Greater than 1.10 V
Explanation
Q = [Zn²⁺]/[Cu²⁺] = 0.01/1 = 0.01 < 1. Nernst: E = E° − (0.0592/2)log(0.01) = 1.10 − (0.0296)(−2) = 1.10 + 0.0592 > 1.10 V.