In the electrochemical cell : Zn | ZnSO4 (0.01 M) || CuSO4 (1.0 M) | Cu, the emf of this Daniell cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changes to E2. From the following, which one is the relationship between E1 and E2? (Given, 2.303RTF = 0.059)
- E1 < E2
- E1 > E2
- E2 = 0 E1
- E1 = E2
Answer
(2) E1 > E2
Solution
Cell reaction: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s), n = 2
Ecell = E cell - 0.0592 [Zn2+][Cu2+]
E1 = E cell - 0.0592 0.011.0
= E cell + 0.059
E2 = E cell - 0.0592 1.00.01
= E cell - 0.059
∴ E1 > E2
Answer: (2)