Home › JEE Advanced › Chemistry › Electrochemical Series › For the cell: Zn|Zn²⁺(0.1 M)||Cu²⁺(0.01 M)|Cu, a…
For the cell: Zn|Zn²⁺(0.1 M)||Cu²⁺(0.01 M)|Cu, at 25°C, E_cell is:
A1.07 V
B1.13 V
C1.04 V
D1.10 V
Answer & Solution
Correct answer: A. 1.07 V
Nernst: E = E° - (0.0591/2) log([Zn²⁺]/[Cu²⁺]) = 1.10 - 0.02955 × log(0.1/0.01) = 1.10 - 0.02955 × 1 = 1.07 V (approximately).
Related questions
The electrolytes of the two half-cells are joined by a:The two portions of a galvanic cell are also called:In the Daniell cell, zinc dissolves at the anode and copper:A negative standard potential marks a reducing agent that is:Going from top to bottom of the table, the electrode potential:The weakest oxidising agent in the table of potentials is:The most powerful reducing agent in aqueous solution is:The weakest reducing agent in the table of potentials is: