Home › NEET UG › Chemistry › Electrochemical Series › For the cell Zn | Zn²⁺ || Cu²⁺ | Cu at 298 K, th…
For the cell Zn | Zn²⁺ || Cu²⁺ | Cu at 298 K, the Nernst equation expresses E_cell as:
AE°_cell − (0.059/2) log ([Zn²⁺]/[Cu²⁺])
BE°_cell − (0.059) log ([Zn²⁺]/[Cu²⁺])
CE°_cell + (0.059/2) log ([Zn²⁺]/[Cu²⁺])
DE°_cell − (0.059/2) log ([Cu²⁺]/[Zn²⁺])
Answer & Solution
Correct answer: A. E°_cell − (0.059/2) log ([Zn²⁺]/[Cu²⁺])
Cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, n=2. Q = [Zn²⁺]/[Cu²⁺] (products/reactants for ions only). So **E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺])**.
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: