The cell, `Zn | Zn^(2+) (1M) || (1M) | Cu (E_(cell)^(@) = 1.10V)`, was was allowed to be completely discharged at 298 K. The concentration of `Zn^(2+)` `rightarrow CU^(2+)([([Zn^(2+)]/([Cu^(2+)])])` is :

Options

A. `9.65 xx 10^(4)`
B. Antilog `(24.08)`
C. `37.3`
D. `10^(37.3)`

Answer

D

Explanation

Anode
`Zn - 2e rarr Zn^(2+)`
Cathode
`Cu^(2+) +2e rarr Cu`
Net cell reaction
`Zn +Cu^(2+) rarr Zn^(2+) +Cu`
Applying Nernst equation
`E_(cell) = E_(cell)^(@) - (0.059)/(2) log.([ZN^(2+)])/([Cu^(2+)])`
When the cell is completely discharged `E_(cell) = 0`
`0 = 1.1 - (0.059)/(2) log.([ZN^(2+)])/([Cu^(2+)])`
`log.([ZN^(2+)])/([Cu^(2+)]) = (2 xx 1.1)/(0.059) = 37.3`
or `([ZN^(2+)])/([Cu^(2+)]) = 10^(37.3)`