A cell, as shown below, consists of there compartments separated by porous pots. The first contains a cobalt eletrode in `5.0L` of `0.10M Co(NO_(3)),` the second contains `5.0L` of `0.10M KNO_(3),` the third contains an `Ag` electrode in `5.0L` of `0.10M AgNO_(3)`. Assuming that current with in the cell iis carried equally by the negative and positive ions by passign `0.1F` of electricity.

Given `: Co^(2+)+2e^(-) rarr Co" "E^(c-)=-0.28V`
Ag^(o+)+e^(-) rarr Ag" "E^(c-)=0.80V`
`2Ag^(o+)+Corarr Co^(2+)+2Ag" "E^(c-)._(cell)=1.08V`
Finally the `II` compartment contains

Options

A. `I`
B. `I,II,`
C. `I,II,III`
D. `II,III,IV`

Answer

d

Explanation

Since `Co^(2+)` ions from `I` compartment do not move to `III` compartment, so `Co^(2+)` ions are not present there.
As explained in Question, there is a movement of `NO_(3)^(c-)` ions, some concentration of `NO^(c-)._(3)` ions is left in `III` compartment. There is also a movement of `K^(o+)` ions from `II` compartment to both `I` and `III` compartments.
There is decrease in positive charge due to reduction of `Ag(o+)` to `Ag` in `III` compartment.
Hence, ions present in `III` compartment are`: Ag^(o+), NO_(3)^(c-),` and `K^(O+)`.
Effect of electrode`:` Oxidation of `Co` to `Co^(2+)` at anode and reduction of `Ag^(o+)` to `Ag` at cathode occurs by the passage of `0.1 F` of electricity.
Effect by the migration of ions through porous barries `:`
Since charge carried by positive or negative ions is equal therefore transport number of each ion is equal, `i.e.,t_(c)o+=t_(a)c- =(1)/(1+1)=0.5` . Hence, number of ions transported from one compartment to another`-0.5xx0.1F=0.05Eq.`