Consider the cell Ag(s)|AgBr(s)Br^(c-)(a...

Consider the cell `Ag(s)|AgBr(s)Br^(c-)(aq)||AgCl(s),Cl^(c-)(aq)|Ag(s)` at `298 K`. The `K_(sp)` of `AgBr` and `AgCl`, respectively are `5xx10^(-13)` and `1xx10^(-10)` . At what ratio of `[Br^(c-)]` and `[Cl^(c-)]` ions, `EMF_(cell)` would be zero ?

`200:1`

`1:200`

`1:100`

`1:500`

Text Solution

Verified by Experts

The correct Answer is:

Cell reaction `:`
`Ag(s) +Ag^(o+)(` from `AgCl) rarr Ag(s)+Ag^(o+)(` from `AgBr)`
`E_(cell)=E^(c-)._(cell)-(0.059)/(1)log .([Ag^(o+)]_(AgBr))/([Ag^(o+)]_(AgCl))`
`=0-0.059log .((K_(sp)(AgBr))/([Br^(c-)])xx([Cl^(c-)])/(K_(sp(AgCl))))`
For `E_(cell)=0`
`(K_(sp)(AgBr)xx[Cl^(c-)])/(K_(sp)(AgCl)xx[Br^(c-)])=1`
`([Cl^(c-)])/([Br^(c-)])xx(5xx10^(-13))/(1xx10^(-10))implies[(Cl^(c-))/[Br^(c-))]=(1)/(200)`
`[(Br^(c-))/(Cl^(c-))]=200:1`

Topper's Solved these Questions

ELECTROCHEMISTRY
CENGAGE CHEMISTRY|Exercise Ex 3.2 (Objective)|26 Videos
ELECTROCHEMISTRY
CENGAGE CHEMISTRY|Exercise Ex 3.3 (Objective)|10 Videos
ELECTROCHEMISTRY
CENGAGE CHEMISTRY|Exercise Solved Examples(Electrolysis And Electrolytic Cells)|12 Videos
D AND F BLOCK ELEMENTS
CENGAGE CHEMISTRY|Exercise Archives Subjective|29 Videos
GENERAL PRINCIPLES AND PROCESS OF ISOLATION OF ELEMENTS
CENGAGE CHEMISTRY|Exercise Archives (Subjective)|14 Videos

Consider the cell Ag|AgBr(s)Br^(-)||AgCl(s)Cl^(-)|Ag at 298 K the solubility product of AgCl and AgBr are 1xx10^(-10) and 5xx10^(-13) respectively. What should be the ratio of concentration of Br^(-) and Cl^(-) by which emf of the cell becomes zero?

200

`(1)/(200)`

`(1)/(300)`

none of these

Ag_((s))|AgBr_((s))|Br^(-)(0.011M)||Cl^(-)(0.2M)|AgCl_((s))|Ag_((s)) if K_(sp) of AgBr & AgCl at 25^(@)C are 10^(-12) & 10^(10) respectively find which is/are true for above cell at 25^(@)C(log2=0.3)

As `E_(Ag^(+)//Ag)^(@)` is not given so `E_("cell")` cannot be determined

`E_("cell")` is `0.0413` volt

`/_\Glt0`

`E^(@)=0`

For the cell (at 298K) Ag(s) | AgCl(s) | Cl^(-)(aq)|| AgNO_(3)(aq) | Ag(s) Which of the following is correct?

The cell emf will be zero when `[Ag^(+)]_(a)` = `[Ag^(+)]_(c)` ([Ag^(+)] in anodic compartment = `[Ag^(+)]` in cathodic compartment).

The amount of AgCl(s). Precipitate in anodic compartment will decrease with the working of the cell.

The concentration of `[Ag^(+)]` = constant, in anodic compartment during working of cell .

`E_(cell) = E_(Ag^(+)|Ag)^(@) - (0.059/1)log (1/[Cl^(-)]_(a))`