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Excess of solid AgCl is added to a 0.1 M...

Excess of solid `AgCl` is added to a `0.1 M` solution of `Br^(c-)` ions. `E^(c-)` for half cell is `:`
`AgBr+e^(-) rarr Ag+Br^(c-),E^(c-)=0.095V`
`AgCl+e^(-) rarr Ag+Cl^(c-), E^(c-)=0.222V`
The value of `[Br^(c-)]` ion at equilibrium is `:`
`[` Given `: Antilog (2.152)=142]`

A

`0.0007M`

B

`0.013M`

C

`0.99M`

D

`0.099M`

Text Solution

Verified by Experts

The correct Answer is:
d

`E^(c-)._9cell)=(E^(c-)._(red))_(c)-(E^(c-)._(red))_(a)=(0.222-0.0950V`
`0.127V`
At equilibrium, `E_(cell)=0`.
`E_(cell)=E^(c-)._(cell)-(0.059)/(1) log .((x))/((0.1-x)),,,(i)`
Substitute the value of `E_(cell)` and `E^(c-)._(cell)` in `Eq. (i)`,
`(0.127)/(0.059)=log.(x)/(0.1-x)`
Antilog `(2.152)=(x)/(0.1-x)`
Antilog `2.152=142(` Given`)`
`:. 142(0.1-x)=x`
Solve for `x`,
`:. x=0.099 M`
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