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An aqueous solution containing 0.1M Fe^(...

An aqueous solution containing `0.1M Fe^(3+)` and `0.01 M Fe^(2+)` was titrated with a concentrated solution of `NaOH` at `30^(@)C` , so that changes in volumes were negligible. Assuming that the new species formed during titration are `Fe(OH)_(3)` and `Fe(OH)_(2)` only.
Given `E^(c-)._(Fe^(3+)|Fe^(2+))=0.80V, `
`K_(spFe(OH)_(3))=10^(-37), ` and `K_(sp Fe(OH)_(2))=10^(-19)`
The redox potential of `Fe^(3+)|Fe^(2+)` electrode at `pH=6` is

A

`0.8V`

B

`0.5V`

C

`0.2V`

D

`0.1V`

Text Solution

Verified by Experts

The correct Answer is:
a

`Fe^(3+)+e^(-) rarr Fe^(2+)" "E^(c-)=0.80V`
When `Fe(OH)_(3)` precipitates , `K_(sp)=[Fe^(3+)[overset(c-)(O)H]^(3)`
`:. [Fe^(3+)]=(K_(sp))/([overset(c-)(O)H]^(3))=(10^(-37))/([overset(c-)(O)H]^(3))`
When `Fe(OH)_(2)` precipitates , `K_(sp)=[Fe^(3+)[overset(c-)(O)H]^(2)`
`:. [Fe^(2+)]=(K_(sp))/([overset(c-)(O)H]^(2))=(10^(-19))/([overset(c-)(O)H]^(2))`
at `pH=2`,
`E_(cell)=E^(c-)._(cell)-(0.06)/(1) log .([Fe^(2+)])/([Fe^(3+)])`
`=0.80-0.06log.(0.01)/(0.01)`
`=0.080`
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Knowledge Check

  • In a solution of 0.04M FeCl_(2) and 0.01 M FeCl_(3) , how large may be its pH of without being precipitation of either Fe(OH)_(2) or Fe(OH)_(3) ? [Given K_(sp) Fe(OH)_(2) = 16 xx 10^(-6) and K_(sp) Fe(OH)_(3) = 8 xx 10^(-26) ]

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    `6.3`
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