Find the reduction potential of `AsO_(4)^(3-)|AsO_(2)^(c-)` in a solution when `18mL` of `0.1 N` solution of `NaI` is added to `20mL` of `0.1N Na_(3)AsO_(4)` solution at `pH=5.` The standard reduction potential of `AsO_(4)^(3-)|AsO_(2)^(c-)=-0.70V`.

Explanation

Half cell reaction `:`
`4H^(o+)+AsO_(4)^(3-)+2e^(-) rarr AsO_(2)^(c-)+2H_(2)O`
`E_((AsO_(4)^(3-)|AsO_(4)^(3-)))=E^(c-)._((AsO_(4)^(3-)|AsO_(2)^(c-)))`
`-(0.059)/(2) log .([AsO_(2)^(c-)])/([AsO_(4)^(3-)][H^(o+)]^(4))`
`mEq ` of `AsO_(4)^(3-)=20xx0.1=2`
`mEq` of `I^(c-)=18xx0.1=1.8`
Therefore, `1.8 m Eq` of `I^(c-)` reacts with `1.8 mEq` of `AsO_(4)^(3-)` to give `1.8 mEq` of `AsO_(2)^(c-)`
So `mEq` of `AsO_(4)^(3-)` left `2-1.8=0.2 meQ`.
Total volume of solution `=(18+20)mL=38mL`
Hence, `[AsO_(4)^(3-)]=(0.2mEq)/(38mL),[AsO_(2)^(c-)]=(1.8mEq)/(38mL)`
`pH=5implies[H^(o+]=10^(-5)M`
Substituting the values in Eq. `(i)`
`E_((AsO_(4)^(3-)|AsO_(2)^(c-)))=E^(c-)._((AsO_(4)^(3-)|AsO_(2)^(c-)))`
`-(0.059)/(2) log . ([AsO_(2)^(c-)])/([AsO_(4)^(3-)][H^(o+)]^(4))`
`=-0.70V-(0.059)/(2)log .((1.8//38))/(((0.2)/(38))(10^(-5))^(4))`
`=-0.70V-(0.059)/(2)log .((1.8)/(0.2xx10^(-20)))`
`=-0.70V-(0.059)/(2)[log 9 + log 10^(20)]`
`=-0.70V-(0.059)/(2)[2log3+ 20] (Take 0.059=0.06log3~~0.48)`
`=-0.70V-0.30[2xx0.48+20]`
`=-0.70V-0.03xx20.96`
`=(-0.7-0.6288)V=-1.328V`
`:. `Hence,`E_((AsO_(4)^(3-)|AsO_(2)^(c-)))=-1.328V`