1.

The standard reduction potentials for two reactions are given below: AgCl(s)+e^(-)toAg(s)+Cl^(-)(aq),E^(@)=0.22V ltBrgt Ag^(+)(aq)+e^(-)toAg(s),E^(@)=0.80V Calculate the solubility product of AgCl under standard conditions of temperature (298K).

Answer»

SOLUTION :Subtracting second eqn. from the FIRST eqn., we get
`AgCl(s)hArrAg^(+)(aq)+Cl^(-)(aq),E^(@)=-0.58V`
Applying nernst EQUATION,
`E=E^(@)-(0.0591)/(1)"LOG"([Ag^(+)][Cl^(-)])/([AgCl(s)])`
Putting `AgCl(s)=1` and at equilibrium `E=0`
`E^(@)=0.0591log[Ag^(+)][Cl^(-)]=0.0591logK_(sp)`
`-0.58=0.0591" log "K_(sp)" or log "K_(sp)=-9.8139=overline(10).1861`
or `K_(sp)="Antilog "overline(10).1861=1.535xx10^(-10)`.


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