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Given the following E^@ values at 25^@C ...

Given the following `E^@` values at `25^@C` calcualte `K_(sp)` for silver bromide, AgBr.
`Ag^+(aq)+e=Ag(s), E_1^0=0.80V`
`AgBr(s)+e=Ag(s)+Br^(-) (aq), E_2^0=0.07V`
Also calculate `Delta G^@` at `25^@C` for the process
`AgBr(s) leftrightarrow Ag^+ (aq) + Br^(-) (aq)`

Text Solution

Verified by Experts

Reduction potential `E_1^0 gt E_2^0` so cell reaction is
`Ag^+ (aq)+Br^(-) (aq)=AgBr(s)`
for which,
`E_(cell)^@=E_1^0-E_2^0=0.80-0.07=0.73V`
We have
`E^@=(2.303 RT)/(nF) logk`
`=0.73=(2.303 times 8.314 times298)/(1 times 96500)log k`
`logk=12.3515`
or `K=2.246 times 10^12`
Thus for the eqb, AgBr(s) `leftrightarrow Ag^+ (aq)+Br^(-) (aq)`
`K_(sp)=1/K=1/((2.246 times 10^12))=4.45 times 10^-13`
Further for the reaction `Ag^+ (aq)+ Br^(-) (aq) leftrightarrow AgBr(s)`
`Delta G^@=-2.303 RT logk `
`=-2.303times 8.314 times 298 times 12.3515`
`=-70475.7 J//mol e`
`=-70.47//mo l e`
`=-70.47kJ//mol e`
`therefore for AgBr(s) leftrightarrow Ag^+ (aq)+Br^(-) (aq)`
`Delta G^@=+70.47kJ//mol e`
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