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Given the following E^@ values at 25^@Ccalcualte 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) |
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Answer» Solution :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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