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Calculate the solubility product of AgCl from the two half reactions and standardelectrode potentials at 25^@CAg^(+) + e to Ag(s)E^@ = 0.799VAgCl + e to Ag(s) + Cl^(-) E^@ = 0.222V |
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Answer» SOLUTION :`AGCL to Ag^(+) Cl^(-) , E^@ = 0.222-0.799` ` E^@ = 0.0591 LOG [Ag^+] Cl^-] = 0.0591 log K_(sp)` `1.66 XX 10^(-10)` |
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