From the data of two half-cell reactions:
AgCl(s) + e– → Ag(s) + Cl– (aq) E0 = +0.22 V
Ag+ (aq) + e– → Ag(s) E0 = +0.80 V
The solubility product of AgCl at 298 K, is calculated to be
1
1.5 × 10-10
2
2.1 × 10-7
3
3.0 × 10-3
4
1.2 × 10-5
5
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