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A silver wire dipped in 0.1M HCl solution saturated with AgCl develops oxidation potential of -0.25V. If Eo Ag+/Ag=-0.799V then Ksp of AgCl


Aastha Shetty 25th May, 2019
Answer (1)
yamini 27th May, 2019

Hello Aastha,

Eel = E0el – 0.0592/ n log[Ag+]

Substitute the values in the above expression

-0.25V = -0.799V – 0.0592/1 log[Ag+]

log[Ag+] = -9.2736

[Ag+]= 5.3* 10-10

Hence the solubility product will be

Ksp = [Ag+][Cl-] = 5.3* 10-10 * 0.1

= 5.3 * 10-11

Hope this helps you

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