ok to the third one. at the electrode accordin to me this reaction will occur Ag+ to Ag so Eo=0.8V conc of Ag+ = root of 90*10^-9 by Ksp = 9.45*10^-9 so nernst equation will be E=Eo-0.059log(1/Ag+) so E = 0.8 - 0.059log(1/9.45*10^-9) = 0.8-0.52 = 0.28V please correct me if i am wrong or nudge me if you have any doubts
Moderation Team
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