A) 9.804
B) - 9.804
C) - 4.902
D) 10.004
Correct Answer: B
Solution :
Anode half reaction \[Ag\xrightarrow{{}}A{{g}^{+}}+{{e}^{-}}\] Cathode half reaction \[A{{g}^{+}}\xrightarrow{{}}Ag+{{e}^{-}}\] \[NaCl\] is added to precipitate \[A{{g}^{+}}\] and \[AgCl\], thus \[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{n}\log \frac{[A{{g}^{+}}]\,[Cl]}{[A{{g}^{+}}]}\] \[{{E}_{cell}}=0-\frac{0.0591}{2}\log \,\,{{K}_{sp}}\] \[\because [A{{g}^{+}}]=1\] (Given)] \[0.29\times 2=-0.0591\,\,\log \,\,{{K}_{sp}}\] \[{{\log }_{10}}\,\,{{K}_{sp}}=-\frac{0.29\times 2}{0.0591}=-4.906\times 2\] = - 9.804You need to login to perform this action.
You will be redirected in
3 sec