Question

Consider two half-cells based on the reaction $A g^{+}(a q)+e^{-} \rightarrow A g(s)$ The left half-cell contains $A g^{+}$ions at unit concentration, and the right half-cell initially had the same concentration of $A g^{+}$ions, but just enough $NaCl_{(a q)}$ had been added to completely precipitate the $A g_{(a q)}^{+}$as $AgCl$. If the emf of the cell is $0.29\, V$, the $\log _{10} K_{s p}$ would have been

• A. 9.804
• B. -9.804
• C. -4.902
• D. 10.004

Answer 1

Answer: (B)

-9.804

Answer 2

Anode half reaction $A g \rightarrow A g^{+}+e^{-}$
Cathode half reaction $Ag ^{+} \rightarrow Ag +e^{-} NaCl$
is added to precipitate $Ag ^{+}$and $A g C l$,
thus $E_{\text {cell }}=E_{\text {cell }}^{o}-\frac{0.0591}{n} \log \frac{\left[ Ag ^{+}\right][ Cl ]}{\left[ Ag ^{+}\right]}$
$E_{\text{cell}}=0-\frac{0.0591}{2} \log K_{s p}$
$\because\left[A g^{+}\right]=1$ (Given)]
$0.29 \times 2=-0.0591\, \log K_{s p}$
$\log _{10} K_{s p}=-\frac{0.29 \times 2}{0.0591}$
$=-4.906 \times 2=-9.804$