Question

The emf of cell $Ag|AgI\left(s\right)\left(0.05M\right)KI||0.05MAgN{O}_2|Ag$ is $0.788V$. The solubility product of $AgI$ is:

• A. $1.1\times {10}^{-15}$
• B. $1.4\times {10}^{-15}$
• C. $1.1\times {10}^{-17}$
• D. none of these

Answer 1

Answer: (A)

$1.1\times {10}^{-15}$

$\left(A\right)1.1\times {10}^{-15}$

$E_{cell}=E_{oxidisedproduct}+E_{reducedproduct}$

$E_{cell}=E_{Ag/A{g}^{+}}-\frac{0.059}{1}\log \left[A^{+}\right]$

$+E_{A{g}^{+}/Ag}+\frac{0.059}{1}\log \left[A{g}^{+}\right]$

$0.788=0.059\log \frac{0.05}{\left[A{g}^{+}\right]}$

$\therefore [A{g}^{+}{]}_{LHS}=2.203\times {10}^{-15}$

$\therefore Ksp=2.203\times {10}^{-15}\times 0.05$

$Ksp=1.1\times {10}^{-15}$.