ElectrochemistryHard
Question
The EMF of the cell: Ag, AgCl in 0.1 M – KCl||satd. NH4NO3||0.1 M – AgNO3, Ag is 0.42 V at 25°C. 0.1 M – KCl is 50% dissociated and 0.1 M – AgNO3 is 40% dissociated. The solubility product of AgCl is (2.303 RT/F = 0.06)
Options
A.1.0 × 10−10
B.2.0 × 10−9
C.1.0 × 10−9
D.2.0 × 10−10
Solution
On assuming concentration cell, the net cell reaction is Ag+ (C1M, Right) ? Ag+ (C2M, left)
$\text{Now, }\text{C}_{1} = 0.1 \times \frac{40}{100} = 0.04\text{ M} $$${\text{and }C_{2} = \frac{K_{sp}}{\left\lbrack Cl^{-} \right\rbrack} = \frac{K_{sp}}{0.1 \times \frac{50}{100}} = \frac{K_{sp}}{0.05}\text{ M} }{\text{Now, }\text{E}_{cell} = 0 - \frac{0.06}{1}.\log\frac{C_{2}}{C_{1}} }{\text{Or, 0.42 = } - \frac{0.06}{1}.\log\frac{K_{sp}/0.05}{0.04} \Rightarrow K_{sp} = 2 \times 10^{- 10}}$$
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