ElectrochemistryHard
Question
What is the solubility product of a saturated solution of Ag2CrO4 in water at 298 K if the EMF of the cell:
Ag|Ag+ (satd. Ag2CrO4)||Ag+ (0.1 M)|Ag is 0.162 V at 298 K? [2.303 RT/F = 0.06, log 2 = 0.3]
Options
A.2.0 × 10−4
B.3.2 × 10−11
C.8.0 × 10−12
D.4.0 × 10−12
Solution
Net cell reaction: $Ag^{+}\left( C_{1} = 0.1\text{ M} \right) \rightarrow Ag^{+}\left( C_{2} = 2\left( \frac{K_{sp}}{4} \right)^{1/3}M \right)$
$\text{Now, }E_{cell} = 0 - \frac{0.06}{1}.\log\frac{C_{2}}{C_{1}} $$${\Rightarrow 0.162 = - 0.06.\log\frac{\left( 2K_{sp} \right)^{1/3}}{0.1} }{\therefore K_{sp} = 4 \times 10^{- 12}}$$
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