Electrochemistry Result Question 40
40. Find the solubility product of a saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$ in water at $298$ $K$ , if the emf of the cell $\mathrm{Ag} \mid \mathrm{Ag}^{+}$ (Saturated $\mathrm{Ag}_2 \mathrm{CrO}_4$ solution. ) $| \mathrm{Ag}^{+}(0.1 \mathrm{M}) \mid \mathrm{Ag}$ is $0.164$ $V$ at $298 $ $K$ .
(1998, 6M)
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Answer:
Correct Answer: 40. $( 2.45 \times 10^{-12} )$
Solution:
$ \begin{aligned} & E=0-\frac{0.0592}{1} \log \frac{\left[\mathrm{Ag}^{+}\right]{\text {anode }}}{\left[\mathrm{Ag}^{+}\right]{\text {cathode }}} \\ & \Rightarrow \quad 0.164=-0.0592 \log \frac{\left[\mathrm{Ag}^{+}\right]{\text {anode }}}{0.10} \\ & \Rightarrow \quad\left[\mathrm{Ag}^{+}\right]{\text {anode }}=1.7 \times 10^{-4} \mathrm{M} \end{aligned} $
In saturated $\mathrm{Ag}_2 \mathrm{CrO}_4$ solution present in anode chamber :
$ \begin{aligned} & \mathrm{Ag}_2 \mathrm{CrO}_4(s) \rightleftharpoons \underset{1.7 \times 10^{-4} \mathrm{M}}{2 \mathrm{Ag}^{+}}+\underset{\frac{1.7}{2} \times 10^{-4} \mathrm{M}}{\mathrm{CrO}4^{2-}} \\ & K{\text {sp }}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CrO}_4^{2-}\right] \\ & =\left(1.7 \times 10^{-4}\right)^2\left(\frac{1.7}{2} \times 10^{-4}\right) \\ & =2.45 \times 10^{-12} \\ \end{aligned} $
BETA
