Electrochemistry 2 Question 41
40. Find the solubility product of a saturated solution of $\mathrm{Ag}{2} \mathrm{CrO}{4}$ in water at $298 \mathrm{~K}$, 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 $0.164 \mathrm{~V}$ at $298 \mathrm{~K}$.
$(1998,6 \mathrm{M})$
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Solution:
- $E=0-\frac{0.0592}{1} \log \frac{\left[\mathrm{Ag}^{+}\right]{\text {anode }}}{\left[\mathrm{Ag}^{+}\right]{\text {cathode }}}$
$$ \Rightarrow \quad 0.164=-0.0592 \log \frac{\left[\mathrm{Ag}^{+}\right]_{\text {anode }}}{0.10} $$
$$ \Rightarrow \quad\left[\mathrm{Ag}^{+}\right]_{\text {anode }}=1.7 \times 10^{-4} \mathrm{M} $$
In saturated $\mathrm{Ag}{2} \mathrm{CrO}{4}$ solution present in anode chamber :
$$ \begin{aligned} \mathrm{Ag}{2} \mathrm{CrO}{4}(s) & \underset{1.7 \times 10^{-4} \mathrm{M}}{2 \mathrm{Ag}^{+}}+\underset{\frac{1.7}{2} \times 10^{-4} \mathrm{M}}{\mathrm{CrO}{4}^{2-}} \ K{\text {sp }}= & {\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CrO}_{4}^{2-}\right] } \ = & \left(1.7 \times 10^{-4}\right)^{2}\left(\frac{1.7}{2} \times 10^{-4}\right) \ = & 2.45 \times 10^{-12} \end{aligned} $$
