Question

What is 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 Ag Is $0.164\mathrm{V}$ At $298\mathrm{K}$?

Answer 1

Cell potential:

• In an electrochemical cell, the cell potential,${\mathrm{E}}_{\mathrm{cell}}$, is the gap between two half cells.
• The capacity of electrons to pass by one-half cell to another causes the potential difference.
• Because the chemical process is a redox reaction, electrons can travel between electrodes.
• When one chemical is oxidised while another is reduced, a redox reaction occurs.
• The component loses one or more electrons during oxidation and consequently becomes positively charged.
• During reduction, on the other hand, the material acquires electrons and becomes negatively charged.
• This pertains to cell potential measurement because the difference between the potential for the reducing agent to get oxidized and the potential for the oxidising agent to become reduced determines the cell potential.

Formula for Cell potential:

• ${\mathrm{E}}_{\mathrm{cell}}=\frac{0.059}{1}\log \frac{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{RHS}}}{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{LHS}}}$

Step 1: Analyzing the given quantities

• $$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}}=0.164 \\ {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{RHS}}=0.1 \end{gathered}$$

Step 2: Applying values in the formula

• $$\begin{gathered} 0.164=\frac{0.059}{1}\log \frac{0.1}{{\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{LHS}}} \\ \Rightarrow {\left[{\mathrm{Ag}}^{+}\right]}_{\mathrm{LHS}}=1.66\times {10}^{-4}\mathrm{M} \end{gathered}$$

Step 3: finding solubility product ${\mathrm{K}}_{\mathrm{sp}\left({\mathrm{Ag}}_2{\mathrm{CrO}}_4\right)}$

• $$\begin{gathered} \left[{\mathrm{CrO}}_4^{2-}\right]=\frac{1.66\times {10}^{-4}}{2} \\ {\mathrm{K}}_{\mathrm{sp}\left({\mathrm{Ag}}_2{\mathrm{CrO}}_4\right)}={\left[{\mathrm{Ag}}^{+}\right]}^2\left[{\mathrm{CrO}}_4^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}\left({\mathrm{Ag}}_2{\mathrm{CrO}}_4\right)}={(1.66\times {10}^{-4})}^2\left(\frac{(1.66\times {10}^{-4}}{2}\right) \\ {\mathrm{K}}_{\mathrm{sp}\left({\mathrm{Ag}}_2{\mathrm{CrO}}_4\right)}=2.287\times {10}^{-12}{\mathrm{mol}}^3{\mathrm{L}}^{-3} \end{gathered}$$

Thus the solubility product is ${\mathrm{K}}_{\mathrm{sp}\left({\mathrm{Ag}}_2{\mathrm{CrO}}_4\right)}=2.287\times {10}^{-12}{\mathrm{mol}}^3{\mathrm{L}}^{-3}$