Question

Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant for AgCl at 298 K is 1.77 × 10 − 10 and the standard reduction potential of the half‑reaction Ag + ( aq ) + e − − ⇀ ↽ − Ag ( s ) is + 0.799 V .

Answer 1

Answer: The value of $E^{o}$ for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          $AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$

Its solubility product will be as follows.

       $K_{sp} = [Ag^{+}][Cl^{-}]$

Cell reaction for this equation is as follows.

     $Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)$

Reduction half-reaction: $Ag^{+} + 1e^{-} \rightarrow Ag(s)$,  $E^{o}_{Ag^{+}/Ag} = 0.799 V$

Oxidation half-reaction: $Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}$,   $E^{o}_{AgCl/Ag}$ = ?

Cell reaction: $Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)$

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, $K_{sp} = [Ag^{+}][Cl^{-}]$

                                               = $1.77 \times 10^{-10}$

Therefore, according to the Nernst equation

           $E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

At equilibrium, $E_{cell}$ = 0.00 V

Putting the given values into the above formula as follows.

         $E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}$

        $0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}$    

       $E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}$

                  = $0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}$

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           $E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}$

       $0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}$

       $0.577 V = 0.799 V - E^{o}_{AgCl/Ag}$

       $E^{o}_{AgCl/Ag}$ = 0.222 V

Thus, we can conclude that value of $E^{o}$ for the half-cell reaction is 0.222 V.