Question

The Ksp of copper(II) ferrocyanide (Cu2[Fe(CN)6]) is 1.3 × 10−16 at 25°C. Determine the potential of a concentration cell in which one half-cell consists of a copper electrode in 1.00 M copper(II) nitrate, and the other consists of a copper electrode in a saturated solution of Cu2[Fe(CN)6].
Ferrocyanide, ([Fe(CN)6]4−), is a complex ion.

Answer 1

Explanation:

Expression for $K_{sp}$ of the given reaction is as follows.

           $K_{sp} = [Cu^{2+}]^{2}[Fe(CN)_{6}]$

Let us assume that the concentration of given species is "s". As the value of $K_{sp}$ is given as $1.3 \times 10^{-16}$.

              $K_{sp} = [Cu^{2+}]^{2}[Fe(CN)_{6}]$

          $1.3 \times 10^{-16} = s^{2} \times s$      

               $s^{3} = 1.3 \times 10^{-16}$    

                     s = $3.19 \times 10^{-6}$

Therefore, concentration of $Cu^{2+}$ will be calculated as follows.

            $Cu^{2+}$ = 2s

                         = $2 \times 3.19 \times 10^{-6}$        

                        = $6.38 \times 10^{-6}$ M

Now, we will calculate the value of $E_{cell}$ as follows.

                  $E_{cell} = E^{o}_{cell} - \frac{0.0591}{2} \times log \frac{6.38 \times 10^{-6}}{1}$

                            = $0 - \frac{0.0591}{2} \times log \frac{6.38 \times 10^{-6}}{1}$

                            = 0.1535 V

Thus, we can conclude that the potential of given cell is 0.1535 V.