Question

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,1atm)|H+(aq,1.0M)|Au3+(aq,?M)|Au(s). What is the concentration of Au3+ in the solution if Ecell is 1.27 V ? Express your answer using two significant figures.

Answer 1

Explanation:

The standard reduction potential $(E_{cell})$ is given as 1.27 V.

Concentration of $H^{+}$ = 1.0 M

Hence, reduction reaction is as follows.

       $Au^{3+}(aq) + 3e^{-} \rightarrow 2Au(s)$

Oxidation reaction is as follows.

       $H_{2}(g) \rightarrow H^{+}(aq) + e^{-}$

Therefore, overall net chemical equation will be as follows.

       $2Au^{3+}(aq) + 3H_{2}(g) \rightarrow 2Au(s) + 6H^{+}(aq)$

As, its is known that standard electrode potential ($E^{o}_{cell}$) for hydrogen is equal to zero.

And, standard electrode potential for $Au^{3+}$ is $E_{o}$ equal 1.50 V.

Hence, formula to calculate standard cell potential is as follows.

           $E^{o}_{cell}$ = $(E_{o})_{reduction}$ - $(E_{o})_{oxidation}$

                               = 1.50 V - 0 V

                                = 1.50 V

Therefore, concentration of $Au^{3+}$ is calculated as follows.

            $E_{cell} = E^{o}_{cell} - \frac{0.059}{2}log \frac{[H^{+}]^{6}}{[Au^{3+}]^{2}}$

Hence, substitute the given values as follows.

             $E_{cell} = E^{o}_{cell} - \frac{0.059}{2}log \frac{[H^{+}]^{6}}{[Au^{3+}]^{2}}$

                     1.27 = 1.50 - \frac{0.059}{6}log \frac{(1)^{6}}{[Au^{3+}]^{2}}[/tex]

                   $log\frac{1}{[Au^{3+}]^{2}}$ = 23

Taking antilog on both the sides, the above equation will be as follows.

               $\frac{1}{[Au^{3+}]^{2}}$ = $10^{23}$

                                  = $1 \times 10^{23}$

                 $\frac{1}{[Au^{3+}]^{2}}$ = $3.16 \times 10^{11}$

                          $[Au^{3+}]$ = $3.16 \times 10^{-12}$ M

Hence, we can conclude that the concentration for $[Au^{3+}]$ is $3.16 \times 10^{-12}$ M.