Question

What is the value of the equilibrium constant at 25 oC for the reaction between the pair: I2(s) and Br-(aq) Give your answer using E-notation with NO decimal places (e.g., 2 x 10-2 would be 2E-2; and 2.12 x 10-2 would also be 2E-2.). Use the reduction potentials for I2(s) is 0.535 V and for Br2(l) is +1.065 V.

Answer 1

Explanation:

Formula to calculate standard electrode potential is as follows.

          $E^{o}_{cell} = E^{0}_{cathode} - E^{0}_{anode}$

                             = 0.535 - 1.065

                             = - 0.53 V

Also, it is known that relation between $E^{o}_{cell}$ and K is as follows.

            $E^{o}_{cell} = \frac{RT}{nF} \times ln K$

                 ln K = $\frac{nFE^{0}_{cell}}{RT}$      

Substituting the given values into the above formula as follows.

                 ln K = $\frac{nFE^{0}_{cell}}{RT}$    

                        =  $\frac{2 \times 96485 C mol^{-1} \times -0.53 V}{8.314 l atm/mol K \times 298 K} \times \frac{1 J}{1 V C}$  

                ln K = -41.28

                    K = $e^{-41.28}$    

                        = $1 \times 10^{-18}$

Thus, we can conclude that the value of the equilibrium constant for the given reaction is $1 \times 10^{-18}$.