Question

Calculate the value of E_cell at 298 K for the following cell:

`(Al)/(Al^(3+)) (0.01M) || Sn^(2+) ((0.015 M))/(Sn)`

`E°  _(Al^(3+))/(AI)= -1.66 " Volt and " E° _(Sn^(2+)) /(Sn) = -0.14` volt

Answer 1

`(Al)/(Al^(3+)) (0.01M) || Sn^(2+) ((0.015 M))/(Sn)`

Given,    `E° _((AI^(3+))/(AI)) = -1.66V,`

              `E° _((Sn^(2+))/(Sn)) = -0.14V,`

                      `E°_(cell) = E°_R - E°_L`

                                 =-0.14 - (-1.66)

                                 =-0.14 + 1.66=1.52 V

Cell reaction is: 

2AI + 3Sn²⁺ → 2AI³⁺ + 3Sn and n = 6 

According to Nernst equation, at 298 K 

                      `E_(cell) = E°_(cell) - 0.059/n log  ([AI^(3+)]^2)/([Sn^(2+)]^3)`

                            `=1.52 - 0.059/6 log  ([0.01]^2)/([0.015]^3) = 1.52 -0.01447`

                    `E_(cell) = 1.50553` V.