Question

The equilibrium constant of a cell reaction:

\[\ce{Sn^{4+}(aq) + Al(s) -> Al^{3+} + Sn^{2+}(aq)}\] is 4.617 × 10¹⁸⁴, at 25°C.

1. Calculate the standard emf of the cell.
   (Given: log 4.617 × 10¹⁸⁴ = 184.6644)
2. What will be the E⁰ of the half cell \[\ce{Al^{3+}/Al}\], if the E⁰ of the half cell \[\ce{Sn^{4+}/Sn^{2+}}\] is 0.15 V.

Answer 1

a. `E_(cell) = E_(cell)^0 - (2.303 "RT")/("nF")*"log Kc"`

At 298 K

`E_(cell) = E_(cell)^0 - (0.0591)/("n")*"log Kc"`

At equilibrium, E_cell = 0, n = 6

`E_(cell)^0 = (0.0591)/("n")*"log Kc"`

= `0.059/6 log 4.617 xx 10^184`

= 0.00983 × 184.6644

= 1.8152

b. `E_(cell)^0 = E_(Sn^(4+)//Sn^(2+))^0 - E_(Al^(3+)//Al)^0`

1.81 = `-0.15 - E_(Al^(3+)//Al)^0`

`E_(Al^(3+)//Al)^0` = − 1.66 V