Question

Represent the cell in which the following reaction takes place.The value of E˚ for the cell is 1.260 V. What is the value of E_cell?

\[\ce{2Al (s) + 3Cd^{2+} (0.1M) -> 3Cd (s) + 2Al^{3+} (0.01M)}\]

Answer 1

Al(s) /Cd²⁺ (0.1M) // Al³⁺ (0.01M) /Cd(s)

\[\ce{2Al(s) + 3Cd^{2+} (0.1M) -> 3Cd (s) + 2Al^{3+} (0.01M)}\]

Ecell = E°cell `(- 0.059)/n log  [Al^(3+}]^2/([Cd^(2+)]^3`

Ecell = `1.26 - 0.059/6 log  (0.01)^2/(0.1)^3`

= `1.26 - 0.059/6 (- 1)`

= 1.26 + 0.009

= 1.269 V