Question

E°_cell for the given redox reaction is 2.71 V
Mg_(s) + Cu²⁺ _{(0.01 M)} → Mg²⁺ (0.001 M) + Cu_(s)

Calculate E_cell for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and
(ii) greater than 2.71 V

Answer 1

E_cell = `"E"°_"cell" - (0.0591)/(n)log  ["Mg"^(2+)] // ["Cu"^(2+)]`

= `2.71 - (0.0591)/(2)log  (0.001)/(0.01)`

=`2.71 - 0.0295 xx (-1)`

= `2.7395  "V"`

i) When an external potential is less than 2.71 then current will flow from copper to magnesium.
ii) When an external potential is greater than 2.71 then current will flow from magnesium to copper.