`E_("Cell")` for the given redox reaction is 2.71 V
`Mg_((s))+Cu_((0.01 M))^(2+)to Mg_((0.011M))^(2+)+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

Explanation

`E_("cell")=E_("Cell")^(@)-(0.059)/(n)log K_(c) `
`=E_("Cell")^(@)-(0.059)/(2)"log"(10^(-3))/(10^(-2))`
`=2.71+0.0295`
`E_("cell")=2.7395V`
(i) Cu to Mg / Cathode to anode / Same direction
(ii) Mg to Cu / Anode to cathode / Opposite direction