1.

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

Answer»

Solution :`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


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