InterviewSolution
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InterviewSolution
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`E_(cell)^(@)` for the given redox reaction is 2.71 V `Mg_((s))+Cu_((0.01 M))^(2+)toMg_((0.001 M))^(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. |
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Answer» Cell reaction `Mg(s)+Cu^(2+)toMg^(2+)+Cu` `E_(cell)=E_(cell)^(0)-(0.0591)/nlog""([Mg^(2+)])/([Cu^(2+)])=2.71-(0.0591)/2log""((0.001)/(0.01))=2.71-(0.0591)/2log""(1/10)` `=2.71-(0.0591)/2xx(-1)=2.71+(0.0591)/2=2.739 V`. (i) The electrons flow from Mg to Cu i.e., the direction of flow of current is from Cu to Mg. (ii) The electrons flow from Cu to Mg i.e., the direction of current is from Mg to Cu. |
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