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Question
E°cell for the given redox reaction is 2.71V
\[\ce{Mg_{(s)} +Cu^{2+}_{(0.01 M)}->Mg^{2+}_{(0.001M)}+Cu_{(s)}}\]
Calculate Ecell 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
Solution
\[\ce{Mg_{(s)} +Cu^{2+}_{(aq)}->Mg^{2+}_{(aq)}+Cu_{(s)}}\]
Q=`(["Mg"^(2+)]["Cu"])/(["Mg"]["Cu"^2] `
`=((0.001)(1))/((1)(0.01))=0.1`
Using the Nernst equation,
Ecell = E°cell `-0.0591/"n" "log Q"`
Ecell`=2.71-0.0591/2 "log Q"`
Ecell=2.74V
(i) Since the voltage applied externally is less than cell E then the direction of flow of current is from cathode to anode
(ii) When the external voltage applied exceeds cell E, the direction of flow of current is from anode to cathode. If the voltage is applied is greater than 2.74V, the direction of current is from anode to cathode.
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