Represent the cell in which the following reaction takes place:
`Mg(s)+Cu^(2+)(0.0001M) to Mg^(2+)(0.1M)+Cu(s)`.
Calculate its E, if `E^(@)` is `2.71V`.

Explanation

`Mg(s)+Cu^(2+)(aq)toMg^(2+)(aq)+Cu(s)`
Oxidation: `Mg(s) to Mg^(2+)(aq)+2e^(-)`
Reduction: `Cu^(2+)(aq) +2e^(-) to Cu(s)`
Overall : `Mg(s)+Cu^(2+)(aq)toMg^(2+)(aq)+Cu(s)`
Here, n=2
Nernst equation (at 298 K) is
`E^(@)=E^(@)-(0.0591V)/(n)logQ_(C)`
`=2.71V-(0.0591V)/(2)"log"([Mg^(2+)])/([Cu^(2+)])`
`=2.71V-0.0295V" log "(0.1)/(0.0001)`
`=2.71V-0.0295Vlog10^(3)`
`=2.71V-3xx0.0295V`
`=2.71V-0.0885V`
`=(2.71-0.09)V=2.62V`