Represent the cell in which the followig reaction takes place,
`Mg_((s))+2Ag_((0.0001M))^(+)to Mg_((0.130M))^(2+)+2Ag_((s))` Calculate its `E_(cell)` if `E_(cell)^(Theta)=3.17V`

Explanation

In this cell `Mg_((s))|Mg^(2+)` is anode because Mg get oxidized and also `Ag^(+)` is reduced in Ag so on cathode `Ag^(+)|Ag` will be on right side. Between these two electrodes salt bridge is denoted by two verticle line || and hence the symbolic representation of cell will be
`Mg_((s))|Mg_((0.130M))^(2+)||Ag_((0.0001M))^(+)|Ag_((s))`
Calculation for `E_(cell)` potential is as follows:
`E_(cell)=E_(cell)^(Theta)-(RT)/(nF)ln([Mg^(2+)])/([Ag^(+)]^(2))`
`therefore E_(cell)=3.17V-(0.0591)/(2)"log"([Mg^(2+)])/([Ag^(+)]^(2))`
`=3.17V-(0.0591)/(2)"log"(0.130)/((0.0001)^(2))`
`=3.17-0.2955log(1.33xx10^(+7))`
`=3.17-0.2955(+7.1139)`
`=3.17-0.2099=3.17-0.21=2.96V`