The following reaction takes place in a galvanic cell .
`Mg(s) + 2Ag^(-1) (0.00001 M) to Mg^(2+) (0.130 M) + 2Ag(s)`
Calculate the emf of the cell.
Given `E_(Mg^(2+)//Mg)^(0) = - 2.37V " and "E_(Ag^(+)//Ag)^(0) = 0.80 V`.

Explanation

The cell would be

`E_("cell")^(0) = E_(Ag^(+)//Ag)^(0) - E_(Mg^(2+)//Mg)^(0) `
`= 0.80 - (-2.57)`
` = 3.17 V`
`E_("cell")= E_("cell")^(0) - (0.059)/2 log. ([Mg^(2+)])/([Ag^(+)]^(2))`
` = 3.17 - (0.059)/2 log. (0.130)/((10^(-4))^(2))`
` = 3.17 - 0.02955 log (0.130 xx 10^(+7))`
` = 3.17 - 0.02955 xx (7.1139)`
` = 3.17 - 0.21 = 2.96 V`