The following chemical reaction occurring in an electrochemical cell :
`Mg(s)+2Ag^(+)(0.0001 M) to Mg^(2+)(0.10 M)+2Ag(s)`
Given `E_(Mg^(2+)//Mg)^(@)=-2.36" V", E_(Ag^(+)//Ag)^(@)=+0.81" V"`
For the cell, calculate/write :
(i) `E^(@)` value for the electrode `2Ag^(+)//2Ag`
(ii) Standard cell potential `(E^(@))`
(iii) Cell potential (E)
(iv) Give the symbolic representation of the above cell
(v) Will the above cell reaction be spontaneous ?

Explanation

(i) `E^(@)` value of the electrode , `2Ag^(+)//2Ag` will remain the same i.e., 0.81 V. It does not change.
(ii) `" " E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=0.81-(-2.36)=3.17" V"`
(iii) `" " E_(cell)=E_(cell)^(@)=((0.0591" V"))/(2)"log"([Mg^(2+)(aq)])/([Ag^(+)(aq)]^(2))`
`=(3.17" V")-((0.0591" V"))/(2)"log"((0.1))/((0.0001)^(2)`
`=(.17" V")-(0.2068" V")=2.9632" V" `
(iv) Cell notation :` Mg(s)//Mg^(2+)(0.10" M") || 2Ag^(+)(0.0001" M")//2Ag(s)`
(v) Since Mg is placed below Ag in the activity series, it is a stronger reducing agent. Therefore, the cell reaction is spontaneous.