Calculate the standard electrode potential of the `Ni^(2+)//Ni` electrode , if the cell potential potential of the cell,
`Ni//N^(2+)(0.01 M)//Cu is 0.59" V ". "Given" E_(Cu^(2+)//Cu)^(@)=+0.34 " V "`

Explanation

`Ni |Ni^(2+) (0.01 M) ||Cu^(2+) (0.1 M) |Cu`
The cell reaction for the cell is :
`Ni (s) + Cu^(2+) (aq) to Cu(s) + Ni^(2+) (aq)`
`E = E_(cell)^(Theta) - (0.059)/(n "log" ([Ni^(2+) (aq)])/([Cu^(2+) (aq)])`
`0.59 = E_(cell)^(Theta) - (0.059)/(2) "log" (0.01)/(0.1) ` (Here n = 2)
`0.59 = E_(cell)^(Theta) - (0.059)/(2) log (1)/(10)`
`0.59 = E_(cell)^(Theta) + 0.0295`
`therefore " " E_(cell)^(Theta) = 0.59 - 0.0295 = 0.5605`
Now , `E_(cell)^(Theta) = E_((Cu^(2+) |Cu)) - E_((Ni^(2+) |Ni))^(Theta)`
`0.5605 = 0.34 - E_((Ni^(2+) |Ni))^(Theta)`
`therefore E_((Ni^(2+) |Ni))^(Theta) = 0.34 - 0.5605 = - 0.2205 V`.