1.

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 "`

Answer» Correct Answer - `-0.2205 V`
Cell reaction : `Ni(s)+Cu^(2+)(0.1" M ") to Ni^(2+)(0.01" M ")+Cu(s)`
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Ni^(2+)])/([Cu^(2+)])`
`0.59=E_(cell)^(@)-(0.0591)/(n)"log"(0.01)/(0.1)=E_(cell)^(@)-0.02955" log "(10^(-1))=E_(cell)^(@)+0.02955`
`E_(cell)^(@)=0.59-0.02955=0.5605`
Now, `" " E_(cell)^(@)=E_((Cu^(2+)//Cu))^(@)-E_((Ni^(2+)//Ni))^(@)`
`:. E_((Ni^(2+)//Ni))^(@)=E_((Cu^(2+)//Cu))^(@)-E_(cell)^(@)=0.34-0.5605=0.2205" V "`


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