A cell is constructed using `Pb^(2+), Pb and Ni^(2+), Ni` electrodes. If `E^@` values of Pb and Ni electrode are respectively `-0.13 and -0.25 V`, write (a) the cell reaction and (b) cell notation.

Explanation

Reduction potential of Ni electrode is less and hence it reduces `Pb^(2+)` ions.
The cell reactions is written as, `Ni + Pb^(2+) to Pb + Ni^(2+)`
The oxidation half reaction is , `Ni to Ni^(2+) + 2e^(-)`
The reduction half reaction is , `Pb^(2+) + 2e^(-) to Pb`.
The cell is written as, `Ni , Ni^(2+) (aq)//Pb^(2+)(aq),Pb.`