Question

A cell is prepared by dipping a copper rod in 1M `CuSO_(4)` solution and a nickel rod in 1M `NiSO_(4)` solution. The standard reduction potentials of copper and nickel electrodes are +0.34 V and -0.25 V respectively.
(i) Which electrode will work as anode and which as cathode ?
(ii) What will be the cell reaction ?
(iii) How is the cell represented ?
(iv) Calculate the emf of the cell.

Answer 1

(i) The nickel electrode with smaller `E^(@)` value (-0.25 V) will work as anode while copper with more `E^(@)` value (+0.34) will work as cathode.
(ii) The cell reaction may be written as :
`{:("At anode: " Ni(s)toNi^(2+)(aq)+2e^(-)),("At cathode: " Cu^(2)(aq)+2e^(-)toCu(s)),("Cell reaction: " bar(Ni(s)+Cu^(2)(aq)toNi^(2+)(aq)+Cu(s))):}`
(iii) The cell may be represented as :
`Ni(s)|Ni^(2+)(aq)||Cu^(2+)(aq)|Cu(s)`
(iv) emf of cell `=E_(cathode)^(@)-E_(anode)^(@)=(+0.34)-(-0.25)=0.59 V`