Question

A cell is prepared by dipping a copper rod in 1 M `CuSO_(4)` solution and a nickel rod in 1 M `NiSO_(4)` solution. The standard reduction potentials of copper electrode and nickel electrode are 0.34 volt and -0.25 volt respectively.
(a) What will be the cell reaction?
(b) What will be the standard EMF of the cell?
(c) Which electrode will be positive?
(d) How will the cell be represented?

Answer 1

(a) The cell reaction can be:
either `Ci+CuSO_(4)toNiSO_(4)+Cu," "i.e.," "Ni+Cu^(2+)toNi^(2+)+Cu`
or `Cu+NiSO_(4)toCu^(2+)+Ni," i.e., "Cu+Ni^(2+)toCu^(2+)+Ni`
The correct reaction will be the one which gives positive E.M.F. if the first reactioin is the cell reaction, we will
have `NitoNi^(2+)+2e^(-), "std. oxdn. pot"=-(-0.25)" volt"`
`underline(" "Cu^(2+)+2e^(-)toCu," Std. redn. pot."=+0.34" volt ")`
Overall reaction: `Ni+Cu^(2+)toNi^(2+)+Cu," Std. EMF=+0.59 volt"`
Thus, EMF is positive for this reaction. hence, this is the correct cell reaction. (if the other reaction is tried, EMF will come out to be -0.59 volt)
(b) Standard EMF of the cell as calculated above=0.59volt.
(c) Since oxidation takes place at the nickel electrode, thereofre, it acts as anode or since electrons are produced at this electrode, it acts as a negative pole. obviously, copper electrode will act as cathode or a positive pole.
(d) By convention, the cell will be represented as : `Ni|NiSO_(4)(1M)||CuSO_(4)(1M)|Cu`.