Question

A voltaic cell consists of an Ni/Ni2+ half-cell and a Co/Co2+ half-cell. Calculate Ecell when [Ni2+] = 0.0019 M and [Co2+] = 1.265 M. You should use the reduction potentials for Ni2+ is -0.247 V and for Co2+ is -0.277 V.

Answer 1

Given cell is Ni/Ni2+ half-cell and a Co/Co2+ half-cell

Co(s) + Ni⁺² (aq) ----------> Co⁺² (aq) + Ni (s)

E_cell = E^o_cell - RT/nF In {[Co²⁺] / [Ni²⁺]} (solids cannot be taken into consideration)

E_cell = E^o_cell - 0.059/n Iog {[Co²⁺] / [Ni²⁺]}  where n = no of electrons transferred

At cathode : Ni²⁺ (aq) + 2e^- -------------> Ni (s)   E^o = -0.247 V

At anode : Co (s) --------------> Co²⁺(aq) + 2e^- E^o = + 0.277

E^o_cell = E^o_reduction of reaction at cathode+E^o_oxidation of reaction at anode

= -0.247 + 0.277 V

= + 0.03 V

E^o_cell = + 0.03 V

no of electrons transferred n = 2

Given that  [Co²⁺]= 1.265 M , [Ni²⁺]= 0.0019 M

E_cell = E^o_cell - 0.059/n Iog {[Co²⁺] / [Ni²⁺]}

= (0.03) - 0.059/2 log [1.265/0.0019]

= -0.05 V

Therefore, E_cell = -0.05 V