Question

# A voltaic cell consists of an Ni/Ni2+ half-cell and a Co/Co2+ half-cell. Calculate Ecell when [Ni2+]...

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.

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

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

Ecell = Eocell - RT/nF In {[Co2+] / [Ni2+]} (solids cannot be taken into consideration)

Ecell = Eocell - 0.059/n Iog {[Co2+] / [Ni2+]}  where n = no of electrons transferred

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

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

Eocell = Eoreduction of reaction at cathode+Eooxidation of reaction at anode

= -0.247 + 0.277 V

= + 0.03 V

Eocell = + 0.03 V

no of electrons transferred n = 2

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

Ecell = Eocell - 0.059/n Iog {[Co2+] / [Ni2+]}

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

= -0.05 V

Therefore, Ecell = -0.05 V