Calculate Ecell when conc. of Ni2+ = 0.078 M and of Co2+ = 0.492 M.

Question

A voltaic cell consists of an Ni/Ni²⁺ half-cell and a Co/Co²⁺ half-cell. Calculate E_cell when conc. of Ni²⁺ = 0.078 M and of Co²⁺ = 0.492 M.

The reduction potential for Ni²⁺ is -0.247 V and for Co²⁺ is -0.277 V.

The correct is  0.00633 

Answer 1

Ni²⁺ + Co(s) ==> Co²⁺ + Ni(s)

Eºcell = -0.247 -(-0.277) = 0.03 V

Q = [Co2+]/[Ni2+] = 0.492 M/0.078 M = 6.3

Ecell = Eºcell - (RT/nF) ln Q and when T is 25º we have Ecell = Eºcell - (0.0592/n) log Q

Ecell = 0.03 - 0.0296 log 6.3

Ecell = 0.00633