The cell potential for this reaction is ________ V when the Zn2+ concentration is 2.73 M and the Pb2+concentration is 0.001043 M at 298 K.

Question

The standard cell potential (E°_cell) for the reaction below is +0.63 V. 

Pb²⁺ (aq) + Zn (s) → Zn²⁺ (aq) + Pb (s)

 

Answer 1

Nernst equation (at 298K):

Ecell = Eºcell - 0.0592/n log Q

Ecell = ?

Eºcell = 0.63V

n = mols electrons transferred = 2

log Q = [Zn2+]/[Pb2+] = log (2.73/0.001043) = 3.42

Ecell = 0.63 - 0.0592/2 (3.42) = 0.63 - 0.10

Ecell = 0.53 V