For the electrochemical cell Pb(s) + Cu²⁺(aq) → Pb²⁺(aq) + Cu(s) (E° = 0.466 V, [Cu²⁺] = 2.50 M), what is the value of E when [Pb²⁺] = 0.055 M? Assume T is 298 K.

Answer 1

Nernst equation:

Ecell = Eºcell - RT/nF ln Q and at 298K and converting to log₁₀ we have..

Ecell = Eºcell - 0.0592 / n log Q

Ecell = ?

Eºcell = 0.466 V

n = 2 mols electrons

Q = [Pb²⁺] / [Cu²⁺] = 0.055 / 2.50 = 0.022

Solving for Ecell

Ecell = 0.466 - (0.0592/2) log 0.022

Ecell = 0.466 - (0.0296 * -1.66)

Ecell = 0.466 - (-0.0491)

Ecell = 0.466 + 0.0491

Ecell = 0.515 V