Chemistry lab 10

Question

When the Pb²⁺ concentration is 1.40 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 2.337V. What is the Mg²⁺ concentration?

Pb²⁺(aq) + Mg(s) Pb(s) + Mg²⁺(aq)

Answer 1

Arrhenius equation:

Ecell = Eºcell - RT/nF ln Q and when @ 298K and using log₁₀ we have...

Ecell = Eºcell - 0.0592 / n log Q

Ecell = 2.337

Eºcell = -0.13 - (-2.37) = +2.24 V (use standard reduction potentials for Pb and Mg from a table)

Q = [Mg²⁺] / [Pb²⁺] = [Mg²⁺] / 1.40

n = 2 moles of electrons

2.337 = 2.24 - 0.0592/2 log [Mg²⁺] / 1.40

0.097 = -0.0296 log [Mg²⁺] / 1.40

log [Mg²⁺] / 1.40 = -3.277

[Mg²⁺] / 1.40 = 5.28x10^-4

[Mg²⁺] = 7.40x10^-4 M

(be sure to check all of the math)