Chem post lab 10

Question

When the Hg²⁺ concentration is 1.25 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 2.577V. What is the Al³⁺ concentration?

3Hg²⁺(aq) + 2Al(s)3Hg(l) + 2Al³⁺(aq)

Answer 1

Arrhenius equation: Ecell = Eºcell - RT/nF ln Q

@298K and using log₁₀ this converts to

Ecell = Eºcell - 0.0592 / n log Q

3Hg²⁺(aq) + 2Al(s) ==> 3Hg(l) + 2Al³⁺(aq)

Ecell = 2.577 V

Eºcell = 2.51 V (see below for calculation of Eºcell)

n = 6 moles of electrons

Q = [Al³⁺]² / [Hg²⁺]³ = [Al³⁺]² / [1.25]³

Calculation of Eºcell:

Al³⁺ + 3e- ==> Al(s) Eº = -1.66 V (looked up in a table of standard reduction potentials)

Hg²⁺ + 2e- ==> Hg(l) Eº = +0.85 V

3Hg²⁺(aq) + 2Al(s) ==> 3Hg(l) + 2Al³⁺(aq) Eº = 0.85 + 1.66 = 2.51 V

Ecell = Eºcell - 0.0592 / n log Q

2.577 = 2.51 - 0.0592 / 6 log [Al³⁺]² / [1.25]³

2.577 = 2.51 - 0.00987 log [Al³⁺]² / [1.95]

0.067 = - 0.00987 log [Al³⁺]² / [1.95]

log [Al³⁺]² / [1.95] = -6.79

[Al³⁺]² / [1.95] = 1.63x10^-7

[Al3+]² = 3.18x10^-7

[Al³⁺] = 5.63x10^-4 M

(be sure to check all of the math)