Question

Calculate E cell for the new conditons: How does the voltage change and why? (please explain using Le Chateliers principles)

Mg(s) , Mg+2(0.500M) ; Al+3 (2.500M), Al(s)

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Answer 1

Calculate E cell for the new conditons: How does the voltage change and why? (please explain using Le Chateliers principles)

Mg(s) , Mg+2(0.500M) ; Al+3 (2.500M), Al(s)

Mg2+ + 2 e− ⇌ Mg(s) −2.372

Al3+ + 3 e− ⇌ Al(s) −1.662

Initially:

E° = Ecathode - Eanode

cathode = the higher potential, so choose Al

anode = the lower potential, so MG

E° = Ecathode - Eanode = -1.662 - - 2.372 = 0.71 V

n= 2x3 = 6 electrons are being transferred

now...

When the cell is NOT under standard conditions, i.e. 1M of each reactants at T = 25°C and P = 1 atm; then we must use Nernst Equation.

The equation relates E°cell, number of electrons transferred, charge of 1 mol of electron to Faraday and finally, the Quotient retio between products/reactants

The Nernst Equation:

Ecell = E0cell - (RT/nF) x lnQ

In which:

Ecell = non-standard value

E° or E0cell or E°cell or EMF = Standard EMF: standard cell potential
R is the gas constant (8.3145 J/mol-K)
T is the absolute temperature = 298 K
n is the number of moles of electrons transferred by the cell's reaction
F is Faraday's constant = 96485.337 C/mol or typically 96500 C/mol
Q is the reaction quotient, where

Q = [C]^c * [D]^d / [A]^a*[B]^b

pure solids and pure liquids are not included. Also note that if we use partial pressure (for gases)

Q = P-A^a / (P-B)^b

substitute in Nernst Equation:

Ecell = E° - (RT/nF) x lnQ

Ecell = 0.71 - (8.314*298)/(6*96500) x lnQ

Q = [Mg+2]^3 / [Al+3]^2

Q = (0.5^3) / (2.5^2) = 0.02

Ecell = 0.71  - (8.314*298)/(6*96500) * ln(0.02)

Ecell = 0.7267 V

clearly, this favour Ecell, since there is plenty of reactants ( Al3+ ) vs products ( mg2+)