Question

Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C.

2Al(s) + 3Mg2+ (aq) ® 2 Al3+(aq) + 3Mg(s)

Answer 1

First, identify the half equations.

red: Mg(II) + 2e to Mg
ox: Al to Al(III) + 3e

then, look up the standard reduction potentials for the two half equations. the potential difference is:

E(Mg/Mg^2+) - E(Al/Al^3+) = -2.37 - (-1.66) = -0.71 Volts

the standard potential diff is negative; this tells us that the forward reaction is not spontaneous. of course, you'd say, Mg is higher on the activity series than Al.

the equilibrium constant is related to the standard gibbs free energy change for this reaction, which in turn is related to the potential diff.

K = exp(-delta G / RT) = exp (n F * potential diff / RT)

F is Faraday's constant, or the charge of one mol of electrons in coulombs. what's n? it depends on the balanced equation, so let's look at it again:

2Al(s) + 3Mg2+ (aq) ® 2 Al3+(aq) + 3Mg(s)

in plain words, two mol of Al atoms donate 6 mols of electrons to 3 mols of Mg^2+ ions to make two mol of Al^3+ and 3 mols of Mg atoms.

so the reaction consists of a transfer of 6 mols of electrons, and n=6 mols

so K = exp (6 mol * 96485 Coulomb/mol * -0.71 Volt / RT)

R is the gas constant (8.3145 J per mol per K), T is the temperature in *Kelvins*, and Volt is really Joule per Coulomb, so the units will cleanly cancel out.