Question

Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. 2 Al(s) + 3 Mg2+(aq) → 2 Al3+(aq) + 3 Mg(s) A) 1.1 × 1072 B) 8.9 × 10-73 C) 1.1 × 10-72 D) 1.0 × 1024 E) 4.6 × 1031

Answer 1

Answer: Thus the value of equilibrium constant for the balanced redox reaction at 25°C is $8.9\times 10^{-73}$

Explanation:

$2Al(s)+3Mg^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Mg(s)$

Here Al undergoes oxidation by loss of electrons, thus act as anode. magnesium undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Mg^{2+}/Mg]}= -2.37V$

$E^0_{[Al^{3+}/Al]}=-1.66V$

$E^0=E^0_{[Mg^{2+}/Mg]}- E^0_{[Al^{3+}/Al]}$

$E^0=-2.37V-(-1.66)=-0.71$

The standard emf of a cell is related to Gibbs free energy by following relation:

$\Delta G=-nFE^0$

$\Delta G$ = gibbs free energy

n= no of electrons gained or lost = 6

F= faraday's constant

$E^0$ = standard emf

$\Delta G=-6\times 96500\times (-0.71)=411090J$

The Gibbs free energy is related to equilibrium constant by following relation:

$\Delta G=-2.303RTlog K$

R = gas constant = 8.314 J/Kmol

T = temperature in kelvin =$25^0C=25+273=298K$

K = equilibrium constant

$\Delta G=-2.303RTlog K$

$411090=-2.303\times 8.314\times 298\times logK$

$K=8.9\times 10^{-73}$

Thus the value of equilibrium constant for the balanced redox reaction at 25°C is $8.9\times 10^{-73}$