In the rusting of iron in acidic moist air, the anodic reaction is Fe -> Fe^2+ + 2e- with E deg(Fe^2+/Fe) = -0.44 V and the cathodic reaction is O2 + 4H+ + 4e- -> 2H2O with E deg = +1.23 V. Find the magnitude of the standard Gibbs energy change, in kJ, for the corrosion reaction written per mole of iron, i.e. Fe + 1/2 O2 + 2H+ -> Fe^2+ + H2O (nearest integer). Take F = 96500 C mol^-1.

JEE Advanced levelNumerical answerDifficulty 3 of 3ec172

In the rusting of iron in acidic moist air, the anodic reaction is $Fe Fe X^{2 +} + 2 e X^{-}$ with $E^{\circ} (Fe X^{2 +} / Fe) = - 0.44 V$ and the cathodic reaction is $O X_{2} + 4 H X^{+} + 4 e X^{-} 2 H X_{2} O$ with $E^{\circ} = + 1.23 V$ . Find the magnitude of the standard Gibbs energy change, in kJ, for the corrosion reaction written per mole of iron, i.e. $Fe + \frac{1}{2} O X_{2} + 2 H X^{+} Fe X^{2 +} + H X_{2} O$ (nearest integer). Take $F = 96500 C mo l^{- 1}$ .

Numerical answer type, answer in kJ, tolerance 1.

The corrosion cell has

E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = 1.23 - (- 0.44) = 1.67 V

. Per mole of iron,

n = 2

Δ_{r} G^{\circ} = - n F E^{\circ} = - 2 \times 96500 \times 1.67 = - 3.223 \times 1 0^{5} J = - 322 kJ

, so the magnitude is

322 kJ

. The large negative value shows why rusting is spontaneous.